Integrand size = 25, antiderivative size = 204 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\frac {6 a \left (a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {6 a \left (a^2+2 b^2\right ) \tan (e+f x)}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \tan (e+f x)\right )}{5 d^2 f \sqrt {d \sec (e+f x)}} \]
6/5*a*(a^2+2*b^2)*(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(1/2*arctan(tan (f*x+e)))*EllipticE(sin(1/2*arctan(tan(f*x+e))),2^(1/2))*(sec(f*x+e)^2)^(1 /4)/d^2/f/(d*sec(f*x+e))^(1/2)-6/5*a*(a^2+2*b^2)*tan(f*x+e)/d^2/f/(d*sec(f *x+e))^(1/2)-2/5*cos(f*x+e)^2*(b-a*tan(f*x+e))*(a+b*tan(f*x+e))^2/d^2/f/(d *sec(f*x+e))^(1/2)-2/5*(2*b*(a^2+2*b^2)-a*(3*a^2+5*b^2)*tan(f*x+e))/d^2/f/ (d*sec(f*x+e))^(1/2)
Time = 3.94 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {d \sec (e+f x)} \left (-b \left (9 a^2+17 b^2\right ) \cos (e+f x)-3 a^2 b \cos (3 (e+f x))+b^3 \cos (3 (e+f x))+12 a \left (a^2+2 b^2\right ) \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+a^3 \sin (e+f x)-3 a b^2 \sin (e+f x)+a^3 \sin (3 (e+f x))-3 a b^2 \sin (3 (e+f x))\right )}{10 d^3 f} \]
(Sqrt[d*Sec[e + f*x]]*(-(b*(9*a^2 + 17*b^2)*Cos[e + f*x]) - 3*a^2*b*Cos[3* (e + f*x)] + b^3*Cos[3*(e + f*x)] + 12*a*(a^2 + 2*b^2)*Sqrt[Cos[e + f*x]]* EllipticE[(e + f*x)/2, 2] + a^3*Sin[e + f*x] - 3*a*b^2*Sin[e + f*x] + a^3* Sin[3*(e + f*x)] - 3*a*b^2*Sin[3*(e + f*x)]))/(10*d^3*f)
Time = 0.37 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3994, 495, 27, 675, 225, 212}
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.
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3994 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \int \frac {(a+b \tan (e+f x))^3}{\left (\tan ^2(e+f x)+1\right )^{9/4}}d(b \tan (e+f x))}{b d^2 f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 495 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\frac {2}{5} b^2 \int \frac {(a+b \tan (e+f x)) \left (\left (\frac {3 a^2}{b^2}+4\right ) b^2-a b \tan (e+f x)\right )}{2 b^2 \left (\tan ^2(e+f x)+1\right )^{5/4}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{5 \left (\tan ^2(e+f x)+1\right )^{5/4}}\right )}{b d^2 f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\frac {1}{5} \int \frac {(a+b \tan (e+f x)) \left (3 a^2-b \tan (e+f x) a+4 b^2\right )}{\left (\tan ^2(e+f x)+1\right )^{5/4}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{5 \left (\tan ^2(e+f x)+1\right )^{5/4}}\right )}{b d^2 f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 675 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\frac {1}{5} \left (-3 a \left (a^2+2 b^2\right ) \int \frac {1}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-\frac {4 b^2 \left (a^2+2 b^2\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}+\frac {2 a b \left (3 a^2+5 b^2\right ) \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{5 \left (\tan ^2(e+f x)+1\right )^{5/4}}\right )}{b d^2 f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 225 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\frac {1}{5} \left (-3 a \left (a^2+2 b^2\right ) \left (\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}-\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{5/4}}d(b \tan (e+f x))\right )-\frac {4 b^2 \left (a^2+2 b^2\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}+\frac {2 a b \left (3 a^2+5 b^2\right ) \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{5 \left (\tan ^2(e+f x)+1\right )^{5/4}}\right )}{b d^2 f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 212 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\frac {1}{5} \left (-3 a \left (a^2+2 b^2\right ) \left (\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )\right )-\frac {4 b^2 \left (a^2+2 b^2\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}+\frac {2 a b \left (3 a^2+5 b^2\right ) \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{5 \left (\tan ^2(e+f x)+1\right )^{5/4}}\right )}{b d^2 f \sqrt {d \sec (e+f x)}}\) |
((Sec[e + f*x]^2)^(1/4)*((-2*(a + b*Tan[e + f*x])^2*(b^2 - a*b*Tan[e + f*x ]))/(5*(1 + Tan[e + f*x]^2)^(5/4)) + ((-4*b^2*(a^2 + 2*b^2))/(1 + Tan[e + f*x]^2)^(1/4) + (2*a*b*(3*a^2 + 5*b^2)*Tan[e + f*x])/(1 + Tan[e + f*x]^2)^ (1/4) - 3*a*(a^2 + 2*b^2)*(-2*b*EllipticE[ArcTan[Tan[e + f*x]]/2, 2] + (2* b*Tan[e + f*x])/(1 + Tan[e + f*x]^2)^(1/4)))/5))/(b*d^2*f*Sqrt[d*Sec[e + f *x]])
3.6.99.3.1 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)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) , x] - Simp[a Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ a, 0] && PosQ[b/a]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (a*d - b*c*x)*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1)*Simp[a* d^2*(n - 1) - b*c^2*(2*p + 3) - b*c*d*(n + 2*p + 2)*x, x], x], x] /; FreeQ[ {a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 1] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
Int[((d_) + (e_.)*(x_))*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[a*(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + (- Simp[(c*d*f - a*e*g)*x*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Simp[(a* e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)) Int[(a + c*x^2)^(p + 1), x], x]) / ; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] && !(IntegerQ[p] && NiceSqrtQ [(-a)*c])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP art[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/2])) Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] && !IntegerQ[m] && IntegerQ[n]
Result contains complex when optimal does not.
Time = 26.00 (sec) , antiderivative size = 1289, normalized size of antiderivative = 6.32
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1289\) |
| default | \(\text {Expression too large to display}\) | \(1602\) |
2/5*a^3/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)/d^2*(3*I*cos(f*x+e)*Elliptic E(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f *x+e)+1))^(1/2)-3*I*cos(f*x+e)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/( cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+6*I*(1/(cos(f*x+e)+ 1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f* x+e)),I)-6*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot (f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)+3*I*sec(f*x+e)*EllipticE(I*(csc(f*x+e )-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2 )-3*I*sec(f*x+e)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1)) ^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+sin(f*x+e)*cos(f*x+e)^2+sin(f*x+e )*cos(f*x+e)+3*sin(f*x+e))-1/10*b^3/f/(d*sec(f*x+e))^(1/2)/d^2*(5*(-cos(f* x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln((2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^ 2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+ 1))-5*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(2*(2*cos(f*x+e)*(-cos(f*x+e) /(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e) +1)/(cos(f*x+e)+1))-4*cos(f*x+e)^2+5*sec(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1 )^2)^(1/2)*ln((2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f *x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))-5*sec(f*x+e)*( -cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(2*(2*cos(f*x+e)*(-cos(f*x+e)/(cos(f *x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (-i \, a^{3} - 2 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, a^{3} + 2 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (5 \, b^{3} \cos \left (f x + e\right ) + {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, d^{3} f} \]
-1/5*(3*sqrt(2)*(-I*a^3 - 2*I*a*b^2)*sqrt(d)*weierstrassZeta(-4, 0, weiers trassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 3*sqrt(2)*(I*a^3 + 2*I*a*b^2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f *x + e) - I*sin(f*x + e))) + 2*(5*b^3*cos(f*x + e) + (3*a^2*b - b^3)*cos(f *x + e)^3 - (a^3 - 3*a*b^2)*cos(f*x + e)^2*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^3*f)
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]