Integrand size = 45, antiderivative size = 233 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^{5/2} (40 A+38 B+25 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^3 (24 A-54 B-49 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (24 A+42 B+31 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac {a (6 B+5 C) \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d} \]
1/8*a^(5/2)*(40*A+38*B+25*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^( 1/2))/d+1/12*a*(6*B+5*C)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)*sec(d*x+c)^(1/2 )/d+1/3*C*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+1/24*a^3*(2 4*A-54*B-49*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+1/24*a ^2*(24*A+42*B+31*C)*sin(d*x+c)*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)/d
Time = 3.15 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.70 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^3 \left (15 (8 A+5 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)-114 B \arcsin \left (\sqrt {\sec (c+d x)}\right ) \tan (c+d x)+\sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \left (48 A \sin (c+d x)+\left (24 A+66 B+75 C+2 (6 B+17 C) \sec (c+d x)+8 C \sec ^2(c+d x)\right ) \tan (c+d x)\right )\right )}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x] ^2))/Sqrt[Sec[c + d*x]],x]
(a^3*(15*(8*A + 5*C)*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 114*B*A rcSin[Sqrt[Sec[c + d*x]]]*Tan[c + d*x] + Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*(48*A*Sin[c + d*x] + (24*A + 66*B + 75*C + 2*(6*B + 17*C)*Sec[c + d*x] + 8*C*Sec[c + d*x]^2)*Tan[c + d*x])))/(24*d*Sqrt[1 - Sec[c + d*x]]*S qrt[a*(1 + Sec[c + d*x])])
Time = 1.43 (sec) , antiderivative size = 246, 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, 4576, 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.
\(\displaystyle \int \frac {(a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4576 |
\(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^{5/2} (a (6 A-C)+a (6 B+5 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx}{3 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^{5/2} (a (6 A-C)+a (6 B+5 C) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (6 A-C)+a (6 B+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 4506 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 (8 A-2 B-3 C) a^2+(24 A+42 B+31 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{4} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 (8 A-2 B-3 C) a^2+(24 A+42 B+31 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{4} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 (8 A-2 B-3 C) a^2+(24 A+42 B+31 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 4506 |
\(\displaystyle \frac {\frac {1}{4} \left (\int \frac {\sqrt {\sec (c+d x) a+a} \left ((24 A-54 B-49 C) a^3+3 (40 A+38 B+25 C) \sec (c+d x) a^3\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((24 A-54 B-49 C) a^3+3 (40 A+38 B+25 C) \sec (c+d x) a^3\right )}{\sqrt {\sec (c+d x)}}dx+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((24 A-54 B-49 C) a^3+3 (40 A+38 B+25 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 4503 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 a^3 (40 A+38 B+25 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 a^3 (40 A+38 B+25 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 4288 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (\frac {2 a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^3 (40 A+38 B+25 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}+\frac {1}{4} \left (\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}+\frac {1}{2} \left (\frac {6 a^{7/2} (40 A+38 B+25 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )\right )}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\) |
Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/S qrt[Sec[c + d*x]],x]
(C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(3*d) + ((a ^2*(6*B + 5*C)*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x]) /(2*d) + ((a^3*(24*A + 42*B + 31*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d + ((6*a^(7/2)*(40*A + 38*B + 25*C)*ArcSinh[(Sqrt[a]* Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^4*(24*A - 54*B - 49*C)*S qrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/2)/4)/(6*a)
3.6.98.3.1 Defintions of rubi rules used
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[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m , n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(729\) vs. \(2(201)=402\).
Time = 1.96 (sec) , antiderivative size = 730, normalized size of antiderivative = 3.13
method | result | size |
parts | \(\frac {A \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (5 \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+5 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+5 \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}+5 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+4 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right )}{2 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\sec \left (d x +c \right )}}-\frac {B \,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 {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+19 \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-22 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-4 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \tan \left (d x +c \right )\right )}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}-\frac {C \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (75 \cos \left (d x +c \right )^{2} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+75 \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-150 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-68 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-16 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \tan \left (d x +c \right )\right )}{48 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(730\) |
default | \(\text {Expression too large to display}\) | \(867\) |
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)^(1/2 ),x,method=_RETURNVERBOSE)
1/2*A/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(1/2)*(5*ar ctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/ 2))*(-1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+5*(-1/(cos(d*x+c)+1))^(1/2)*arcta n(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))* cos(d*x+c)+5*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(-1/(cos (d*x+c)+1))^(1/2))*(-1/(cos(d*x+c)+1))^(1/2)+5*(-1/(cos(d*x+c)+1))^(1/2)*a rctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/ 2))+4*sin(d*x+c)+2*tan(d*x+c))-1/8*B/d*a^2*(a*(1+sec(d*x+c)))^(1/2)*sec(d* x+c)^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2)*(19*cos(d*x+c)*arctan( 1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))+1 9*cos(d*x+c)*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos( d*x+c)+1))^(1/2))-22*sin(d*x+c)*(-1/(cos(d*x+c)+1))^(1/2)-4*(-1/(cos(d*x+c )+1))^(1/2)*tan(d*x+c))-1/48*C/d*a^2*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^( 3/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2)*(75*cos(d*x+c)^2*arctan(1/2* (-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))+75*co s(d*x+c)^2*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d* x+c)+1))^(1/2))-150*(-1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*sin(d*x+c)-68*sin (d*x+c)*(-1/(cos(d*x+c)+1))^(1/2)-16*(-1/(cos(d*x+c)+1))^(1/2)*tan(d*x+c))
Time = 0.49 (sec) , antiderivative size = 524, normalized size of antiderivative = 2.25 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\left [\frac {3 \, {\left ({\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{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 (48 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, A + 22 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, B + 17 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, 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 )}}}{96 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {3 \, {\left ({\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (48 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, A + 22 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, B + 17 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, 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 )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
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 )^(1/2),x, algorithm="fricas")
[1/96*(3*((40*A + 38*B + 25*C)*a^2*cos(d*x + c)^3 + (40*A + 38*B + 25*C)*a ^2*cos(d*x + c)^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*(48*A*a^2*cos(d*x + c)^3 + 3*(8*A + 22*B + 25*C)*a^2*cos(d*x + c)^2 + 2*(6*B + 17*C)*a^2*cos(d*x + c) + 8*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)^3 + d*co s(d*x + c)^2), 1/48*(3*((40*A + 38*B + 25*C)*a^2*cos(d*x + c)^3 + (40*A + 38*B + 25*C)*a^2*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d* x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^ 2 - a*cos(d*x + c) - 2*a)) + 2*(48*A*a^2*cos(d*x + c)^3 + 3*(8*A + 22*B + 25*C)*a^2*cos(d*x + c)^2 + 2*(6*B + 17*C)*a^2*cos(d*x + c) + 8*C*a^2)*sqrt ((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*co s(d*x + c)^3 + d*cos(d*x + c)^2)]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 17788 vs. \(2 (201) = 402\).
Time = 3.54 (sec) , antiderivative size = 17788, normalized size of antiderivative = 76.34 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Too large to display} \]
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 )^(1/2),x, algorithm="maxima")
1/96*(24*(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*sin(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 + 1 0*(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)^2 + 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...
\[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
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 )^(1/2),x, algorithm="giac")
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]
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))^(1/2),x)