3.3.94 \(\int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx\) [294]

Optimal. Leaf size=55 \[ \frac {2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}} \]

[Out]

-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)

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Rubi [A]
time = 0.04, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2696, 2721, 2719} \begin {gather*} \frac {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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tan[e + f*x]]/Sqrt[d*Sec[e + f*x]],x]

[Out]

(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]])

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)}}{\sqrt {d \sec (e+f x)}} \, dx &=\frac {\sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)} \, dx}{\sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}\\ &=\frac {\sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)} \, dx}{\sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}\\ &=\frac {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)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.60, size = 62, normalized size = 1.13 \begin {gather*} -\frac {2 b \, _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)}}{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]]/Sqrt[d*Sec[e + f*x]],x]

[Out]

(-2*b*Hypergeometric2F1[-1/4, 1/4, 3/4, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(1/4))/(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 = 0.50, size = 551, normalized size = 10.02

method	result	size
risch	\(-\frac {i \sqrt {2}\, \sqrt {-\frac {i b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}}{f \sqrt {\frac {d \,{\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}}+\frac {i \left (\frac {2 i \left (-i b d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d b \right )}{b d \sqrt {{\mathrm e}^{i \left (f x +e \right )} \left (-i b d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d b \right )}}-\frac {\sqrt {{\mathrm e}^{i \left (f x +e \right )}+1}\, \sqrt {-2 \,{\mathrm e}^{i \left (f x +e \right )}+2}\, \sqrt {-{\mathrm e}^{i \left (f x +e \right )}}\, \left (-2 \EllipticE \left (\sqrt {{\mathrm e}^{i \left (f x +e \right )}+1}, \frac {\sqrt {2}}{2}\right )+\EllipticF \left (\sqrt {{\mathrm e}^{i \left (f x +e \right )}+1}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {-i b d \,{\mathrm e}^{3 i \left (f x +e \right )}+i d b \,{\mathrm e}^{i \left (f x +e \right )}}}\right ) \sqrt {2}\, \sqrt {-\frac {i b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {-i d \,{\mathrm e}^{i \left (f x +e \right )} b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}}{f \sqrt {\frac {d \,{\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}\)	\(371\)
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 )}}\, \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 {-\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 )+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 )}}\, \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 )-\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 ) \sqrt {2}-\sqrt {2}\right ) \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {2}}{f \sqrt {\frac {d}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )}\)	\(551\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/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)*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)
-(-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)+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)*EllipticE(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))-(-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)*2^(1/2)-2^(1/2))*(b*sin(f*x+e
)/cos(f*x+e))^(1/2)/(d/cos(f*x+e))^(1/2)/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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 70, normalized size = 1.27 \begin {gather*} \frac {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 )}{d 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))^(1/2),x, algorithm="fricas")

[Out]

(I*sqrt(-2*I*b*d)*weierstrassZeta(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*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 )}}}{\sqrt {d \sec {\left (e + f x \right )}}}\, 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))**(1/2),x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}} \,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))^(1/2),x)

[Out]

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

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