3.4.1 \(\int \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2} \, dx\) [301]

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

[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 = 0.08, 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 = {2691, 2696, 2721, 2720} \begin {gather*} \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)}} \end {gather*}

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 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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 \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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.83, size = 105, normalized size = 1.19 \begin {gather*} \frac {b \sqrt {d \sec (e+f x)} \left (\sec ^{\frac {3}{2}}(e+f x)-\frac {\, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {1+\sec (e+f x)}}{\sqrt {2}}\right ) \sqrt {b \tan (e+f x)}}{f \sec ^{\frac {3}{2}}(e+f x)} \end {gather*}

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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Maple [C] Result contains complex when optimal does not.
time = 0.39, size = 211, normalized size = 2.40

method	result	size
default	\(\frac {\left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {\frac {d}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (i \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right ) \sqrt {2}-\sqrt {2}\right ) \sqrt {2}}{2 f \left (\cos \left (f x +e \right )-1\right ) \sin \left (f x +e \right )}\)	\(211\)

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,method=_RETURNVERBOSE)

[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*(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*c
os(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*sin(f*x+e)*cos(f*x+e)+cos(f*x+e)*2^(1/2)-2^(1/2))/(cos(
f*x+e)-1)/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((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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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 98, normalized size = 1.11 \begin {gather*} -\frac {\sqrt {-2 i \, b d} b {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2 i \, b d} b {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{2 \, f} \end {gather*}

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]

-1/2*(sqrt(-2*I*b*d)*b*weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2*I*b*d)*b*weierstrassP
Inverse(4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*b*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)))/f

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

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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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((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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,\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))^(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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