Integrand size = 43, antiderivative size = 227 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(19 A-75 B+163 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(A-9 B+17 C) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(21 A-93 B+197 C) \tan (c+d x)}{24 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(15 A-39 B+95 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d} \]
1/32*(19*A-75*B+163*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*(A-B+C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*se c(d*x+c))^(5/2)-1/16*(A-9*B+17*C)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x +c))^(3/2)-1/24*(21*A-93*B+197*C)*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)+ 1/48*(15*A-39*B+95*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^3/d
Time = 3.04 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.75 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\left (12 \sqrt {2} (19 A-75 B+163 C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x)+2 \sqrt {1-\sec (c+d x)} \left (-27 A+147 B-299 C+(-39 A+255 B-503 C) \sec (c+d x)+32 (3 B-5 C) \sec ^2(c+d x)+32 C \sec ^3(c+d x)\right )\right ) \tan (c+d x)}{96 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \]
Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a* Sec[c + d*x])^(5/2),x]
((12*Sqrt[2]*(19*A - 75*B + 163*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] *Cos[(c + d*x)/2]^4*Sec[c + d*x]^2 + 2*Sqrt[1 - Sec[c + d*x]]*(-27*A + 147 *B - 299*C + (-39*A + 255*B - 503*C)*Sec[c + d*x] + 32*(3*B - 5*C)*Sec[c + d*x]^2 + 32*C*Sec[c + d*x]^3))*Tan[c + d*x])/(96*d*Sqrt[1 - Sec[c + d*x]] *(a*(1 + Sec[c + d*x]))^(5/2))
Time = 1.41 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4572, 27, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {\sec ^3(c+d x) (2 a (A+3 B-3 C)+a (3 A-3 B+11 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec ^3(c+d x) (2 a (A+3 B-3 C)+a (3 A-3 B+11 C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (2 a (A+3 B-3 C)+a (3 A-3 B+11 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int -\frac {\sec ^2(c+d x) \left (4 a^2 (A-9 B+17 C)-a^2 (15 A-39 B+95 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (A-9 B+17 C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {\sec ^2(c+d x) \left (4 a^2 (A-9 B+17 C)-a^2 (15 A-39 B+95 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (A-9 B+17 C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^2 (A-9 B+17 C)-a^2 (15 A-39 B+95 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (A-9 B+17 C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^3 (15 A-39 B+95 C)-2 a^3 (21 A-93 B+197 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a (15 A-39 B+95 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A-9 B+17 C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {\sec (c+d x) \left (a^3 (15 A-39 B+95 C)-2 a^3 (21 A-93 B+197 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a (15 A-39 B+95 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A-9 B+17 C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^3 (15 A-39 B+95 C)-2 a^3 (21 A-93 B+197 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a (15 A-39 B+95 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A-9 B+17 C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {-\frac {-\frac {3 a^3 (19 A-75 B+163 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^3 (21 A-93 B+197 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (15 A-39 B+95 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A-9 B+17 C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {3 a^3 (19 A-75 B+163 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^3 (21 A-93 B+197 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (15 A-39 B+95 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A-9 B+17 C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {6 a^3 (19 A-75 B+163 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^3 (21 A-93 B+197 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (15 A-39 B+95 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A-9 B+17 C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {3 \sqrt {2} a^{5/2} (19 A-75 B+163 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^3 (21 A-93 B+197 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (15 A-39 B+95 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}-\frac {a (A-9 B+17 C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
-1/4*((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/ 2)) + (-1/2*(a*(A - 9*B + 17*C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec [c + d*x])^(3/2)) - ((-2*a*(15*A - 39*B + 95*C)*Sqrt[a + a*Sec[c + d*x]]*T an[c + d*x])/(3*d) - ((3*Sqrt[2]*a^(5/2)*(19*A - 75*B + 163*C)*ArcTan[(Sqr t[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d - (4*a^3*(21*A - 93*B + 197*C)*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(4*a^2)) /(8*a^2)
3.6.28.3.1 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_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(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*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(532\) vs. \(2(200)=400\).
Time = 1.28 (sec) , antiderivative size = 533, normalized size of antiderivative = 2.35
| 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 (6 A \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-6 B \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+6 C \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+21 A \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-45 B \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+69 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+57 A \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}}-225 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 )+489 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}}-60 A \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+300 B \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-668 C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+33 A \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-249 B \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+465 C \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{96 a^{3} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )}\) | \(533\) |
| parts | \(\frac {A \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (2 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+11 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+19 \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 )\right )}{32 d \,a^{3}}-\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 (2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+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 ) \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}+83 \cot \left (d x +c \right )-83 \csc \left (d x +c \right )\right )}{32 d \,a^{3}}+\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 (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 )}\) | \(582\) |
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, method=_RETURNVERBOSE)
1/96/a^3/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(6*A*(1-cos(d*x+ c))^7*csc(d*x+c)^7-6*B*(1-cos(d*x+c))^7*csc(d*x+c)^7+6*C*(1-cos(d*x+c))^7* csc(d*x+c)^7+21*A*(1-cos(d*x+c))^5*csc(d*x+c)^5-45*B*(1-cos(d*x+c))^5*csc( d*x+c)^5+69*C*(1-cos(d*x+c))^5*csc(d*x+c)^5+57*A*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)-225*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))+489*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)-60*A*(1-cos(d*x+c))^3*csc(d*x+c)^3+300*B*(1-cos(d*x+c))^3*csc(d*x +c)^3-668*C*(1-cos(d*x+c))^3*csc(d*x+c)^3+33*A*(-cot(d*x+c)+csc(d*x+c))-24 9*B*(-cot(d*x+c)+csc(d*x+c))+465*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 = 593, normalized size of antiderivative = 2.61 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {3 \, \sqrt {2} {\left ({\left (19 \, A - 75 \, B + 163 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (19 \, A - 75 \, B + 163 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A - 75 \, B + 163 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (19 \, A - 75 \, B + 163 \, 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 (27 \, A - 147 \, B + 299 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A - 255 \, B + 503 \, C\right )} \cos \left (d x + c\right )^{2} - 32 \, {\left (3 \, B - 5 \, C\right )} \cos \left (d x + c\right ) - 32 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}}, -\frac {3 \, \sqrt {2} {\left ({\left (19 \, A - 75 \, B + 163 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (19 \, A - 75 \, B + 163 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A - 75 \, B + 163 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (19 \, A - 75 \, B + 163 \, 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 (27 \, A - 147 \, B + 299 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A - 255 \, B + 503 \, C\right )} \cos \left (d x + c\right )^{2} - 32 \, {\left (3 \, B - 5 \, C\right )} \cos \left (d x + c\right ) - 32 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{96 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}}\right ] \]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5 /2),x, algorithm="fricas")
[-1/192*(3*sqrt(2)*((19*A - 75*B + 163*C)*cos(d*x + c)^4 + 3*(19*A - 75*B + 163*C)*cos(d*x + c)^3 + 3*(19*A - 75*B + 163*C)*cos(d*x + c)^2 + (19*A - 75*B + 163*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*((27*A - 147*B + 299*C)*cos(d*x + c)^3 + (39*A - 255*B + 503*C)*cos(d*x + c)^2 - 32*(3*B - 5*C)*cos(d*x + c) - 32*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c ))*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d* cos(d*x + c)^2 + a^3*d*cos(d*x + c)), -1/96*(3*sqrt(2)*((19*A - 75*B + 163 *C)*cos(d*x + c)^4 + 3*(19*A - 75*B + 163*C)*cos(d*x + c)^3 + 3*(19*A - 75 *B + 163*C)*cos(d*x + c)^2 + (19*A - 75*B + 163*C)*cos(d*x + c))*sqrt(a)*a rctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a )*sin(d*x + c))) + 2*((27*A - 147*B + 299*C)*cos(d*x + c)^3 + (39*A - 255* B + 503*C)*cos(d*x + c)^2 - 32*(3*B - 5*C)*cos(d*x + c) - 32*C)*sqrt((a*co s(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3* d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))]
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a*(sec( c + d*x) + 1))**(5/2), x)
Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5 /2),x, algorithm="maxima")
Time = 1.68 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {{\left ({\left (3 \, {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} - B a^{5} + C a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {\sqrt {2} {\left (7 \, A a^{5} - 15 \, B a^{5} + 23 \, C a^{5}\right )}}{a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {4 \, \sqrt {2} {\left (15 \, A a^{5} - 75 \, B a^{5} + 167 \, C a^{5}\right )}}{a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {3 \, \sqrt {2} {\left (11 \, A a^{5} - 83 \, B a^{5} + 155 \, C a^{5}\right )}}{a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {3 \, \sqrt {2} {\left (19 \, A - 75 \, B + 163 \, C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{96 \, d} \]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5 /2),x, algorithm="giac")
1/96*(((3*(2*sqrt(2)*(A*a^5 - B*a^5 + C*a^5)*tan(1/2*d*x + 1/2*c)^2/(a^6*s gn(cos(d*x + c))) + sqrt(2)*(7*A*a^5 - 15*B*a^5 + 23*C*a^5)/(a^6*sgn(cos(d *x + c))))*tan(1/2*d*x + 1/2*c)^2 - 4*sqrt(2)*(15*A*a^5 - 75*B*a^5 + 167*C *a^5)/(a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 3*sqrt(2)*(11*A*a^ 5 - 83*B*a^5 + 155*C*a^5)/(a^6*sgn(cos(d*x + c))))*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*sqr t(2)*(19*A - 75*B + 163*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. \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]