Integrand size = 25, antiderivative size = 198 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt {d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f} \]
2/21*a*(7*a^2-6*b^2)*d^2*(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(1/2*arc tan(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)+2/21*a*(7*a^2-6*b^2)*d^2*(d*sec(f*x+e))^ (1/2)*tan(f*x+e)/f+2/9*b*d^2*sec(f*x+e)^2*(d*sec(f*x+e))^(1/2)*(a+b*tan(f* x+e))^2/f+2/315*b*d^2*sec(f*x+e)^2*(d*sec(f*x+e))^(1/2)*(154*a^2-28*b^2+65 *a*b*tan(f*x+e))/f
Time = 4.21 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.79 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=-\frac {2 d (d \sec (e+f x))^{3/2} \left (63 b \left (-3 a^2+b^2\right ) \cos ^2(e+f x)-15 a \left (7 a^2-6 b^2\right ) \cos ^{\frac {9}{2}}(e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )-15 a \left (7 a^2-6 b^2\right ) \cos ^3(e+f x) \sin (e+f x)-\frac {5}{2} b^2 (14 b+27 a \sin (2 (e+f x)))\right ) (a+b \tan (e+f x))^3}{315 f (a \cos (e+f x)+b \sin (e+f x))^3} \]
(-2*d*(d*Sec[e + f*x])^(3/2)*(63*b*(-3*a^2 + b^2)*Cos[e + f*x]^2 - 15*a*(7 *a^2 - 6*b^2)*Cos[e + f*x]^(9/2)*EllipticF[(e + f*x)/2, 2] - 15*a*(7*a^2 - 6*b^2)*Cos[e + f*x]^3*Sin[e + f*x] - (5*b^2*(14*b + 27*a*Sin[2*(e + f*x)] ))/2)*(a + b*Tan[e + f*x])^3)/(315*f*(a*Cos[e + f*x] + b*Sin[e + f*x])^3)
Time = 0.36 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3994, 497, 27, 25, 676, 211, 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.
| \(\displaystyle \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3dx\) |
| \(\Big \downarrow \) 3994 |
| \(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \int (a+b \tan (e+f x))^3 \sqrt [4]{\tan ^2(e+f x)+1}d(b \tan (e+f x))}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 497 |
| \(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {2}{9} b^2 \int -\frac {(a+b \tan (e+f x)) \left (\left (4-\frac {9 a^2}{b^2}\right ) b^2-13 a b \tan (e+f x)\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{2 b^2}d(b \tan (e+f x))+\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2-\frac {1}{9} \int -\left ((a+b \tan (e+f x)) \left (9 a^2+13 b \tan (e+f x) a-4 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}\right )d(b \tan (e+f x))\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {1}{9} \int (a+b \tan (e+f x)) \left (9 a^2+13 b \tan (e+f x) a-4 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}d(b \tan (e+f x))+\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 676 |
| \(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {1}{9} \left (\frac {9}{7} a \left (7 a^2-6 b^2\right ) \int \sqrt [4]{\tan ^2(e+f x)+1}d(b \tan (e+f x))+\frac {4}{5} b^2 \left (11 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{5/4}+\frac {26}{7} a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{5/4}\right )+\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {1}{9} \left (\frac {9}{7} a \left (7 a^2-6 b^2\right ) \left (\frac {1}{3} \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))+\frac {2}{3} b \sqrt [4]{\tan ^2(e+f x)+1} \tan (e+f x)\right )+\frac {4}{5} b^2 \left (11 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{5/4}+\frac {26}{7} a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{5/4}\right )+\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 229 |
| \(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {1}{9} \left (\frac {9}{7} a \left (7 a^2-6 b^2\right ) \left (\frac {2}{3} b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )+\frac {2}{3} b \tan (e+f x) \sqrt [4]{\tan ^2(e+f x)+1}\right )+\frac {4}{5} b^2 \left (11 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{5/4}+\frac {26}{7} a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{5/4}\right )+\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
(d^2*Sqrt[d*Sec[e + f*x]]*((2*b^2*(a + b*Tan[e + f*x])^2*(1 + Tan[e + f*x] ^2)^(5/4))/9 + ((4*b^2*(11*a^2 - 2*b^2)*(1 + Tan[e + f*x]^2)^(5/4))/5 + (2 6*a*b^3*Tan[e + f*x]*(1 + Tan[e + f*x]^2)^(5/4))/7 + (9*a*(7*a^2 - 6*b^2)* ((2*b*EllipticF[ArcTan[Tan[e + f*x]]/2, 2])/3 + (2*b*Tan[e + f*x]*(1 + Tan [e + f*x]^2)^(1/4))/3))/7)/9))/(b*f*(Sec[e + f*x]^2)^(1/4))
3.6.94.3.1 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)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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 = 619.10 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.76
| method | result | size |
| default | \(-\frac {2 d^{2} \sqrt {d \sec \left (f x +e \right )}\, \left (105 i \cos \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a^{3}-90 i \cos \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a \,b^{2}+105 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a^{3}-90 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a \,b^{2}-105 \tan \left (f x +e \right ) a^{3}+90 \tan \left (f x +e \right ) a \,b^{2}-189 \left (\sec ^{2}\left (f x +e \right )\right ) a^{2} b -135 \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right ) a \,b^{2}+63 b^{3} \left (\sec ^{2}\left (f x +e \right )\right )-35 \left (\sec ^{4}\left (f x +e \right )\right ) b^{3}\right )}{315 f}\) | \(349\) |
| parts | \(-\frac {2 a^{3} \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-\tan \left (f x +e \right )\right )}{3 f}+\frac {2 b^{3} \left (\frac {\left (d \sec \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {d^{2} \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5}\right )}{f \,d^{2}}+\frac {6 a^{2} b \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}+\frac {2 i a \,b^{2} \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (2 F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+2 \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+2 i \tan \left (f x +e \right )-3 i \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )\right )}{7 f}\) | \(370\) |
-2/315*d^2/f*(d*sec(f*x+e))^(1/2)*(105*I*cos(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^3-90*I*cos(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+105*I*(1/(cos(f*x+e)+1 ))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(cos(f*x+e)/(cos(f*x+e)+1) )^(1/2)*a^3-90*I*(1/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+ e)),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2-105*tan(f*x+e)*a^3+90*tan(f *x+e)*a*b^2-189*sec(f*x+e)^2*a^2*b-135*tan(f*x+e)*sec(f*x+e)^2*a*b^2+63*b^ 3*sec(f*x+e)^2-35*sec(f*x+e)^4*b^3)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.02 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\frac {-15 i \, \sqrt {2} {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 15 i \, \sqrt {2} {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (35 \, b^{3} d^{2} + 63 \, {\left (3 \, a^{2} b - b^{3}\right )} d^{2} \cos \left (f x + e\right )^{2} + 15 \, {\left (9 \, a b^{2} d^{2} \cos \left (f x + e\right ) + {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{2} \cos \left (f x + e\right )^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{315 \, f \cos \left (f x + e\right )^{4}} \]
1/315*(-15*I*sqrt(2)*(7*a^3 - 6*a*b^2)*d^(5/2)*cos(f*x + e)^4*weierstrassP Inverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 15*I*sqrt(2)*(7*a^3 - 6*a* b^2)*d^(5/2)*cos(f*x + e)^4*weierstrassPInverse(-4, 0, cos(f*x + e) - I*si n(f*x + e)) + 2*(35*b^3*d^2 + 63*(3*a^2*b - b^3)*d^2*cos(f*x + e)^2 + 15*( 9*a*b^2*d^2*cos(f*x + e) + (7*a^3 - 6*a*b^2)*d^2*cos(f*x + e)^3)*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^4)
Timed out. \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\text {Timed out} \]
\[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \,d x } \]
\[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \,d x } \]
Timed out. \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \]