Integrand size = 25, antiderivative size = 167 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=-\frac {2 b^3}{d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 b \left (3 a^2-b^2\right ) \cos ^2(e+f x)}{5 d^2 f \sqrt {d \sec (e+f x)}}+\frac {6 a \left (a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{5 d^2 f \sqrt {d \sec (e+f x)}}+\frac {2 a \left (a^2-3 b^2\right ) \cos (e+f x) \sin (e+f x)}{5 d^2 f \sqrt {d \sec (e+f x)}} \] Output:
-2*b^3/d^2/f/(d*sec(f*x+e))^(1/2)-2/5*b*(3*a^2-b^2)*cos(f*x+e)^2/d^2/f/(d* sec(f*x+e))^(1/2)+6/5*a*(a^2+2*b^2)*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)+2/5*a*(a^2-3*b^2) *cos(f*x+e)*sin(f*x+e)/d^2/f/(d*sec(f*x+e))^(1/2)
Time = 2.68 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {d \sec (e+f x)} \left (-b \left (9 a^2+17 b^2\right ) \cos (e+f x)-3 a^2 b \cos (3 (e+f x))+b^3 \cos (3 (e+f x))+12 a \left (a^2+2 b^2\right ) \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+a^3 \sin (e+f x)-3 a b^2 \sin (e+f x)+a^3 \sin (3 (e+f x))-3 a b^2 \sin (3 (e+f x))\right )}{10 d^3 f} \] Input:
Integrate[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(5/2),x]
Output:
(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 = 1.19, 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 |
\(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\frac {1}{5} \left (-3 a \left (a^2+2 b^2\right ) \left (\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )\right )-\frac {4 b^2 \left (a^2+2 b^2\right )}{\sqrt [4]{\tan ^2(e+f x)+1}}+\frac {2 a b \left (3 a^2+5 b^2\right ) \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{5 \left (\tan ^2(e+f x)+1\right )^{5/4}}\right )}{b d^2 f \sqrt {d \sec (e+f x)}}\) |
Input:
Int[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(5/2),x]
Output:
((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]])
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 = 27.08 (sec) , antiderivative size = 652, normalized size of antiderivative = 3.90
method | result | size |
parts | \(\frac {2 a^{3} \left (\sin \left (f x +e \right ) \left (\cos \left (f x +e \right )^{2}+\cos \left (f x +e \right )+3\right )-3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right )\right )}{5 f \left (1+\cos \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}\, d^{2}}+\frac {b^{3} \left (\sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-\cos \left (f x +e \right )+1}{1+\cos \left (f x +e \right )}\right ) \left (-5-5 \sec \left (f x +e \right )\right )+\sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-2 \cos \left (f x +e \right )+2}{1+\cos \left (f x +e \right )}\right ) \left (5+5 \sec \left (f x +e \right )\right )+4 \cos \left (f x +e \right )^{2}-20\right )}{10 f \,d^{2} \sqrt {d \sec \left (f x +e \right )}}+\frac {6 a \,b^{2} \left (\sin \left (f x +e \right ) \left (-\cos \left (f x +e \right )^{2}-\cos \left (f x +e \right )+2\right )-2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right )\right )}{5 f \left (1+\cos \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}\, d^{2}}-\frac {6 a^{2} b}{5 f \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}\) | \(652\) |
default | \(\text {Expression too large to display}\) | \(792\) |
Input:
int((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/5*a^3/f/(1+cos(f*x+e))/(d*sec(f*x+e))^(1/2)/d^2*(sin(f*x+e)*(cos(f*x+e)^ 2+cos(f*x+e)+3)-3*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^( 1/2)*(2+cos(f*x+e)+sec(f*x+e))*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)+3*I* (1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(cs c(f*x+e)-cot(f*x+e)),I)*(2+cos(f*x+e)+sec(f*x+e)))+1/10*b^3/f/d^2/(d*sec(f *x+e))^(1/2)*((-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*ln((2*cos(f*x+e)*(-cos( f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)-cos( f*x+e)+1)/(1+cos(f*x+e)))*(-5-5*sec(f*x+e))+(-cos(f*x+e)/(1+cos(f*x+e))^2) ^(1/2)*ln(2*(2*cos(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*(-cos(f*x +e)/(1+cos(f*x+e))^2)^(1/2)-cos(f*x+e)+1)/(1+cos(f*x+e)))*(5+5*sec(f*x+e)) +4*cos(f*x+e)^2-20)+6/5*a*b^2/f/(1+cos(f*x+e))/(d*sec(f*x+e))^(1/2)/d^2*(s in(f*x+e)*(-cos(f*x+e)^2-cos(f*x+e)+2)-2*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f *x+e)/(1+cos(f*x+e)))^(1/2)*(2+cos(f*x+e)+sec(f*x+e))*EllipticF(I*(csc(f*x +e)-cot(f*x+e)),I)+2*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)) )^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(2+cos(f*x+e)+sec(f*x+e)))- 6/5*a^2*b/f/(d*sec(f*x+e))^(5/2)
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (-i \, a^{3} - 2 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, a^{3} + 2 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (5 \, b^{3} \cos \left (f x + e\right ) + {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, d^{3} f} \] Input:
integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(5/2),x, algorithm="fricas")
Output:
-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)
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((a+b*tan(f*x+e))**3/(d*sec(f*x+e))**(5/2),x)
Output:
Integral((a + b*tan(e + f*x))**3/(d*sec(e + f*x))**(5/2), x)
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(5/2),x, algorithm="maxima")
Output:
integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(5/2), x)
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(5/2),x, algorithm="giac")
Output:
integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(5/2), x)
Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \] Input:
int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(5/2),x)
Output:
int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(5/2), x)
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {d}\, \left (\left (\int \frac {\sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right )^{3}}d x \right ) a^{3}+\left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )^{3}}{\sec \left (f x +e \right )^{3}}d x \right ) b^{3}+3 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{3}}d x \right ) a \,b^{2}+3 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{3}}d x \right ) a^{2} b \right )}{d^{3}} \] Input:
int((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(5/2),x)
Output:
(sqrt(d)*(int(sqrt(sec(e + f*x))/sec(e + f*x)**3,x)*a**3 + int((sqrt(sec(e + f*x))*tan(e + f*x)**3)/sec(e + f*x)**3,x)*b**3 + 3*int((sqrt(sec(e + f* x))*tan(e + f*x)**2)/sec(e + f*x)**3,x)*a*b**2 + 3*int((sqrt(sec(e + f*x)) *tan(e + f*x))/sec(e + f*x)**3,x)*a**2*b))/d**3