Integrand size = 42, antiderivative size = 261 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(163 B-283 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(13 B-21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(85 B-157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(475 B-787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d} \]
1/32*(163*B-283*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^ (1/2))/a^(5/2)/d*2^(1/2)+1/4*(B-C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+ c))^(5/2)+1/16*(13*B-21*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3 /2)-1/120*(985*B-1729*C)*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)-1/80*(85* B-157*C)*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)+1/240*(475*B -787*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^3/d
Time = 3.27 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.68 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\left (30 \sqrt {2} (163 B-283 C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x)+\sqrt {1-\sec (c+d x)} \left (-1495 B+2671 C-5 (503 B-911 C) \sec (c+d x)-32 (25 B-49 C) \sec ^2(c+d x)+160 (B-C) \sec ^3(c+d x)+96 C \sec ^4(c+d x)\right )\right ) \tan (c+d x)}{240 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \]
((30*Sqrt[2]*(163*B - 283*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[( c + d*x)/2]^4*Sec[c + d*x]^2 + Sqrt[1 - Sec[c + d*x]]*(-1495*B + 2671*C - 5*(503*B - 911*C)*Sec[c + d*x] - 32*(25*B - 49*C)*Sec[c + d*x]^2 + 160*(B - C)*Sec[c + d*x]^3 + 96*C*Sec[c + d*x]^4))*Tan[c + d*x])/(240*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))
Time = 1.86 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.10, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4560, 3042, 4507, 27, 3042, 4507, 27, 3042, 4509, 27, 3042, 4498, 27, 3042, 4489, 3042, 4282, 216}
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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\sec ^5(c+d x) (B+C \sec (c+d x))}{(a \sec (c+d x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (8 a (B-C)-a (5 B-13 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (8 a (B-C)-a (5 B-13 C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (8 a (B-C)-a (5 B-13 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^3(c+d x) \left (6 a^2 (13 B-21 C)-a^2 (85 B-157 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^3(c+d x) \left (6 a^2 (13 B-21 C)-a^2 (85 B-157 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (6 a^2 (13 B-21 C)-a^2 (85 B-157 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4509 |
| \(\displaystyle \frac {\frac {\frac {2 \int -\frac {\sec ^2(c+d x) \left (4 a^3 (85 B-157 C)-a^3 (475 B-787 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {\sec ^2(c+d x) \left (4 a^3 (85 B-157 C)-a^3 (475 B-787 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^3 (85 B-157 C)-a^3 (475 B-787 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {\frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^4 (475 B-787 C)-2 a^4 (985 B-1729 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {\sec (c+d x) \left (a^4 (475 B-787 C)-2 a^4 (985 B-1729 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^4 (475 B-787 C)-2 a^4 (985 B-1729 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {-\frac {-\frac {15 a^4 (163 B-283 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^4 (985 B-1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {-\frac {-\frac {15 a^4 (163 B-283 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^4 (985 B-1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {\frac {-\frac {-\frac {-\frac {30 a^4 (163 B-283 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^4 (985 B-1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}-\frac {-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}-\frac {\frac {15 \sqrt {2} a^{7/2} (163 B-283 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^4 (985 B-1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
((B - C)*Sec[c + d*x]^4*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ( (a*(13*B - 21*C)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3 /2)) + ((-2*a^2*(85*B - 157*C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) - ((-2*a^2*(475*B - 787*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) - ((15*Sqrt[2]*a^(7/2)*(163*B - 283*C)*ArcTan[(Sqrt[a]*Tan[ c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d - (4*a^4*(985*B - 1729*C) *Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(5*a))/(4*a^2))/(8*a^2 )
3.5.1.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B , e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b *(m + 1), 0] && !LtQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* ((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int [Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) *Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a *B, 0] && !LtQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + 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] && LtQ[m, -2^(-1)] && G tQ[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)*d* Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[d/(b*(m + n)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[b*B*(n - 1) + (A*b*(m + n) + a*B*m)*Csc[e + f*x], x], x], x] /; Fr eeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[n, 1]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 1.05 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.62
| method | result | size |
| default | \(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (30 B \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}-30 C \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+315 B \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-435 C \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+2445 B \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}-4245 C \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-3685 B \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+6733 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+5665 B \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-9865 C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-2325 B \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+4365 C \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{480 a^{3} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\) | \(423\) |
| parts | \(\frac {B \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (6 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+69 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+489 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-668 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-465 \cot \left (d x +c \right )+465 \csc \left (d x +c \right )\right )}{96 d \,a^{3} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )}-\frac {C \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (30 \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+435 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+4245 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}-6733 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+9865 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+4365 \cot \left (d x +c \right )-4365 \csc \left (d x +c \right )\right )}{480 d \,a^{3} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\) | \(454\) |
int(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,me thod=_RETURNVERBOSE)
1/480/a^3/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(30*B*(1-cos(d* x+c))^9*csc(d*x+c)^9-30*C*(1-cos(d*x+c))^9*csc(d*x+c)^9+315*B*(1-cos(d*x+c ))^7*csc(d*x+c)^7-435*C*(1-cos(d*x+c))^7*csc(d*x+c)^7+2445*B*ln(csc(d*x+c) -cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2*csc (d*x+c)^2-1)^(5/2)-4245*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(5/2)*ln(csc(d *x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))-3685*B*(1-cos(d* x+c))^5*csc(d*x+c)^5+6733*C*(1-cos(d*x+c))^5*csc(d*x+c)^5+5665*B*(1-cos(d* x+c))^3*csc(d*x+c)^3-9865*C*(1-cos(d*x+c))^3*csc(d*x+c)^3-2325*B*(-cot(d*x +c)+csc(d*x+c))+4365*C*(-cot(d*x+c)+csc(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x +c)^2-1)^2
Time = 0.32 (sec) , antiderivative size = 596, normalized size of antiderivative = 2.28 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {15 \, \sqrt {2} {\left ({\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (1495 \, B - 2671 \, C\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (503 \, B - 911 \, C\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (25 \, B - 49 \, C\right )} \cos \left (d x + c\right )^{2} - 160 \, {\left (B - C\right )} \cos \left (d x + c\right ) - 96 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{960 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}}, -\frac {15 \, \sqrt {2} {\left ({\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (1495 \, B - 2671 \, C\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (503 \, B - 911 \, C\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (25 \, B - 49 \, C\right )} \cos \left (d x + c\right )^{2} - 160 \, {\left (B - C\right )} \cos \left (d x + c\right ) - 96 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{480 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 ),x, algorithm="fricas")
[1/960*(15*sqrt(2)*((163*B - 283*C)*cos(d*x + c)^5 + 3*(163*B - 283*C)*cos (d*x + c)^4 + 3*(163*B - 283*C)*cos(d*x + c)^3 + (163*B - 283*C)*cos(d*x + c)^2)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((1495*B - 2671*C)*cos(d*x + c)^4 + 5*(503*B - 911*C)*cos(d*x + c)^3 + 32*(25*B - 49*C)*cos(d*x + c)^ 2 - 160*(B - C)*cos(d*x + c) - 96*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c ))*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d* cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2), -1/480*(15*sqrt(2)*((163*B - 283*C )*cos(d*x + c)^5 + 3*(163*B - 283*C)*cos(d*x + c)^4 + 3*(163*B - 283*C)*co s(d*x + c)^3 + (163*B - 283*C)*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(2)*sqrt ((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*((1495*B - 2671*C)*cos(d*x + c)^4 + 5*(503*B - 911*C)*cos(d*x + c)^3 + 32*(25*B - 49*C)*cos(d*x + c)^2 - 160*(B - C)*cos(d*x + c) - 96*C)*sqrt((a *cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a ^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)]
\[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 ),x, algorithm="maxima")
Time = 1.59 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {{\left ({\left ({\left (15 \, {\left (\frac {2 \, {\left (\sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2}} + \frac {21 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 29 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3685 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 6733 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, {\left (1133 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 1973 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, {\left (155 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 291 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {15 \, {\left (163 \, \sqrt {2} B - 283 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{480 \, d} \]
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 ),x, algorithm="giac")
1/480*((((15*(2*(sqrt(2)*B*a^2*sgn(cos(d*x + c)) - sqrt(2)*C*a^2*sgn(cos(d *x + c)))*tan(1/2*d*x + 1/2*c)^2/a^2 + (21*sqrt(2)*B*a^2*sgn(cos(d*x + c)) - 29*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - (3685 *sqrt(2)*B*a^2*sgn(cos(d*x + c)) - 6733*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a ^2)*tan(1/2*d*x + 1/2*c)^2 + 5*(1133*sqrt(2)*B*a^2*sgn(cos(d*x + c)) - 197 3*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - 15*(155*s qrt(2)*B*a^2*sgn(cos(d*x + c)) - 291*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2) *tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d* x + 1/2*c)^2 + a)) - 15*(163*sqrt(2)*B - 283*sqrt(2)*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)))/(sqrt(-a)*a^2 *sgn(cos(d*x + c))))/d
Timed out. \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]