Integrand size = 25, antiderivative size = 146 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx=\frac {2 a \left (a^2+6 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sec ^2(e+f x)^{3/4}}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (a^2-2 b^2\right )+a b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}} \]
2/3*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)))*EllipticF(sin(1/2*arctan(tan(f*x+e))),2^(1/2))*(sec(f*x+e)^2)^(3 /4)/f/(d*sec(f*x+e))^(3/2)-2/3*(b-a*tan(f*x+e))*(a+b*tan(f*x+e))^2/f/(d*se c(f*x+e))^(3/2)-2/3*b*sec(f*x+e)^2*(2*a^2-4*b^2+a*b*tan(f*x+e))/f/(d*sec(f *x+e))^(3/2)
Time = 3.80 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx=\frac {\sec ^2(e+f x) \left (-3 a^2 b+7 b^3+\left (-3 a^2 b+b^3\right ) \cos (2 (e+f x))+2 a \left (a^2+6 b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+a^3 \sin (2 (e+f x))-3 a b^2 \sin (2 (e+f x))\right )}{3 f (d \sec (e+f x))^{3/2}} \]
(Sec[e + f*x]^2*(-3*a^2*b + 7*b^3 + (-3*a^2*b + b^3)*Cos[2*(e + f*x)] + 2* a*(a^2 + 6*b^2)*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] + a^3*Sin[2*( e + f*x)] - 3*a*b^2*Sin[2*(e + f*x)]))/(3*f*(d*Sec[e + f*x])^(3/2))
Time = 0.36 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3994, 495, 27, 676, 229}
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.
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3994 |
| \(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \int \frac {(a+b \tan (e+f x))^3}{\left (\tan ^2(e+f x)+1\right )^{7/4}}d(b \tan (e+f x))}{b f (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 495 |
| \(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {2}{3} b^2 \int \frac {(a+b \tan (e+f x)) \left (\left (\frac {a^2}{b^2}+4\right ) b^2-3 a b \tan (e+f x)\right )}{2 b^2 \left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{3 \left (\tan ^2(e+f x)+1\right )^{3/4}}\right )}{b f (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {1}{3} \int \frac {(a+b \tan (e+f x)) \left (a^2-3 b \tan (e+f x) a+4 b^2\right )}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{3 \left (\tan ^2(e+f x)+1\right )^{3/4}}\right )}{b f (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 676 |
| \(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {1}{3} \left (a \left (a^2+6 b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))-4 b^2 \left (a^2-2 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}-2 a b^3 \tan (e+f x) \sqrt [4]{\tan ^2(e+f x)+1}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{3 \left (\tan ^2(e+f x)+1\right )^{3/4}}\right )}{b f (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 229 |
| \(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {1}{3} \left (2 a b \left (a^2+6 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )-4 b^2 \left (a^2-2 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}-2 a b^3 \tan (e+f x) \sqrt [4]{\tan ^2(e+f x)+1}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{3 \left (\tan ^2(e+f x)+1\right )^{3/4}}\right )}{b f (d \sec (e+f x))^{3/2}}\) |
((Sec[e + f*x]^2)^(3/4)*((-2*(a + b*Tan[e + f*x])^2*(b^2 - a*b*Tan[e + f*x ]))/(3*(1 + Tan[e + f*x]^2)^(3/4)) + (2*a*b*(a^2 + 6*b^2)*EllipticF[ArcTan [Tan[e + f*x]]/2, 2] - 4*b^2*(a^2 - 2*b^2)*(1 + Tan[e + f*x]^2)^(1/4) - 2* a*b^3*Tan[e + f*x]*(1 + Tan[e + f*x]^2)^(1/4))/3))/(b*f*(d*Sec[e + f*x])^( 3/2))
3.6.98.3.1 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)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) )*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], 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 = 23.96 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.21
| method | result | size |
| default | \(-\frac {2 \left (i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a^{3}+6 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a \,b^{2}+i \sec \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a^{3}+6 i \sec \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a \,b^{2}+3 \cos \left (f x +e \right ) a^{2} b -\cos \left (f x +e \right ) b^{3}-a^{3} \sin \left (f x +e \right )+3 a \,b^{2} \sin \left (f x +e \right )-3 \sec \left (f x +e \right ) b^{3}\right )}{3 d f \sqrt {d \sec \left (f x +e \right )}}\) | \(323\) |
| parts | \(\text {Expression too large to display}\) | \(1146\) |
-2/3/d/f/(d*sec(f*x+e))^(1/2)*(I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos (f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*a^3+6*I*(1/(cos(f *x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)- cot(f*x+e)),I)*a*b^2+I*sec(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(co s(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*a^3+6*I*sec(f*x+ e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I* (csc(f*x+e)-cot(f*x+e)),I)*a*b^2+3*cos(f*x+e)*a^2*b-cos(f*x+e)*b^3-a^3*sin (f*x+e)+3*a*b^2*sin(f*x+e)-3*sec(f*x+e)*b^3)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx=\frac {\sqrt {2} {\left (-i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2} {\left (i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (3 \, b^{3} - {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, d^{2} f} \]
1/3*(sqrt(2)*(-I*a^3 - 6*I*a*b^2)*sqrt(d)*weierstrassPInverse(-4, 0, cos(f *x + e) + I*sin(f*x + e)) + sqrt(2)*(I*a^3 + 6*I*a*b^2)*sqrt(d)*weierstras sPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(3*b^3 - (3*a^2*b - b^ 3)*cos(f*x + e)^2 + (a^3 - 3*a*b^2)*cos(f*x + e)*sin(f*x + e))*sqrt(d/cos( f*x + e)))/(d^2*f)
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]