Integrand size = 42, antiderivative size = 261 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(15 B-19 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(651 B-799 C) \tan (c+d x)}{105 a d \sqrt {a+a \sec (c+d x)}}+\frac {(63 B-67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}-\frac {(7 B-11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}-\frac {(273 B-397 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d} \]
-1/4*(15*B-19*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1 /2))/a^(3/2)/d*2^(1/2)+1/2*(B-C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c) )^(3/2)+1/105*(651*B-799*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/70*(63 *B-67*C)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/14*(7*B-11*C )*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/210*(273*B-397*C)*( a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d
Time = 0.90 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.78 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-210 \sqrt {2} (15 B-19 C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)+\frac {1}{4} (2751 B-2339 C+24 (217 B-213 C) \cos (c+d x)+60 (63 B-67 C) \cos (2 (c+d x))+1512 B \cos (3 (c+d x))-1608 C \cos (3 (c+d x))+1029 B \cos (4 (c+d x))-1201 C \cos (4 (c+d x))) \sqrt {1-\sec (c+d x)} \sec ^4(c+d x)\right ) \tan (c+d x)}{420 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \]
((-210*Sqrt[2]*(15*B - 19*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[( c + d*x)/2]^2*Sec[c + d*x] + ((2751*B - 2339*C + 24*(217*B - 213*C)*Cos[c + d*x] + 60*(63*B - 67*C)*Cos[2*(c + d*x)] + 1512*B*Cos[3*(c + d*x)] - 160 8*C*Cos[3*(c + d*x)] + 1029*B*Cos[4*(c + d*x)] - 1201*C*Cos[4*(c + d*x)])* Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^4)/4)*Tan[c + d*x])/(420*d*Sqrt[1 - Se c[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))
Time = 1.83 (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, 4509, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\sec ^5(c+d x) (B+C \sec (c+d x))}{(a \sec (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (8 a (B-C)-a (7 B-11 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (8 a (B-C)-a (7 B-11 C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (8 a (B-C)-a (7 B-11 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4509 |
| \(\displaystyle \frac {\frac {2 \int -\frac {\sec ^3(c+d x) \left (6 a^2 (7 B-11 C)-a^2 (63 B-67 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {\sec ^3(c+d x) \left (6 a^2 (7 B-11 C)-a^2 (63 B-67 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (6 a^2 (7 B-11 C)-a^2 (63 B-67 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4509 |
| \(\displaystyle \frac {-\frac {\frac {2 \int -\frac {\sec ^2(c+d x) \left (4 a^3 (63 B-67 C)-a^3 (273 B-397 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {\sec ^2(c+d x) \left (4 a^3 (63 B-67 C)-a^3 (273 B-397 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^3 (63 B-67 C)-a^3 (273 B-397 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a^2 (63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^4 (273 B-397 C)-2 a^4 (651 B-799 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (273 B-397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {\int \frac {\sec (c+d x) \left (a^4 (273 B-397 C)-2 a^4 (651 B-799 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (273 B-397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^4 (273 B-397 C)-2 a^4 (651 B-799 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a^2 (273 B-397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {105 a^4 (15 B-19 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^4 (651 B-799 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (273 B-397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {105 a^4 (15 B-19 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^4 (651 B-799 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (273 B-397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {-\frac {210 a^4 (15 B-19 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^4 (651 B-799 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (273 B-397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {-\frac {-\frac {2 a^2 (63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}-\frac {-\frac {2 a^2 (273 B-397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}-\frac {\frac {105 \sqrt {2} a^{7/2} (15 B-19 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^4 (651 B-799 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}}{7 a}-\frac {2 a (7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
((B - C)*Sec[c + d*x]^4*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ( (-2*a*(7*B - 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d* x]]) - ((-2*a^2*(63*B - 67*C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + a *Sec[c + d*x]]) - ((-2*a^2*(273*B - 397*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) - ((105*Sqrt[2]*a^(7/2)*(15*B - 19*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d - (4*a^4*(651*B - 799*C)*Ta n[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(5*a))/(7*a))/(4*a^2)
3.4.93.3.1 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 = 1.04 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.62
| method | result | size |
| default | \(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (105 B \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}-105 C \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+1575 B \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {7}{2}}-1995 C \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {7}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-2772 B \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+3508 C \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+6342 B \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-7238 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-5460 B \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+6580 C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+1785 B \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-1785 C \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{420 a^{2} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{3}}\) | \(423\) |
| parts | \(-\frac {B \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (5 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+75 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}-127 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+175 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+85 \cot \left (d x +c \right )-85 \csc \left (d x +c \right )\right )}{20 d \,a^{2} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}+\frac {C \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (105 \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+1995 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {7}{2}}-3508 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+7238 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-6580 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-1785 \cot \left (d x +c \right )+1785 \csc \left (d x +c \right )\right )}{420 d \,a^{2} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{3}}\) | \(454\) |
int(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x,me thod=_RETURNVERBOSE)
-1/420/a^2/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(105*B*(1-cos( d*x+c))^9*csc(d*x+c)^9-105*C*(1-cos(d*x+c))^9*csc(d*x+c)^9+1575*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)^(7/2)-1995*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(7/2)*ln( csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))-2772*B*(1-c os(d*x+c))^7*csc(d*x+c)^7+3508*C*(1-cos(d*x+c))^7*csc(d*x+c)^7+6342*B*(1-c os(d*x+c))^5*csc(d*x+c)^5-7238*C*(1-cos(d*x+c))^5*csc(d*x+c)^5-5460*B*(1-c os(d*x+c))^3*csc(d*x+c)^3+6580*C*(1-cos(d*x+c))^3*csc(d*x+c)^3+1785*B*(-co t(d*x+c)+csc(d*x+c))-1785*C*(-cot(d*x+c)+csc(d*x+c)))/((1-cos(d*x+c))^2*cs c(d*x+c)^2-1)^3
Time = 0.33 (sec) , antiderivative size = 539, normalized size of antiderivative = 2.07 \[ \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))^{3/2}} \, dx=\left [\frac {105 \, \sqrt {2} {\left ({\left (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{3}\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 (1029 \, B - 1201 \, C\right )} \cos \left (d x + c\right )^{4} + 12 \, {\left (63 \, B - 67 \, C\right )} \cos \left (d x + c\right )^{3} - 28 \, {\left (3 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (7 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 60 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{840 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}}, \frac {105 \, \sqrt {2} {\left ({\left (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{3}\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 (1029 \, B - 1201 \, C\right )} \cos \left (d x + c\right )^{4} + 12 \, {\left (63 \, B - 67 \, C\right )} \cos \left (d x + c\right )^{3} - 28 \, {\left (3 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (7 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 60 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{420 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\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))^(3/2 ),x, algorithm="fricas")
[1/840*(105*sqrt(2)*((15*B - 19*C)*cos(d*x + c)^5 + 2*(15*B - 19*C)*cos(d* x + c)^4 + (15*B - 19*C)*cos(d*x + c)^3)*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*co s(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1) ) + 4*((1029*B - 1201*C)*cos(d*x + c)^4 + 12*(63*B - 67*C)*cos(d*x + c)^3 - 28*(3*B - 7*C)*cos(d*x + c)^2 + 12*(7*B - 3*C)*cos(d*x + c) + 60*C)*sqrt ((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3), 1/420*(105*sqrt(2)*((15*B - 19*C)*cos(d*x + c)^5 + 2*(15*B - 19*C)*cos(d*x + c)^4 + (15*B - 19*C)*co s(d*x + c)^3)*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*((1029*B - 1201*C)*cos(d*x + c )^4 + 12*(63*B - 67*C)*cos(d*x + c)^3 - 28*(3*B - 7*C)*cos(d*x + c)^2 + 12 *(7*B - 3*C)*cos(d*x + c) + 60*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d *x + c)^3)]
\[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2 ),x, algorithm="maxima")
Time = 1.78 (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))^{3/2}} \, dx=\frac {\frac {105 \, {\left (15 \, \sqrt {2} B - 19 \, \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 \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {{\left ({\left ({\left ({\left (\frac {105 \, {\left (\sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} C a^{5} \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^{3}} - \frac {4 \, {\left (693 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 877 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {14 \, {\left (453 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 517 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {140 \, {\left (39 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 47 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1785 \, {\left (\sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{3}}\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 )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{420 \, d} \]
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2 ),x, algorithm="giac")
1/420*(105*(15*sqrt(2)*B - 19*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*sgn(cos(d*x + c) )) - ((((105*(sqrt(2)*B*a^5*sgn(cos(d*x + c)) - sqrt(2)*C*a^5*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/a^3 - 4*(693*sqrt(2)*B*a^5*sgn(cos(d*x + c)) - 877*sqrt(2)*C*a^5*sgn(cos(d*x + c)))/a^3)*tan(1/2*d*x + 1/2*c)^2 + 14*( 453*sqrt(2)*B*a^5*sgn(cos(d*x + c)) - 517*sqrt(2)*C*a^5*sgn(cos(d*x + c))) /a^3)*tan(1/2*d*x + 1/2*c)^2 - 140*(39*sqrt(2)*B*a^5*sgn(cos(d*x + c)) - 4 7*sqrt(2)*C*a^5*sgn(cos(d*x + c)))/a^3)*tan(1/2*d*x + 1/2*c)^2 + 1785*(sqr t(2)*B*a^5*sgn(cos(d*x + c)) - sqrt(2)*C*a^5*sgn(cos(d*x + c)))/a^3)*tan(1 /2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/ 2*c)^2 + a)))/d
Timed out. \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]