Integrand size = 35, antiderivative size = 259 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {(75 A+283 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A+21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(465 A+1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(45 A+157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(195 A+787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d} \]
-1/32*(75*A+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*(A+C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+ c))^(5/2)-1/16*(5*A+21*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/ 2)+1/120*(465*A+1729*C)*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)+1/80*(45*A +157*C)*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)-1/240*(195*A+ 787*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^3/d
Time = 2.97 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.67 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\left (-30 \sqrt {2} (75 A+283 C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x)+\sqrt {1-\sec (c+d x)} \left (735 A+2671 C+5 (255 A+911 C) \sec (c+d x)+32 (15 A+49 C) \sec ^2(c+d x)-160 C \sec ^3(c+d x)+96 C \sec ^4(c+d x)\right )\right ) \tan (c+d x)}{240 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \]
((-30*Sqrt[2]*(75*A + 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]]*(735*A + 2671*C + 5* (255*A + 911*C)*Sec[c + d*x] + 32*(15*A + 49*C)*Sec[c + d*x]^2 - 160*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 = 1.73 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.10, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4573, 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.
\(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4573 |
\(\displaystyle -\frac {\int \frac {\sec ^4(c+d x) (8 a C-a (5 A+13 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\sec ^4(c+d x) (8 a C-a (5 A+13 C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (8 a C-a (5 A+13 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4507 |
\(\displaystyle -\frac {\frac {\int \frac {\sec ^3(c+d x) \left (6 a^2 (5 A+21 C)-a^2 (45 A+157 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\int \frac {\sec ^3(c+d x) \left (6 a^2 (5 A+21 C)-a^2 (45 A+157 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (6 a^2 (5 A+21 C)-a^2 (45 A+157 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4509 |
\(\displaystyle -\frac {\frac {\frac {2 \int -\frac {\sec ^2(c+d x) \left (4 a^3 (45 A+157 C)-a^3 (195 A+787 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (45 A+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {-\frac {\int \frac {\sec ^2(c+d x) \left (4 a^3 (45 A+157 C)-a^3 (195 A+787 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (45 A+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^3 (45 A+157 C)-a^3 (195 A+787 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a^2 (45 A+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4498 |
\(\displaystyle -\frac {\frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^4 (195 A+787 C)-2 a^4 (465 A+1729 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (195 A+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {-\frac {-\frac {\int \frac {\sec (c+d x) \left (a^4 (195 A+787 C)-2 a^4 (465 A+1729 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (195 A+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^4 (195 A+787 C)-2 a^4 (465 A+1729 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a^2 (195 A+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4489 |
\(\displaystyle -\frac {\frac {-\frac {-\frac {15 a^4 (75 A+283 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^4 (465 A+1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (195 A+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {-\frac {-\frac {15 a^4 (75 A+283 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^4 (465 A+1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (195 A+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4282 |
\(\displaystyle -\frac {\frac {-\frac {-\frac {-\frac {30 a^4 (75 A+283 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^4 (465 A+1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (195 A+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {-\frac {2 a^2 (45 A+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}-\frac {-\frac {2 a^2 (195 A+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}-\frac {\frac {15 \sqrt {2} a^{7/2} (75 A+283 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^4 (465 A+1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}}{4 a^2}+\frac {a (5 A+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
-1/4*((A + C)*Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) - ((a*(5*A + 21*C)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^ (3/2)) + ((-2*a^2*(45*A + 157*C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) - ((-2*a^2*(195*A + 787*C)*Sqrt[a + a*Sec[c + d*x]]*Tan [c + d*x])/(3*d) - ((15*Sqrt[2]*a^(7/2)*(75*A + 283*C)*ArcTan[(Sqrt[a]*Tan [c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d - (4*a^4*(465*A + 1729*C )*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(5*a))/(4*a^2))/(8*a^ 2)
3.2.100.3.1 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_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) *(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*C sc[e + f*x])^n*Simp[b*C*n + A*b*(2*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, C, n}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.89 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.57
method | result | size |
parts | \(-\frac {A \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+75 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}+83 \cot \left (d x +c \right )-83 \csc \left (d x +c \right )\right )}{32 d \,a^{3}}-\frac {C \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (30 \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+435 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+4245 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}-6733 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+9865 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+4365 \cot \left (d x +c \right )-4365 \csc \left (d x +c \right )\right )}{480 d \,a^{3} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\) | \(407\) |
default | \(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (30 A \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+30 C \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+195 A \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+435 C \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+1125 A \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )+4245 C \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-1725 A \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-6733 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+2745 A \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+9865 C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-1245 A \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-4365 C \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{480 a^{3} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\) | \(423\) |
-1/32*A/d/a^3*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(2*(1-cos(d*x +c))^5*csc(d*x+c)^5+17*(1-cos(d*x+c))^3*csc(d*x+c)^3+75*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)^(1/2)+83*cot(d*x+c)-83*csc(d*x+c))-1/480*C/d/a^3*(-2*a/((1-cos(d*x +c))^2*csc(d*x+c)^2-1))^(1/2)*(30*(1-cos(d*x+c))^9*csc(d*x+c)^9+435*(1-cos (d*x+c))^7*csc(d*x+c)^7+4245*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*cs c(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(5/2)-6733*(1-cos(d *x+c))^5*csc(d*x+c)^5+9865*(1-cos(d*x+c))^3*csc(d*x+c)^3+4365*cot(d*x+c)-4 365*csc(d*x+c))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^2
Time = 0.31 (sec) , antiderivative size = 588, normalized size of antiderivative = 2.27 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {15 \, \sqrt {2} {\left ({\left (75 \, A + 283 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (75 \, A + 283 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (75 \, A + 283 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (75 \, A + 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 (735 \, A + 2671 \, C\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (255 \, A + 911 \, C\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (15 \, A + 49 \, C\right )} \cos \left (d x + c\right )^{2} - 160 \, C \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 (75 \, A + 283 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (75 \, A + 283 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (75 \, A + 283 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (75 \, A + 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 (735 \, A + 2671 \, C\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (255 \, A + 911 \, C\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (15 \, A + 49 \, C\right )} \cos \left (d x + c\right )^{2} - 160 \, C \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 ] \]
[-1/960*(15*sqrt(2)*((75*A + 283*C)*cos(d*x + c)^5 + 3*(75*A + 283*C)*cos( d*x + c)^4 + 3*(75*A + 283*C)*cos(d*x + c)^3 + (75*A + 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*((735*A + 2671*C)*cos(d*x + c) ^4 + 5*(255*A + 911*C)*cos(d*x + c)^3 + 32*(15*A + 49*C)*cos(d*x + c)^2 - 160*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)*((75*A + 283*C)*cos(d*x + c)^5 + 3*(75*A + 283*C)*cos(d*x + c)^4 + 3*(75*A + 283*C)*cos(d*x + c)^3 + (75*A + 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*((735*A + 26 71*C)*cos(d*x + c)^4 + 5*(255*A + 911*C)*cos(d*x + c)^3 + 32*(15*A + 49*C) *cos(d*x + c)^2 - 160*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)]
\[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Time = 1.65 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\frac {{\left ({\left ({\left (15 \, {\left (\frac {2 \, {\left (\sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2}} + \frac {13 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 29 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {1725 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 6733 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, {\left (549 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1973 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, {\left (83 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 291 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {15 \, {\left (75 \, \sqrt {2} A + 283 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{480 \, d} \]
-1/480*((((15*(2*(sqrt(2)*A*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 + (13*sqrt(2)*A*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 - (172 5*sqrt(2)*A*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*(549*sqrt(2)*A*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*(83*sq rt(2)*A*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*(75*sqrt(2)*A + 283*sqrt(2)*C)*log(abs(-sqrt(-a)*ta n(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a^2*s gn(cos(d*x + c))))/d
Timed out. \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]