Integrand size = 25, antiderivative size = 129 \[ \int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx=\frac {2 a \left (a^2-2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 b \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f}+\frac {2 b \sqrt {d \sec (e+f x)} \left (2 \left (7 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{5 f} \]
2*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)))*EllipticF(sin(1/2*arctan(tan(f*x+e))),2^(1/2))*(d*sec(f*x+e))^(1/2 )/f/(sec(f*x+e)^2)^(1/4)+2/5*b*(d*sec(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2/f+2 /5*b*(d*sec(f*x+e))^(1/2)*(14*a^2-4*b^2+3*a*b*tan(f*x+e))/f
Time = 3.91 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.02 \[ \int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx=-\frac {2 \sqrt {d \sec (e+f x)} \left (5 b \left (-3 a^2+b^2\right ) \cos ^3(e+f x)-5 a \left (a^2-2 b^2\right ) \cos ^{\frac {7}{2}}(e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )-\frac {1}{2} b^2 \cos (e+f x) (2 b+5 a \sin (2 (e+f x)))\right ) (a+b \tan (e+f x))^3}{5 f (a \cos (e+f x)+b \sin (e+f x))^3} \]
(-2*Sqrt[d*Sec[e + f*x]]*(5*b*(-3*a^2 + b^2)*Cos[e + f*x]^3 - 5*a*(a^2 - 2 *b^2)*Cos[e + f*x]^(7/2)*EllipticF[(e + f*x)/2, 2] - (b^2*Cos[e + f*x]*(2* b + 5*a*Sin[2*(e + f*x)]))/2)*(a + b*Tan[e + f*x])^3)/(5*f*(a*Cos[e + f*x] + b*Sin[e + f*x])^3)
Time = 0.34 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3994, 497, 27, 25, 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.
| \(\displaystyle \int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3dx\) |
| \(\Big \downarrow \) 3994 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \int \frac {(a+b \tan (e+f x))^3}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 497 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {2}{5} b^2 \int -\frac {(a+b \tan (e+f x)) \left (\left (4-\frac {5 a^2}{b^2}\right ) b^2-9 a b \tan (e+f x)\right )}{2 b^2 \left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))+\frac {2}{5} b^2 \sqrt [4]{\tan ^2(e+f x)+1} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {2}{5} b^2 \sqrt [4]{\tan ^2(e+f x)+1} (a+b \tan (e+f x))^2-\frac {1}{5} \int -\frac {(a+b \tan (e+f x)) \left (5 a^2+9 b \tan (e+f x) a-4 b^2\right )}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {1}{5} \int \frac {(a+b \tan (e+f x)) \left (5 a^2+9 b \tan (e+f x) a-4 b^2\right )}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))+\frac {2}{5} b^2 \sqrt [4]{\tan ^2(e+f x)+1} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 676 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {1}{5} \left (5 a \left (a^2-2 b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))+4 b^2 \left (7 a^2-2 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}+6 a b^3 \tan (e+f x) \sqrt [4]{\tan ^2(e+f x)+1}\right )+\frac {2}{5} b^2 \sqrt [4]{\tan ^2(e+f x)+1} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 229 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {1}{5} \left (10 a b \left (a^2-2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )+4 b^2 \left (7 a^2-2 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}+6 a b^3 \tan (e+f x) \sqrt [4]{\tan ^2(e+f x)+1}\right )+\frac {2}{5} b^2 \sqrt [4]{\tan ^2(e+f x)+1} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
(Sqrt[d*Sec[e + f*x]]*((2*b^2*(a + b*Tan[e + f*x])^2*(1 + Tan[e + f*x]^2)^ (1/4))/5 + (10*a*b*(a^2 - 2*b^2)*EllipticF[ArcTan[Tan[e + f*x]]/2, 2] + 4* b^2*(7*a^2 - 2*b^2)*(1 + Tan[e + f*x]^2)^(1/4) + 6*a*b^3*Tan[e + f*x]*(1 + Tan[e + f*x]^2)^(1/4))/5))/(b*f*(Sec[e + f*x]^2)^(1/4))
3.6.96.3.1 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[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[1/(b *(n + 2*p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^p*Simp[b*c^2*(n + 2*p + 1) - a*d^2*(n - 1) + 2*b*c*d*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, n , p}, x] && If[RationalQ[n], GtQ[n, 1], SumSimplerQ[n, -2]] && NeQ[n + 2*p + 1, 0] && 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 = 21.26 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.31
| method | result | size |
| default | \(\frac {2 \sqrt {d \sec \left (f x +e \right )}\, \left (-5 i \cos \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a^{3}+10 i \cos \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a \,b^{2}-5 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a^{3}+10 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a \,b^{2}+5 \tan \left (f x +e \right ) a \,b^{2}+15 a^{2} b -5 b^{3}+b^{3} \left (\sec ^{2}\left (f x +e \right )\right )\right )}{5 f}\) | \(298\) |
| parts | \(-\frac {2 i a^{3} \left (\cos \left (f x +e \right )+1\right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {d \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}-\frac {b^{3} \sqrt {d \sec \left (f x +e \right )}\, \left (20 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-5 \cos \left (f x +e \right ) \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right )+5 \cos \left (f x +e \right ) \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right )+20 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-4 \sec \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-4 \left (\sec ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right )}{10 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cos \left (f x +e \right )+1\right )}+\frac {6 a^{2} b \sqrt {d \sec \left (f x +e \right )}}{f}+\frac {2 a \,b^{2} \sqrt {d \sec \left (f x +e \right )}\, \left (2 i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\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}}+2 i \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 ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+\tan \left (f x +e \right )\right )}{f}\) | \(560\) |
2/5/f*(d*sec(f*x+e))^(1/2)*(-5*I*cos(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1 /2)*(1/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*a^3+10 *I*cos(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*E llipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*a*b^2-5*I*(cos(f*x+e)/(cos(f*x+e)+1) )^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*a^ 3+10*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*Elliptic F(I*(csc(f*x+e)-cot(f*x+e)),I)*a*b^2+5*tan(f*x+e)*a*b^2+15*a^2*b-5*b^3+b^3 *sec(f*x+e)^2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.27 \[ \int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx=-\frac {5 \, \sqrt {2} {\left (i \, a^{3} - 2 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 \, \sqrt {2} {\left (-i \, a^{3} + 2 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (5 \, a b^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b^{3} + 5 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, f \cos \left (f x + e\right )^{2}} \]
-1/5*(5*sqrt(2)*(I*a^3 - 2*I*a*b^2)*sqrt(d)*cos(f*x + e)^2*weierstrassPInv erse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 5*sqrt(2)*(-I*a^3 + 2*I*a*b^2 )*sqrt(d)*cos(f*x + e)^2*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f *x + e)) - 2*(5*a*b^2*cos(f*x + e)*sin(f*x + e) + b^3 + 5*(3*a^2*b - b^3)* cos(f*x + e)^2)*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^2)
\[ \int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx=\int \sqrt {d \sec {\left (e + f x \right )}} \left (a + b \tan {\left (e + f x \right )}\right )^{3}\, dx \]
\[ \int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx=\int { \sqrt {d \sec \left (f x + e\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \,d x } \]
\[ \int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx=\int { \sqrt {d \sec \left (f x + e\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \,d x } \]
Timed out. \[ \int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx=\int \sqrt {\frac {d}{\cos \left (e+f\,x\right )}}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \]