Integrand size = 35, antiderivative size = 274 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (326 A+283 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{128 d}+\frac {a^3 (326 A+283 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (326 A+283 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (170 A+157 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{240 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (10 A+13 B) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac {a B \sec ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d} \] Output:
1/128*a^(5/2)*(326*A+283*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1 /2))/d+1/128*a^3*(326*A+283*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+ c))^(1/2)+1/192*a^3*(326*A+283*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d *x+c))^(1/2)+1/240*a^3*(170*A+157*B)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*se c(d*x+c))^(1/2)+1/40*a^2*(10*A+13*B)*sec(d*x+c)^(7/2)*(a+a*sec(d*x+c))^(1/ 2)*sin(d*x+c)/d+1/5*a*B*sec(d*x+c)^(7/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c) /d
Leaf count is larger than twice the leaf count of optimal. \(567\) vs. \(2(274)=548\).
Time = 6.58 (sec) , antiderivative size = 567, normalized size of antiderivative = 2.07 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {163 A \sec ^{\frac {3}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2} \sin (c+d x)}{64 d (1+\sec (c+d x))^3}+\frac {283 B \sec ^{\frac {3}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2} \sin (c+d x)}{128 d (1+\sec (c+d x))^3}+\frac {163 A \sec ^{\frac {5}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2} \sin (c+d x)}{96 d (1+\sec (c+d x))^3}+\frac {283 B \sec ^{\frac {5}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2} \sin (c+d x)}{192 d (1+\sec (c+d x))^3}+\frac {17 A \sec ^{\frac {7}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2} \sin (c+d x)}{24 d (1+\sec (c+d x))^3}+\frac {283 B \sec ^{\frac {7}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2} \sin (c+d x)}{240 d (1+\sec (c+d x))^3}+\frac {21 B \sec ^{\frac {9}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2} \sin (c+d x)}{40 d (1+\sec (c+d x))^3}+\frac {A \sec ^{\frac {7}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2} \sin (c+d x)}{4 d (1+\sec (c+d x))^2}+\frac {B \sec ^{\frac {9}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2} \sin (c+d x)}{5 d (1+\sec (c+d x))^2}+\frac {163 A \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) (a (1+\sec (c+d x)))^{5/2} \tan (c+d x)}{64 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^3}+\frac {283 B \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) (a (1+\sec (c+d x)))^{5/2} \tan (c+d x)}{128 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^3} \] Input:
Integrate[Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x ]),x]
Output:
(163*A*Sec[c + d*x]^(3/2)*(a*(1 + Sec[c + d*x]))^(5/2)*Sin[c + d*x])/(64*d *(1 + Sec[c + d*x])^3) + (283*B*Sec[c + d*x]^(3/2)*(a*(1 + Sec[c + d*x]))^ (5/2)*Sin[c + d*x])/(128*d*(1 + Sec[c + d*x])^3) + (163*A*Sec[c + d*x]^(5/ 2)*(a*(1 + Sec[c + d*x]))^(5/2)*Sin[c + d*x])/(96*d*(1 + Sec[c + d*x])^3) + (283*B*Sec[c + d*x]^(5/2)*(a*(1 + Sec[c + d*x]))^(5/2)*Sin[c + d*x])/(19 2*d*(1 + Sec[c + d*x])^3) + (17*A*Sec[c + d*x]^(7/2)*(a*(1 + Sec[c + d*x]) )^(5/2)*Sin[c + d*x])/(24*d*(1 + Sec[c + d*x])^3) + (283*B*Sec[c + d*x]^(7 /2)*(a*(1 + Sec[c + d*x]))^(5/2)*Sin[c + d*x])/(240*d*(1 + Sec[c + d*x])^3 ) + (21*B*Sec[c + d*x]^(9/2)*(a*(1 + Sec[c + d*x]))^(5/2)*Sin[c + d*x])/(4 0*d*(1 + Sec[c + d*x])^3) + (A*Sec[c + d*x]^(7/2)*(a*(1 + Sec[c + d*x]))^( 5/2)*Sin[c + d*x])/(4*d*(1 + Sec[c + d*x])^2) + (B*Sec[c + d*x]^(9/2)*(a*( 1 + Sec[c + d*x]))^(5/2)*Sin[c + d*x])/(5*d*(1 + Sec[c + d*x])^2) + (163*A *ArcSin[Sqrt[1 - Sec[c + d*x]]]*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x]) /(64*d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])^3) + (283*B*ArcSin[Sqrt[1 - Sec[c + d*x]]]*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(128*d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])^3)
Time = 1.53 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4506, 27, 3042, 4506, 27, 3042, 4504, 3042, 4290, 3042, 4290, 3042, 4288, 222}
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.
\(\displaystyle \int \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2} (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4506 |
\(\displaystyle \frac {1}{5} \int \frac {1}{2} \sec ^{\frac {5}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2} (5 a (2 A+B)+a (10 A+13 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{10} \int \sec ^{\frac {5}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2} (5 a (2 A+B)+a (10 A+13 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (5 a (2 A+B)+a (10 A+13 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4506 |
\(\displaystyle \frac {1}{10} \left (\frac {1}{4} \int \frac {1}{2} \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} \left (5 (26 A+21 B) a^2+(170 A+157 B) \sec (c+d x) a^2\right )dx+\frac {a^2 (10 A+13 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} \left (5 (26 A+21 B) a^2+(170 A+157 B) \sec (c+d x) a^2\right )dx+\frac {a^2 (10 A+13 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (5 (26 A+21 B) a^2+(170 A+157 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {a^2 (10 A+13 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4504 |
\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (326 A+283 B) \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (170 A+157 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (10 A+13 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (326 A+283 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^3 (170 A+157 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (10 A+13 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4290 |
\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (326 A+283 B) \left (\frac {3}{4} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (170 A+157 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (10 A+13 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (326 A+283 B) \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (170 A+157 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (10 A+13 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4290 |
\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (326 A+283 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (170 A+157 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (10 A+13 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (326 A+283 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (170 A+157 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (10 A+13 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4288 |
\(\displaystyle \frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{6} a^2 (326 A+283 B) \left (\frac {3}{4} \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (170 A+157 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (10 A+13 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{10} \left (\frac {a^2 (10 A+13 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}+\frac {1}{8} \left (\frac {a^3 (170 A+157 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {5}{6} a^2 (326 A+283 B) \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\right )\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
Input:
Int[Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]
Output:
(a*B*Sec[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + ( (a^2*(10*A + 13*B)*Sec[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x ])/(4*d) + ((a^3*(170*A + 157*B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(3*d*Sqr t[a + a*Sec[c + d*x]]) + (5*a^2*(326*A + 283*B)*((a*Sec[c + d*x]^(5/2)*Sin [c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (3*((Sqrt[a]*ArcSinh[(Sqrt[a]* Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4))/6)/8)/10
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)] Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a , b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 1))) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)) Int[Sqrt[a + b*Csc[e + f* x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ [A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && !LtQ[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)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Time = 0.92 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.30
\[\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (4890 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+4245 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-4890 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-4245 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (4890 \cos \left (d x +c \right )^{3}+3260 \cos \left (d x +c \right )^{2}+1840 \cos \left (d x +c \right )+480\right ) \sqrt {2}\, A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )+\left (4245 \cos \left (d x +c \right )^{4}+2830 \cos \left (d x +c \right )^{3}+2264 \cos \left (d x +c \right )^{2}+1392 \cos \left (d x +c \right )+384\right ) \sqrt {2}\, B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )\right )}{3840 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\]
Input:
int(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)
Output:
1/3840/d*a^2*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(5/2)/(1+cos(d*x+c))/(-1/ (1+cos(d*x+c)))^(1/2)*(4890*A*cos(d*x+c)^3*arctan(1/2*(cot(d*x+c)-csc(d*x+ c)+1)/(-1/(1+cos(d*x+c)))^(1/2))+4245*B*cos(d*x+c)^3*arctan(1/2*(cot(d*x+c )-csc(d*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))-4890*A*cos(d*x+c)^3*arctan(1/2* (-cot(d*x+c)+csc(d*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))-4245*B*cos(d*x+c)^3* arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))+(4890*cos (d*x+c)^3+3260*cos(d*x+c)^2+1840*cos(d*x+c)+480)*2^(1/2)*A*(-2/(1+cos(d*x+ c)))^(1/2)*tan(d*x+c)+(4245*cos(d*x+c)^4+2830*cos(d*x+c)^3+2264*cos(d*x+c) ^2+1392*cos(d*x+c)+384)*2^(1/2)*B*(-2/(1+cos(d*x+c)))^(1/2)*tan(d*x+c)*sec (d*x+c))
Time = 0.21 (sec) , antiderivative size = 555, normalized size of antiderivative = 2.03 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algo rithm="fricas")
Output:
[1/7680*(15*((326*A + 283*B)*a^2*cos(d*x + c)^5 + (326*A + 283*B)*a^2*cos( d*x + c)^4)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d* x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) *sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(15*(326*A + 283*B)*a^2*cos(d*x + c)^4 + 10*(326*A + 283*B)*a^2*cos(d *x + c)^3 + 8*(230*A + 283*B)*a^2*cos(d*x + c)^2 + 48*(10*A + 29*B)*a^2*co s(d*x + c) + 384*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4), 1/3840*(15*( (326*A + 283*B)*a^2*cos(d*x + c)^5 + (326*A + 283*B)*a^2*cos(d*x + c)^4)*s qrt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(-a)*sqrt((a*cos( d*x + c) + a)/cos(d*x + c))/(a*sqrt(cos(d*x + c))*sin(d*x + c))) + 2*(15*( 326*A + 283*B)*a^2*cos(d*x + c)^4 + 10*(326*A + 283*B)*a^2*cos(d*x + c)^3 + 8*(230*A + 283*B)*a^2*cos(d*x + c)^2 + 48*(10*A + 29*B)*a^2*cos(d*x + c) + 384*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(co s(d*x + c)))/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)]
Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(5/2)*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 9242 vs. \(2 (236) = 472\).
Time = 1.11 (sec) , antiderivative size = 9242, normalized size of antiderivative = 33.73 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algo rithm="maxima")
Output:
-1/7680*(10*(1956*(sqrt(2)*a^2*sin(8*d*x + 8*c) + 4*sqrt(2)*a^2*sin(6*d*x + 6*c) + 6*sqrt(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt(2)*a^2*sin(2*d*x + 2*c))* cos(15/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 652*(sqrt(2)*a^2*s in(8*d*x + 8*c) + 4*sqrt(2)*a^2*sin(6*d*x + 6*c) + 6*sqrt(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt(2)*a^2*sin(2*d*x + 2*c))*cos(13/4*arctan2(sin(2*d*x + 2*c ), cos(2*d*x + 2*c))) + 6204*(sqrt(2)*a^2*sin(8*d*x + 8*c) + 4*sqrt(2)*a^2 *sin(6*d*x + 6*c) + 6*sqrt(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt(2)*a^2*sin(2*d *x + 2*c))*cos(11/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2060*(s qrt(2)*a^2*sin(8*d*x + 8*c) + 4*sqrt(2)*a^2*sin(6*d*x + 6*c) + 6*sqrt(2)*a ^2*sin(4*d*x + 4*c) + 4*sqrt(2)*a^2*sin(2*d*x + 2*c))*cos(9/4*arctan2(sin( 2*d*x + 2*c), cos(2*d*x + 2*c))) + 2060*(sqrt(2)*a^2*sin(8*d*x + 8*c) + 4* sqrt(2)*a^2*sin(6*d*x + 6*c) + 6*sqrt(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt(2)* a^2*sin(2*d*x + 2*c))*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 6204*(sqrt(2)*a^2*sin(8*d*x + 8*c) + 4*sqrt(2)*a^2*sin(6*d*x + 6*c) + 6 *sqrt(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt(2)*a^2*sin(2*d*x + 2*c))*cos(5/4*ar ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 652*(sqrt(2)*a^2*sin(8*d*x + 8*c) + 4*sqrt(2)*a^2*sin(6*d*x + 6*c) + 6*sqrt(2)*a^2*sin(4*d*x + 4*c) + 4 *sqrt(2)*a^2*sin(2*d*x + 2*c))*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1956*(sqrt(2)*a^2*sin(8*d*x + 8*c) + 4*sqrt(2)*a^2*sin(6*d*x + 6*c) + 6*sqrt(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt(2)*a^2*sin(2*d*x + 2*c)...
Leaf count of result is larger than twice the leaf count of optimal. 1191 vs. \(2 (236) = 472\).
Time = 5.16 (sec) , antiderivative size = 1191, normalized size of antiderivative = 4.35 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algo rithm="giac")
Output:
1/3840*(15*(326*A*a^(5/2)*sgn(cos(d*x + c)) + 283*B*a^(5/2)*sgn(cos(d*x + 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))^2 - a*(2*sqrt(2) + 3))) - 15*(326*A*a^(5/2)*sgn(cos(d*x + c)) + 283 *B*a^(5/2)*sgn(cos(d*x + 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))^2 + a*(2*sqrt(2) - 3))) + 4*(4890*sqrt(2)* (sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^18*A*a ^(7/2)*sgn(cos(d*x + c)) + 4245*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sq rt(a*tan(1/2*d*x + 1/2*c)^2 + a))^18*B*a^(7/2)*sgn(cos(d*x + c)) - 132030* sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) )^16*A*a^(9/2)*sgn(cos(d*x + c)) - 114615*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1 /2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^16*B*a^(9/2)*sgn(cos(d*x + c)) + 1319880*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/ 2*c)^2 + a))^14*A*a^(11/2)*sgn(cos(d*x + c)) + 1298820*sqrt(2)*(sqrt(a)*ta n(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*B*a^(11/2)*sgn (cos(d*x + c)) - 6888120*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*ta n(1/2*d*x + 1/2*c)^2 + a))^12*A*a^(13/2)*sgn(cos(d*x + c)) - 6176700*sqrt( 2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12* B*a^(13/2)*sgn(cos(d*x + c)) + 18352620*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2 *c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*a^(15/2)*sgn(cos(d*x + c)) + 16394598*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x +...
Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:
int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(5/2) ,x)
Output:
int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(5/2) , x)
\[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) b +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:
int(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)
Output:
sqrt(a)*a**2*(int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)** 5,x)*b + int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4,x)* a + 2*int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4,x)*b + 2*int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x)*a + in t(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x)*b + int(sqr t(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*a)