3.301 \(\int \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=88 \[ \frac {b \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}}{f}-\frac {b^2 \sqrt {\sin (e+f x)} F\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {d \sec (e+f x)}}{f \sqrt {b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 88, 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 = {2611, 2616, 2642, 2641} \[ \frac {b \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}}{f}-\frac {b^2 \sqrt {\sin (e+f x)} F\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {d \sec (e+f x)}}{f \sqrt {b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2611

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 2616

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 2641

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

\begin {align*} \int \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2} \, dx &=\frac {b \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{f}-\frac {1}{2} b^2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx\\ &=\frac {b \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{f}-\frac {\left (b^2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \int \frac {1}{\sqrt {b \sin (e+f x)}} \, dx}{2 \sqrt {b \tan (e+f x)}}\\ &=\frac {b \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{f}-\frac {\left (b^2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {1}{\sqrt {\sin (e+f x)}} \, dx}{2 \sqrt {b \tan (e+f x)}}\\ &=-\frac {b^2 F\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{f \sqrt {b \tan (e+f x)}}+\frac {b \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{f}\\ \end {align*}

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Mathematica [C]  time = 0.76, size = 105, normalized size = 1.19 \[ \frac {b \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)} \left (\sec ^{\frac {3}{2}}(e+f x)-\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\sec (e+f x)+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {2}}\right )}{f \sec ^{\frac {3}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \sec \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )} b \tan \left (f x + e\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \sec \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.68, size = 211, normalized size = 2.40 \[ \frac {\left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {\frac {d}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (i \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}+\cos \left (f x +e \right ) \sqrt {2}-\sqrt {2}\right ) \sqrt {2}}{2 f \left (-1+\cos \left (f x +e \right )\right ) \sin \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2),x)

[Out]

1/2/f*(b*sin(f*x+e)/cos(f*x+e))^(3/2)*(d/cos(f*x+e))^(1/2)*cos(f*x+e)*(I*(-I*(-1+cos(f*x+e))/sin(f*x+e))^(1/2)
*cos(f*x+e)*sin(f*x+e)*(-(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))*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)+cos(f*x+e)*2^(1/2)-2^(1/2))/(-1+
cos(f*x+e))/sin(f*x+e)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \sec \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x))^(3/2)*(d/cos(e + f*x))^(1/2),x)

[Out]

int((b*tan(e + f*x))^(3/2)*(d/cos(e + f*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \sqrt {d \sec {\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(1/2)*(b*tan(f*x+e))**(3/2),x)

[Out]

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

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