Integrand size = 45, antiderivative size = 233 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^{5/2} (8 A+20 B+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^3 (56 A+12 B-27 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (8 A-12 B-21 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{12 d}-\frac {a (4 A-3 C) \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \] Output:
1/4*a^(5/2)*(8*A+20*B+19*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1 /2))/d+1/12*a^3*(56*A+12*B-27*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d* x+c))^(1/2)-1/12*a^2*(8*A-12*B-21*C)*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/ 2)*sin(d*x+c)/d-1/6*a*(4*A-3*C)*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(3/2)*si n(d*x+c)/d+2/3*A*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)
Time = 6.52 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.79 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^3 \left (12 (2 A+5 B) \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)-57 C \arcsin \left (\sqrt {\sec (c+d x)}\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+\sqrt {1-\sec (c+d x)} \left (8 A \sin (c+d x)+\left (8 (8 A+3 B)+3 (4 B+11 C) \sec (c+d x)+6 C \sec ^2(c+d x)\right ) \tan (c+d x)\right )\right )}{12 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x] ^2))/Sec[c + d*x]^(3/2),x]
Output:
(a^3*(12*(2*A + 5*B)*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]^(3/2)*Sin [c + d*x] - 57*C*ArcSin[Sqrt[Sec[c + d*x]]]*Sec[c + d*x]^(3/2)*Sin[c + d*x ] + Sqrt[1 - Sec[c + d*x]]*(8*A*Sin[c + d*x] + (8*(8*A + 3*B) + 3*(4*B + 1 1*C)*Sec[c + d*x] + 6*C*Sec[c + d*x]^2)*Tan[c + d*x])))/(12*d*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.52 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {3042, 4574, 27, 3042, 4506, 27, 3042, 4506, 27, 3042, 4503, 3042, 4288, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^{5/2} (a (5 A+3 B)-a (4 A-3 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^{5/2} (a (5 A+3 B)-a (4 A-3 C) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+3 B)-a (4 A-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 a^2 (8 A+4 B-C)-a^2 (8 A-12 B-21 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {a^2 (4 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{4} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 a^2 (8 A+4 B-C)-a^2 (8 A-12 B-21 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}}dx-\frac {a^2 (4 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a^2 (8 A+4 B-C)-a^2 (8 A-12 B-21 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^2 (4 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{4} \left (\int \frac {\sqrt {\sec (c+d x) a+a} \left ((56 A+12 B-27 C) a^3+3 (8 A+20 B+19 C) \sec (c+d x) a^3\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {a^3 (8 A-12 B-21 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {a^2 (4 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((56 A+12 B-27 C) a^3+3 (8 A+20 B+19 C) \sec (c+d x) a^3\right )}{\sqrt {\sec (c+d x)}}dx-\frac {a^3 (8 A-12 B-21 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {a^2 (4 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((56 A+12 B-27 C) a^3+3 (8 A+20 B+19 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^3 (8 A-12 B-21 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {a^2 (4 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 a^3 (8 A+20 B+19 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^4 (56 A+12 B-27 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^3 (8 A-12 B-21 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {a^2 (4 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 a^3 (8 A+20 B+19 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^4 (56 A+12 B-27 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^3 (8 A-12 B-21 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {a^2 (4 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (\frac {2 a^4 (56 A+12 B-27 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^3 (8 A+20 B+19 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )-\frac {a^3 (8 A-12 B-21 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {a^2 (4 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (\frac {6 a^{7/2} (8 A+20 B+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^4 (56 A+12 B-27 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^3 (8 A-12 B-21 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )-\frac {a^2 (4 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\sec (c+d x)}}\) |
Input:
Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/S ec[c + d*x]^(3/2),x]
Output:
(2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ( -1/2*(a^2*(4*A - 3*C)*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/d + (-((a^3*(8*A - 12*B - 21*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[ c + d*x]]*Sin[c + d*x])/d) + ((6*a^(7/2)*(8*A + 20*B + 19*C)*ArcSinh[(Sqrt [a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^4*(56*A + 12*B - 27* C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/2)/4)/(3 *a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)] Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a , b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a *B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[ e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(481\) vs. \(2(201)=402\).
Time = 11.50 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.07
| method | result | size |
| default | \(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {2}\, A \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (24+24 \sec \left (d x +c \right )\right )+\sqrt {2}\, B \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (60+60 \sec \left (d x +c \right )\right )+\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (57+57 \sec \left (d x +c \right )\right )+\sqrt {2}\, A \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (24+24 \sec \left (d x +c \right )\right )+\sqrt {2}\, B \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (60+60 \sec \left (d x +c \right )\right )+\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (57+57 \sec \left (d x +c \right )\right )+A \left (32 \sin \left (d x +c \right )+256 \tan \left (d x +c \right )\right )+B \left (96 \tan \left (d x +c \right )+48 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )+\left (132 \cos \left (d x +c \right )+24\right ) C \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}\right )}{48 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(482\) |
| parts | \(-\frac {A \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-4 \sin \left (d x +c \right )-32 \tan \left (d x +c \right )+\sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (3+3 \sec \left (d x +c \right )\right )+\sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (3+3 \sec \left (d x +c \right )\right )\right )}{6 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}-\frac {B \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-8 \sin \left (d x +c \right )-4 \tan \left (d x +c \right )+5 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+5 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right )}{4 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\sec \left (d x +c \right )}}+\frac {C \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (19 \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+19 \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (11 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right )\right )}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(527\) |
Input:
int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2 ),x,method=_RETURNVERBOSE)
Output:
1/48/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*(2^(1/ 2)*A*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/( cos(d*x+c)+1))^(1/2))*(24+24*sec(d*x+c))+2^(1/2)*B*(-2/(cos(d*x+c)+1))^(1/ 2)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*(60+60 *sec(d*x+c))+2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-cot(d*x+c)+c sc(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*(57+57*sec(d*x+c))+2^(1/2)*A*(-2/( cos(d*x+c)+1))^(1/2)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(cos(d*x+c) +1))^(1/2))*(24+24*sec(d*x+c))+2^(1/2)*B*(-2/(cos(d*x+c)+1))^(1/2)*arctan( 1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))*(60+60*sec(d*x+c ))+2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)- 1)/(-1/(cos(d*x+c)+1))^(1/2))*(57+57*sec(d*x+c))+A*(32*sin(d*x+c)+256*tan( d*x+c))+B*(96*tan(d*x+c)+48*sec(d*x+c)*tan(d*x+c))+(132*cos(d*x+c)+24)*C*t an(d*x+c)*sec(d*x+c)^2)
Time = 0.29 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.18 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {3 \, {\left ({\left (8 \, A + 20 \, B + 19 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (8 \, A + 20 \, B + 19 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (8 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{48 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {3 \, {\left ({\left (8 \, A + 20 \, B + 19 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (8 \, A + 20 \, B + 19 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (8 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{24 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(3/2),x, algorithm="fricas")
Output:
[1/48*(3*((8*A + 20*B + 19*C)*a^2*cos(d*x + c)^2 + (8*A + 20*B + 19*C)*a^2 *cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos (d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^ 2)) + 4*(8*A*a^2*cos(d*x + c)^3 + 8*(8*A + 3*B)*a^2*cos(d*x + c)^2 + 3*(4* B + 11*C)*a^2*cos(d*x + c) + 6*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), 1/24*(3*((8*A + 20*B + 19*C)*a^2*cos(d*x + c)^2 + (8*A + 20*B + 19*C)*a^2* cos(d*x + c))*sqrt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(- a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))/(a*sqrt(cos(d*x + c))*sin(d*x + c))) + 2*(8*A*a^2*cos(d*x + c)^3 + 8*(8*A + 3*B)*a^2*cos(d*x + c)^2 + 3*( 4*B + 11*C)*a^2*cos(d*x + c) + 6*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c)) ]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x +c)**(3/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 14913 vs. \(2 (201) = 402\).
Time = 3.19 (sec) , antiderivative size = 14913, normalized size of antiderivative = 64.00 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(3/2),x, algorithm="maxima")
Output:
1/48*(4*sqrt(2)*(30*a^2*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(3/2*d*x + 3/2*c) - 30*a^2*cos(3/2*d*x + 3/2*c)*sin(2/3*arct an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 3*sqrt(2)*a^2*log(2*cos (1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*ar ctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*a rctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*ar ctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 3*sqrt(2)*a^2*lo g(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin (1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*co s(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin (1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + 3*sqrt(2) *a^2*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqr t(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt (2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 3* sqrt(2)*a^2*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + 4*a^2*sin(3/2*d*x + 3/2*c) + 30*a^2*sin(1/3*arctan2(sin(3/2*d*x + ...
Leaf count of result is larger than twice the leaf count of optimal. 690 vs. \(2 (201) = 402\).
Time = 3.55 (sec) , antiderivative size = 690, normalized size of antiderivative = 2.96 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(3/2),x, algorithm="giac")
Output:
1/24*(3*(8*A*a^(5/2)*sgn(cos(d*x + c)) + 20*B*a^(5/2)*sgn(cos(d*x + c)) + 19*C*a^(5/2)*sgn(cos(d*x + c)))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sq rt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) - 3*(8*A*a^(5/2) *sgn(cos(d*x + c)) + 20*B*a^(5/2)*sgn(cos(d*x + c)) + 19*C*a^(5/2)*sgn(cos (d*x + c)))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1 /2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 16*(9*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 3*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + (7*sqrt(2)*A*a^4*sgn(cos(d*x + c )) + 3*sqrt(2)*B*a^4*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d* x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) + 12*sqrt(2)*(12*(sqrt(a)* tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*a^(7/2)*sgn (cos(d*x + c)) + 19*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1 /2*c)^2 + a))^6*C*a^(7/2)*sgn(cos(d*x + c)) - 76*(sqrt(a)*tan(1/2*d*x + 1/ 2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a^(9/2)*sgn(cos(d*x + c)) - 171*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 *C*a^(9/2)*sgn(cos(d*x + c)) + 36*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t an(1/2*d*x + 1/2*c)^2 + a))^2*B*a^(11/2)*sgn(cos(d*x + c)) + 89*(sqrt(a)*t an(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^(11/2)*sgn (cos(d*x + c)) - 4*B*a^(13/2)*sgn(cos(d*x + c)) - 9*C*a^(13/2)*sgn(cos(d*x + c)))/((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a ))^4 - 6*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 ...
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/( 1/cos(c + d*x))^(3/2),x)
Output:
int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/( 1/cos(c + d*x))^(3/2), x)
\[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\sqrt {a}\, a^{2} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )d x \right ) b +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}d x \right ) c \right ) \] Input:
int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2 ),x)
Output:
sqrt(a)*a**2*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x) **2,x)*a + 2*int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x), x)*a + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x),x)*b + int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*c + int( sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x),x)*b + 2*int(sqrt(s ec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x),x)*c + int(sqrt(sec(c + d *x))*sqrt(sec(c + d*x) + 1),x)*a + 2*int(sqrt(sec(c + d*x))*sqrt(sec(c + d *x) + 1),x)*b + int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1),x)*c)