Integrand size = 42, antiderivative size = 171 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(7 B-11 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(B-C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(9 B-13 C) \tan (c+d x)}{3 a d \sqrt {a+a \sec (c+d x)}}-\frac {(3 B-7 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d} \]
-1/4*(7*B-11*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)^2*tan(d*x+c)/d/(a+a*sec(d*x+c)) ^(3/2)+1/3*(9*B-13*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/6*(3*B-7*C)* (a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d
Time = 1.88 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-3 \sqrt {2} (7 B-11 C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)+\sqrt {1-\sec (c+d x)} \left (15 B-19 C+12 (B-C) \sec (c+d x)+4 C \sec ^2(c+d x)\right )\right ) \tan (c+d x)}{6 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \]
((-3*Sqrt[2]*(7*B - 11*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sec[c + d*x] + Sqrt[1 - Sec[c + d*x]]*(15*B - 19*C + 12*(B - C) *Sec[c + d*x] + 4*C*Sec[c + d*x]^2))*Tan[c + d*x])/(6*d*Sqrt[1 - Sec[c + d *x]]*(a*(1 + Sec[c + d*x]))^(3/2))
Time = 1.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4560, 3042, 4507, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\sec ^3(c+d x) (B+C \sec (c+d x))}{(a \sec (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (4 a (B-C)-a (3 B-7 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (4 a (B-C)-a (3 B-7 C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a (B-C)-a (3 B-7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^2 (3 B-7 C)-2 a^2 (9 B-13 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 (3 B-7 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {\sec (c+d x) \left (a^2 (3 B-7 C)-2 a^2 (9 B-13 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 (3 B-7 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^2 (3 B-7 C)-2 a^2 (9 B-13 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 (3 B-7 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {-\frac {3 a^2 (7 B-11 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^2 (9 B-13 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 (3 B-7 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {3 a^2 (7 B-11 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^2 (9 B-13 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 (3 B-7 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {-\frac {-\frac {6 a^2 (7 B-11 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^2 (9 B-13 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 (3 B-7 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {-\frac {\frac {3 \sqrt {2} a^{3/2} (7 B-11 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^2 (9 B-13 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 (3 B-7 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
((B - C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ( (-2*(3*B - 7*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) - ((3*Sqrt[2] *a^(3/2)*(7*B - 11*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Se c[c + d*x]])])/d - (4*a^2*(9*B - 13*C)*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(4*a^2)
3.4.95.3.1 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[((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]
Leaf count of result is larger than twice the leaf count of optimal. \(330\) vs. \(2(148)=296\).
Time = 0.94 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.94
| method | result | size |
| default | \(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (3 B \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-3 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+21 B \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-33 C \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-30 B \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+46 C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+27 B \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-27 C \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{12 a^{2} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )}\) | \(331\) |
| parts | \(-\frac {B \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+7 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}+9 \cot \left (d x +c \right )-9 \csc \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {C \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (3 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+33 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-46 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-27 \cot \left (d x +c \right )+27 \csc \left (d x +c \right )\right )}{12 d \,a^{2} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )}\) | \(340\) |
int(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x,me thod=_RETURNVERBOSE)
-1/12/a^2/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(3*B*(1-cos(d*x +c))^5*csc(d*x+c)^5-3*C*(1-cos(d*x+c))^5*csc(d*x+c)^5+21*B*((1-cos(d*x+c)) ^2*csc(d*x+c)^2-1)^(3/2)*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d* x+c)^2-1)^(1/2))-33*C*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)^(3/2)-30*B*(1-cos(d*x+c))^ 3*csc(d*x+c)^3+46*C*(1-cos(d*x+c))^3*csc(d*x+c)^3+27*B*(-cot(d*x+c)+csc(d* x+c))-27*C*(-cot(d*x+c)+csc(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)
Time = 0.31 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.68 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left ({\left (7 \, B - 11 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (7 \, B - 11 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, B - 11 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (B - C\right )} \cos \left (d x + c\right ) + 4 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}, \frac {3 \, \sqrt {2} {\left ({\left (7 \, B - 11 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (7 \, B - 11 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, B - 11 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (B - C\right )} \cos \left (d x + c\right ) + 4 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}\right ] \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2 ),x, algorithm="fricas")
[1/24*(3*sqrt(2)*((7*B - 11*C)*cos(d*x + c)^3 + 2*(7*B - 11*C)*cos(d*x + c )^2 + (7*B - 11*C)*cos(d*x + c))*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*(( 15*B - 19*C)*cos(d*x + c)^2 + 12*(B - C)*cos(d*x + c) + 4*C)*sqrt((a*cos(d *x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*c os(d*x + c)^2 + a^2*d*cos(d*x + c)), 1/12*(3*sqrt(2)*((7*B - 11*C)*cos(d*x + c)^3 + 2*(7*B - 11*C)*cos(d*x + c)^2 + (7*B - 11*C)*cos(d*x + c))*sqrt( a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sq rt(a)*sin(d*x + c))) + 2*((15*B - 19*C)*cos(d*x + c)^2 + 12*(B - C)*cos(d* x + c) + 4*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d *cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))]
\[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{2}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2 ),x, algorithm="maxima")
Time = 1.73 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.44 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {{\left ({\left (\frac {3 \, {\left (\sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a} - \frac {2 \, {\left (15 \, \sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 23 \, \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {27 \, {\left (\sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {3 \, {\left (7 \, \sqrt {2} B - 11 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{12 \, d} \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2 ),x, algorithm="giac")
-1/12*(((3*(sqrt(2)*B*a*sgn(cos(d*x + c)) - sqrt(2)*C*a*sgn(cos(d*x + c))) *tan(1/2*d*x + 1/2*c)^2/a - 2*(15*sqrt(2)*B*a*sgn(cos(d*x + c)) - 23*sqrt( 2)*C*a*sgn(cos(d*x + c)))/a)*tan(1/2*d*x + 1/2*c)^2 + 27*(sqrt(2)*B*a*sgn( cos(d*x + c)) - sqrt(2)*C*a*sgn(cos(d*x + c)))/a)*tan(1/2*d*x + 1/2*c)/((a *tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) - 3*(7*s qrt(2)*B - 11*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))))/d
Timed out. \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]