Integrand size = 25, antiderivative size = 178 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=\frac {2 a \left (a^2-6 b^2\right ) E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{f \sqrt {d \sec (e+f x)}}-\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}} \]
2*a*(a^2-6*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 )/f/(d*sec(f*x+e))^(1/2)-2*a*(a^2-6*b^2)*tan(f*x+e)/f/(d*sec(f*x+e))^(1/2) -2*(b-a*tan(f*x+e))*(a+b*tan(f*x+e))^2/f/(d*sec(f*x+e))^(1/2)-2/3*b*sec(f* x+e)^2*(6*a^2-4*b^2+3*a*b*tan(f*x+e))/f/(d*sec(f*x+e))^(1/2)
Time = 4.90 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=\frac {d \left (6 a \left (a^2-6 b^2\right ) \cos ^{\frac {3}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+b \left (-9 a^2+5 b^2+\left (-9 a^2+3 b^2\right ) \cos (2 (e+f x))+9 a b \sin (2 (e+f x))\right )\right ) (a+b \tan (e+f x))^3}{3 f (d \sec (e+f x))^{3/2} (a \cos (e+f x)+b \sin (e+f x))^3} \]
(d*(6*a*(a^2 - 6*b^2)*Cos[e + f*x]^(3/2)*EllipticE[(e + f*x)/2, 2] + b*(-9 *a^2 + 5*b^2 + (-9*a^2 + 3*b^2)*Cos[2*(e + f*x)] + 9*a*b*Sin[2*(e + f*x)]) )*(a + b*Tan[e + f*x])^3)/(3*f*(d*Sec[e + f*x])^(3/2)*(a*Cos[e + f*x] + b* Sin[e + f*x])^3)
Time = 0.34 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3994, 495, 27, 25, 676, 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.
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}}dx\) |
| \(\Big \downarrow \) 3994 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \int \frac {(a+b \tan (e+f x))^3}{\left (\tan ^2(e+f x)+1\right )^{5/4}}d(b \tan (e+f x))}{b f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 495 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (2 b^2 \int \frac {(a+b \tan (e+f x)) \left (\left (4-\frac {a^2}{b^2}\right ) b^2-5 a b \tan (e+f x)\right )}{2 b^2 \sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\int -\frac {(a+b \tan (e+f x)) \left (a^2+5 b \tan (e+f x) a-4 b^2\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (-\int \frac {(a+b \tan (e+f x)) \left (a^2+5 b \tan (e+f x) a-4 b^2\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-\frac {2 \left (b^2-a b \tan (e+f x)\right ) (a+b \tan (e+f x))^2}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 676 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (-a \left (a^2-6 b^2\right ) \int \frac {1}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-\frac {4}{3} b^2 \left (3 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4}-2 a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{3/4}-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 225 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (-a \left (a^2-6 b^2\right ) \left (\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}-\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{5/4}}d(b \tan (e+f x))\right )-\frac {4}{3} b^2 \left (3 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4}-2 a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{3/4}-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 212 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (-a \left (a^2-6 b^2\right ) \left (\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )\right )-\frac {4}{3} b^2 \left (3 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4}-2 a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{3/4}-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{b f \sqrt {d \sec (e+f x)}}\) |
((Sec[e + f*x]^2)^(1/4)*((-2*(a + b*Tan[e + f*x])^2*(b^2 - a*b*Tan[e + f*x ]))/(1 + Tan[e + f*x]^2)^(1/4) - (4*b^2*(3*a^2 - 2*b^2)*(1 + Tan[e + f*x]^ 2)^(3/4))/3 - 2*a*b^3*Tan[e + f*x]*(1 + Tan[e + f*x]^2)^(3/4) - a*(a^2 - 6 *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))))/(b*f*Sqrt[d*Sec[e + f*x]])
3.6.97.3.1 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 _Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p + 3))/(c*(2*p + 3)) Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g , p}, x] && !LeQ[p, -1]
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 = 21.30 (sec) , antiderivative size = 1114, normalized size of antiderivative = 6.26
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1114\) |
| default | \(\text {Expression too large to display}\) | \(2626\) |
2*a^3/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)*(I*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)*cos (f*x+e)-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)+2*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)-2 *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)+I*sec(f*x+e)*(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)-I*sec(f*x+ e)*(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)+sin(f*x+e))-1/6*b^3/f/(cos(f*x+e)+1)/(d*sec(f*x +e))^(1/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(-12*cos(f*x+e)*(-cos(f*x+ e)/(cos(f*x+e)+1)^2)^(1/2)+3*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))-3*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))-12*(-cos(f*x+e)/( cos(f*x+e)+1)^2)^(1/2)-4*sec(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-4 *sec(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-6*a^2*b/(d*sec(f*x+e)) ^(1/2)/f-6*a*b^2/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)*(2*I*EllipticE(I*(c sc(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)*cos(f*x+e)-2*I*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=-\frac {3 \, \sqrt {2} {\left (-i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (9 \, a b^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b^{3} - 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, d f \cos \left (f x + e\right )} \]
-1/3*(3*sqrt(2)*(-I*a^3 + 6*I*a*b^2)*sqrt(d)*cos(f*x + e)*weierstrassZeta( -4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 3*sqrt (2)*(I*a^3 - 6*I*a*b^2)*sqrt(d)*cos(f*x + e)*weierstrassZeta(-4, 0, weiers trassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - 2*(9*a*b^2*cos(f*x + e)*sin(f*x + e) + b^3 - 3*(3*a^2*b - b^3)*cos(f*x + e)^2)*sqrt(d/cos(f*x + e)))/(d*f*cos(f*x + e))
\[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3}}{\sqrt {d \sec {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\sqrt {d \sec \left (f x + e\right )}} \,d x } \]
\[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\sqrt {d \sec \left (f x + e\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \]