\(\int \frac {(b \tan (e+f x))^{5/2}}{\sqrt {d \sec (e+f x)}} \, dx\) [310]

Optimal result

Integrand size = 25, antiderivative size = 88 \[ \int \frac {(b \tan (e+f x))^{5/2}}{\sqrt {d \sec (e+f x)}} \, dx=-\frac {3 b^2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}+\frac {b (b \tan (e+f x))^{3/2}}{f \sqrt {d \sec (e+f x)}} \]

[Out]

3*b^2*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+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)/f/(d*sec(f*x+e))^(1/2)/sin(f*x+e)^(1/2)+b*(b*tan(f*x+e))^(3/2)/f/(d*sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 88, 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 = {2691, 2696, 2721, 2719} \[ \int \frac {(b \tan (e+f x))^{5/2}}{\sqrt {d \sec (e+f x)}} \, dx=\frac {b (b \tan (e+f x))^{3/2}}{f \sqrt {d \sec (e+f x)}}-\frac {3 b^2 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}} \]

[In]

Int[(b*Tan[e + f*x])^(5/2)/Sqrt[d*Sec[e + f*x]],x]

[Out]

(-3*b^2*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(f*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]]) + (
b*(b*Tan[e + f*x])^(3/2))/(f*Sqrt[d*Sec[e + f*x]])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[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 {b (b \tan (e+f x))^{3/2}}{f \sqrt {d \sec (e+f x)}}-\frac {1}{2} \left (3 b^2\right ) \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx \\ & = \frac {b (b \tan (e+f x))^{3/2}}{f \sqrt {d \sec (e+f x)}}-\frac {\left (3 b^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {b \sin (e+f x)} \, dx}{2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}} \\ & = \frac {b (b \tan (e+f x))^{3/2}}{f \sqrt {d \sec (e+f x)}}-\frac {\left (3 b^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {\sin (e+f x)} \, dx}{2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}} \\ & = -\frac {3 b^2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}+\frac {b (b \tan (e+f x))^{3/2}}{f \sqrt {d \sec (e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.63 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.74 \[ \int \frac {(b \tan (e+f x))^{5/2}}{\sqrt {d \sec (e+f x)}} \, dx=-\frac {b \left (-1+\operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sqrt [4]{\sec ^2(e+f x)}\right ) (b \tan (e+f x))^{3/2}}{f \sqrt {d \sec (e+f x)}} \]

[In]

Integrate[(b*Tan[e + f*x])^(5/2)/Sqrt[d*Sec[e + f*x]],x]

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.21 (sec) , antiderivative size = 471, normalized size of antiderivative = 5.35

method	result	size
default	\(\frac {b^{2} \sqrt {b \tan \left (f x +e \right )}\, \left (6 \cot \left (f x +e \right ) \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-3 \cot \left (f x +e \right ) \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )+6 \csc \left (f x +e \right ) \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-3 \csc \left (f x +e \right ) \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {2}\, \cot \left (f x +e \right )-3 \csc \left (f x +e \right ) \sqrt {2}+\sec \left (f x +e \right ) \csc \left (f x +e \right ) \sqrt {2}\right ) \sqrt {2}}{2 f \sqrt {d \sec \left (f x +e \right )}}\)	\(471\)

[In]

int((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2/f*b^2*(b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(1/2)*(6*cot(f*x+e)*(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*(-I*(co
t(f*x+e)-csc(f*x+e)+I))^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e)))^(1/2)*EllipticE((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2
),1/2*2^(1/2))-3*cot(f*x+e)*(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e)+I))^(1/2)*(-I*(cot
(f*x+e)-csc(f*x+e)))^(1/2)*EllipticF((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/2))+6*csc(f*x+e)*(-I*(I-cot
(f*x+e)+csc(f*x+e)))^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e)+I))^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e)))^(1/2)*EllipticE((
-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/2))-3*csc(f*x+e)*(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*(-I*(cot(f*
x+e)-csc(f*x+e)+I))^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e)))^(1/2)*EllipticF((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/
2*2^(1/2))+2*2^(1/2)*cot(f*x+e)-3*csc(f*x+e)*2^(1/2)+sec(f*x+e)*csc(f*x+e)*2^(1/2))*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 = 114, normalized size of antiderivative = 1.30 \[ \int \frac {(b \tan (e+f x))^{5/2}}{\sqrt {d \sec (e+f x)}} \, dx=\frac {2 \, b^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 3 i \, \sqrt {-2 i \, b d} b^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 i \, \sqrt {2 i \, b d} b^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{2 \, d f} \]

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/2*(2*b^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*sin(f*x + e) - 3*I*sqrt(-2*I*b*d)*b^2*weiers
trassZeta(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e))) + 3*I*sqrt(2*I*b*d)*b^2*weierstrassZ
eta(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e))))/(d*f)

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \tan (e+f x))^{5/2}}{\sqrt {d \sec (e+f x)}} \, dx=\text {Timed out} \]

[In]

integrate((b*tan(f*x+e))**(5/2)/(d*sec(f*x+e))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(b \tan (e+f x))^{5/2}}{\sqrt {d \sec (e+f x)}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\sqrt {d \sec \left (f x + e\right )}} \,d x } \]

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(5/2)/sqrt(d*sec(f*x + e)), x)

Giac [F]

\[ \int \frac {(b \tan (e+f x))^{5/2}}{\sqrt {d \sec (e+f x)}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\sqrt {d \sec \left (f x + e\right )}} \,d x } \]

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(5/2)/sqrt(d*sec(f*x + e)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \tan (e+f x))^{5/2}}{\sqrt {d \sec (e+f x)}} \, dx=\int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \]

[In]

int((b*tan(e + f*x))^(5/2)/(d/cos(e + f*x))^(1/2),x)

[Out]

int((b*tan(e + f*x))^(5/2)/(d/cos(e + f*x))^(1/2), x)