3.3.96 \(\int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx\) [296]

Optimal. Leaf size=95 \[ \frac {4 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}} \]

[Out]

-4/5*(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)/d^2/f/(d*sec(f*x+e))^(1/2)/sin(f*x+e)^(1/2)+2/5*(b*tan(f*x+e))^(3/2)/b/f/(d*sec(f*x+e))
^(5/2)

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Rubi [A]
time = 0.08, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2692, 2696, 2721, 2719} \begin {gather*} \frac {4 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(5*d^2*f*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]]) +
(2*(b*Tan[e + f*x])^(3/2))/(5*b*f*(d*Sec[e + f*x])^(5/2))

Rule 2692

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(a*Sec[e +
f*x])^m)*((b*Tan[e + f*x])^(n + 1)/(b*f*m)), x] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integ
ersQ[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*} \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx &=\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}+\frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx}{5 d^2}\\ &=\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}+\frac {\left (2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {b \sin (e+f x)} \, dx}{5 d^2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}\\ &=\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}+\frac {\left (2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {\sin (e+f x)} \, dx}{5 d^2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}\\ &=\frac {4 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.77, size = 79, normalized size = 0.83 \begin {gather*} -\frac {b \left (-1+\cos (2 (e+f x))+4 \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {3}{4};\sec ^2(e+f x)\right ) \sqrt [4]{-\tan ^2(e+f x)}\right )}{5 d^2 f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.47, size = 571, normalized size = 6.01

method	result	size
default	\(-\frac {\left (-2 \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+4 \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticE \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+\left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}-2 \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )+4 \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticE \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )+\cos \left (f x +e \right ) \sqrt {2}-2 \sqrt {2}\right ) \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {2}}{5 f \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )}\)	\(571\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/f*(-2*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)
-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*cos(f*x
+e)+4*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-si
n(f*x+e))/sin(f*x+e))^(1/2)*EllipticE(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*cos(f*x+e)+2
^(1/2)*cos(f*x+e)^3-2*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I
*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/
2))+4*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-si
n(f*x+e))/sin(f*x+e))^(1/2)*EllipticE(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))+cos(f*x+e)*2
^(1/2)-2*2^(1/2))*(b*sin(f*x+e)/cos(f*x+e))^(1/2)/(d/cos(f*x+e))^(5/2)/cos(f*x+e)^2/sin(f*x+e)*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 121, normalized size = 1.27 \begin {gather*} \frac {2 \, {\left (\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} \sin \left (f x + e\right ) + i \, \sqrt {-2 i \, b 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 ) - i \, \sqrt {2 i \, b 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 )\right )}}{5 \, d^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)^2*sin(f*x + e) + I*sqrt(-2*I*b*d)*wei
erstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e))) - I*sqrt(2*I*b*d)*weierstrassZeta
(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e))))/(d^3*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {b \tan {\left (e + f x \right )}}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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