Integrand size = 37, antiderivative size = 210 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {5 a^{5/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {a^3 (64 A+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (16 A-15 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \] Output:
5*a^(5/2)*C*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/15*a^3* (64*A+15*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/15*a^2* (16*A-15*C)*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d+2/3*a*A*( a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/5*A*(a+a*sec(d*x+c)) ^(5/2)*sin(d*x+c)/d/sec(d*x+c)^(3/2)
Time = 5.06 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.57 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a^3 \left (\frac {75 C \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{\sqrt {1-\sec (c+d x)}}+\frac {6 A \sin (c+d x)+\left (28 A+(86 A+30 C) \sec (c+d x)+15 C \sec ^2(c+d x)\right ) \tan (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)}\right )}{15 d \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x] ^(5/2),x]
Output:
(a^3*((75*C*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/Sqrt[1 - Sec[c + d*x]] + (6*A*Sin[c + d*x] + (28*A + (86*A + 30*C)*Sec[c + d*x] + 15*C*Sec[ c + d*x]^2)*Tan[c + d*x])/Sec[c + d*x]^(3/2)))/(15*d*Sqrt[a*(1 + Sec[c + d *x])])
Time = 1.42 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.378, Rules used = {3042, 4575, 27, 3042, 4505, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^{5/2} (5 a A-a (2 A-5 C) \sec (c+d x))}{2 \sec ^{\frac {3}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^{5/2} (5 a A-a (2 A-5 C) \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (5 a A-a (2 A-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {2}{3} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 a^2 (8 A+5 C)-a^2 (16 A-15 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 a^2 (8 A+5 C)-a^2 (16 A-15 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}}dx+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a^2 (8 A+5 C)-a^2 (16 A-15 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{3} \left (\int \frac {\sqrt {\sec (c+d x) a+a} \left ((64 A+15 C) a^3+75 C \sec (c+d x) a^3\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((64 A+15 C) a^3+75 C \sec (c+d x) a^3\right )}{\sqrt {\sec (c+d x)}}dx-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((64 A+15 C) a^3+75 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (75 a^3 C \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^4 (64 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (75 a^3 C \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^4 (64 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {2 a^4 (64 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {150 a^3 C \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \left (\frac {1}{2} \left (\frac {150 a^{7/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^4 (64 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[((a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2) ,x]
Output:
(2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ( (10*a^2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]] ) + (-((a^3*(16*A - 15*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[ c + d*x])/d) + ((150*a^(7/2)*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*S ec[c + d*x]]])/d + (2*a^4*(64*A + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/( d*Sqrt[a + a*Sec[c + d*x]]))/2)/3)/(5*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[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
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_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co t[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 *(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 2.58 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.15
| method | result | size |
| default | \(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-75 \sec \left (d x +c \right )-75 \sec \left (d x +c \right )^{2}\right )+\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-75 \sec \left (d x +c \right )-75 \sec \left (d x +c \right )^{2}\right )+A \left (24 \sin \left (d x +c \right )+112 \tan \left (d x +c \right )+344 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )+\left (120 \cos \left (d x +c \right )+60\right ) C \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}\right )}{60 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}\) | \(241\) |
| parts | \(\frac {A \left (6 \sin \left (d x +c \right )+28 \tan \left (d x +c \right )+86 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}+\frac {C \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (8 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )+5 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-5 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right )}{4 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\sec \left (d x +c \right )}}\) | \(249\) |
Input:
int((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x,method=_R ETURNVERBOSE)
Output:
1/60/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(5/2)*(2^(1/ 2)*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2/(-1/(cos(d*x+c)+1))^(1/2)*(cot(d *x+c)-csc(d*x+c)+1))*(-75*sec(d*x+c)-75*sec(d*x+c)^2)+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))*(-75*sec(d*x+c)-75*sec(d*x+c)^2)+A*(24*sin(d*x+c)+112*tan(d*x+c)+344 *sec(d*x+c)*tan(d*x+c))+(120*cos(d*x+c)+60)*C*tan(d*x+c)*sec(d*x+c)^2)
Time = 0.11 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.02 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\left [\frac {75 \, {\left (C a^{2} \cos \left (d x + c\right ) + C a^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 28 \, A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (43 \, A + 15 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {75 \, {\left (C a^{2} \cos \left (d x + c\right ) + C a^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 28 \, A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (43 \, A + 15 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, al gorithm="fricas")
Output:
[1/60*(75*(C*a^2*cos(d*x + c) + C*a^2)*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*(6*A*a^2*cos(d*x + c)^3 + 28*A*a^2*cos(d*x + c)^2 + 2*(43*A + 15*C)*a^2*cos(d*x + c) + 15*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) + d) , 1/30*(75*(C*a^2*cos(d*x + c) + C*a^2)*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*sq rt(cos(d*x + c))*sin(d*x + c))) + 2*(6*A*a^2*cos(d*x + c)^3 + 28*A*a^2*cos (d*x + c)^2 + 2*(43*A + 15*C)*a^2*cos(d*x + c) + 15*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) + d)]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 11556 vs. \(2 (180) = 360\).
Time = 0.41 (sec) , antiderivative size = 11556, normalized size of antiderivative = 55.03 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, al gorithm="maxima")
Output:
1/60*(2*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + 15*(8*a^2*cos( 1/2*d*x + 1/2*c)^4*sin(1/2*d*x + 1/2*c) + 16*a^2*cos(1/2*d*x + 1/2*c)^2*si n(1/2*d*x + 1/2*c)^3 + 8*a^2*sin(1/2*d*x + 1/2*c)^5 + 5*(sqrt(2)*a^2*log(2 *cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/ 2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2* c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2* sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c) ^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2) *sin(1/2*d*x + 1/2*c) + 2))*cos(1/2*d*x + 1/2*c)^4 + 10*(sqrt(2)*a^2*log(2 *cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/ 2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2* c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2* sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c) ^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2) *sin(1/2*d*x + 1/2*c) + 2))*cos(1/2*d*x + 1/2*c)^2*sin(1/2*d*x + 1/2*c)...
Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (180) = 360\).
Time = 3.00 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.03 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, al gorithm="giac")
Output:
1/30*(75*C*a^(5/2)*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)))*sgn(cos(d*x + c)) - 75*C*a^(5 /2)*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)))*sgn(cos(d*x + c)) + 60*(3*sqrt(2)*(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^(7/2)*sgn( cos(d*x + c)) - sqrt(2)*C*a^(9/2)*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 + a))^2*a + a^2) + 4*(60*sqrt(2) *A*a^5*sgn(cos(d*x + c)) + 15*sqrt(2)*C*a^5*sgn(cos(d*x + c)) + (80*sqrt(2 )*A*a^5*sgn(cos(d*x + c)) + 30*sqrt(2)*C*a^5*sgn(cos(d*x + c)) + (32*sqrt( 2)*A*a^5*sgn(cos(d*x + c)) + 15*sqrt(2)*C*a^5*sgn(cos(d*x + c)))*tan(1/2*d *x + 1/2*c)^2)*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)^(5/2))/d
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2))/(1/cos(c + d*x))^( 5/2),x)
Output:
int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2))/(1/cos(c + d*x))^( 5/2), x)
\[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\sqrt {a}\, a^{2} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}d x \right ) c \right ) \] Input:
int((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x)
Output:
sqrt(a)*a**2*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x) **3,x)*a + 2*int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)* *2,x)*a + 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)*c + in t(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x),x)*c + 2*int(sqrt (sec(c + d*x))*sqrt(sec(c + d*x) + 1),x)*c)