\(\int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx\) [323]

Optimal result

Integrand size = 25, antiderivative size = 97 \[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=-\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {2 d^2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}} \]

[Out]

-2*d^2/b/f/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2)+2*d^2*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*P
i+1/2*f*x)*EllipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2))*(b*tan(f*x+e))^(1/2)/b^2/f/(d*sec(f*x+e))^(1/2)/sin(f*
x+e)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2688, 2696, 2721, 2719} \[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=-\frac {2 d^2 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}} \]

[In]

Int[(d*Sec[e + f*x])^(3/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(-2*d^2)/(b*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]]) - (2*d^2*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[
e + f*x]])/(b^2*f*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]])

Rule 2688

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^2*(a*Sec[e
+ f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 1)/(b*f*(n + 1))), x] - Dist[a^2*((m - 2)/(b^2*(n + 1))), Int[(a*Sec[e
+ f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && (GtQ[m, 1] || (Eq
Q[m, 1] && EqQ[n, -3/2])) && IntegersQ[2*m, 2*n]

Rule 2696

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[a^(m + n)*((b
*Tan[e + f*x])^n/((a*Sec[e + f*x])^n*(b*Sin[e + f*x])^n)), Int[(b*Sin[e + f*x])^n/Cos[e + f*x]^(m + n), x], x]
/; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[n + 1/2] && IntegerQ[m + 1/2]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {d^2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx}{b^2} \\ & = -\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\left (d^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {b \sin (e+f x)} \, dx}{b^2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}} \\ & = -\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\left (d^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {\sin (e+f x)} \, dx}{b^2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}} \\ & = -\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {2 d^2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.82 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=-\frac {2 d^2 \left (3+\operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sqrt [4]{\sec ^2(e+f x)} \tan ^2(e+f x)\right )}{3 b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \]

[In]

Integrate[(d*Sec[e + f*x])^(3/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(-2*d^2*(3 + Hypergeometric2F1[3/4, 5/4, 7/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(1/4)*Tan[e + f*x]^2))/(3*b*f*
Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]])

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.98 (sec) , antiderivative size = 365, normalized size of antiderivative = 3.76

method	result	size
default	\(\frac {\csc \left (f x +e \right ) \left (2 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {2}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {2}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1\right ) {\left (-\frac {d \left (\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1\right )}{\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1}\right )}^{\frac {3}{2}} \left (1-\cos \left (f x +e \right )\right ) \sqrt {2}}{f {\left (\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1\right )}^{2} {\left (-\frac {b \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1}\right )}^{\frac {3}{2}}}\)	\(365\)

[In]

int((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/f*csc(f*x+e)*(2*(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*2^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e)+I))^(1/2)*(I*(csc(f*
x+e)-cot(f*x+e)))^(1/2)*EllipticE((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/2))-(-I*(I-cot(f*x+e)+csc(f*x+
e)))^(1/2)*2^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e)+I))^(1/2)*(I*(csc(f*x+e)-cot(f*x+e)))^(1/2)*EllipticF((-I*(I-cot
(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/2))-csc(f*x+e)^2*(1-cos(f*x+e))^2-1)*(-d*(csc(f*x+e)^2*(1-cos(f*x+e))^2+1)
/(csc(f*x+e)^2*(1-cos(f*x+e))^2-1))^(3/2)*(1-cos(f*x+e))/(csc(f*x+e)^2*(1-cos(f*x+e))^2+1)^2/(-b/(csc(f*x+e)^2
*(1-cos(f*x+e))^2-1)*(csc(f*x+e)-cot(f*x+e)))^(3/2)*2^(1/2)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.34 \[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=-\frac {2 \, d \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + i \, \sqrt {-2 i \, b d} d \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - i \, \sqrt {2 i \, b d} d \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{b^{2} f \sin \left (f x + e\right )} \]

[In]

integrate((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-(2*d*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)^2 + I*sqrt(-2*I*b*d)*d*sin(f*x + e)*
weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e))) - I*sqrt(2*I*b*d)*d*sin(f*x +
e)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e))))/(b^2*f*sin(f*x + e))

Sympy [F]

\[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((d*sec(f*x+e))**(3/2)/(b*tan(f*x+e))**(3/2),x)

[Out]

Integral((d*sec(e + f*x))**(3/2)/(b*tan(e + f*x))**(3/2), x)

Maxima [F]

\[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(3/2)/(b*tan(f*x + e))^(3/2), x)

Giac [F]

\[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(3/2)/(b*tan(f*x + e))^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]

[In]

int((d/cos(e + f*x))^(3/2)/(b*tan(e + f*x))^(3/2),x)

[Out]

int((d/cos(e + f*x))^(3/2)/(b*tan(e + f*x))^(3/2), x)