3.127 \(\int \frac {(a \sin (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx\)

Optimal. Leaf size=68 \[ -\frac {8 a^2 b (a \sin (e+f x))^{3/2}}{21 f (b \tan (e+f x))^{3/2}}-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f (b \tan (e+f x))^{3/2}} \]

[Out]

-8/21*a^2*b*(a*sin(f*x+e))^(3/2)/f/(b*tan(f*x+e))^(3/2)-2/7*b*(a*sin(f*x+e))^(7/2)/f/(b*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.10, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2598, 2589} \[ -\frac {8 a^2 b (a \sin (e+f x))^{3/2}}{21 f (b \tan (e+f x))^{3/2}}-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f (b \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^(7/2)/Sqrt[b*Tan[e + f*x]],x]

[Out]

(-8*a^2*b*(a*Sin[e + f*x])^(3/2))/(21*f*(b*Tan[e + f*x])^(3/2)) - (2*b*(a*Sin[e + f*x])^(7/2))/(7*f*(b*Tan[e +
f*x])^(3/2))

Rule 2589

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*(a*Sin[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n - 1, 0]

Rule 2598

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> -Simp[(b*(a*Sin[
e + f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] + Dist[(a^2*(m + n - 1))/m, Int[(a*Sin[e + f*x])^(m - 2)*(b*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && IntegersQ[2
*m, 2*n]

Rubi steps

\begin {align*} \int \frac {(a \sin (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx &=-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f (b \tan (e+f x))^{3/2}}+\frac {1}{7} \left (4 a^2\right ) \int \frac {(a \sin (e+f x))^{3/2}}{\sqrt {b \tan (e+f x)}} \, dx\\ &=-\frac {8 a^2 b (a \sin (e+f x))^{3/2}}{21 f (b \tan (e+f x))^{3/2}}-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f (b \tan (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 52, normalized size = 0.76 \[ \frac {a^3 \cos (e+f x) (3 \cos (2 (e+f x))-11) \sqrt {a \sin (e+f x)}}{21 f \sqrt {b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[e + f*x])^(7/2)/Sqrt[b*Tan[e + f*x]],x]

[Out]

(a^3*Cos[e + f*x]*(-11 + 3*Cos[2*(e + f*x)])*Sqrt[a*Sin[e + f*x]])/(21*f*Sqrt[b*Tan[e + f*x]])

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fricas [A]  time = 0.66, size = 71, normalized size = 1.04 \[ \frac {2 \, {\left (3 \, a^{3} \cos \left (f x + e\right )^{4} - 7 \, a^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{21 \, b f \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(7/2)/(b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/21*(3*a^3*cos(f*x + e)^4 - 7*a^3*cos(f*x + e)^2)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))/(b*f
*sin(f*x + e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {7}{2}}}{\sqrt {b \tan \left (f x + e\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(7/2)/(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e))^(7/2)/sqrt(b*tan(f*x + e)), x)

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maple [A]  time = 0.50, size = 60, normalized size = 0.88 \[ \frac {2 \left (3 \left (\cos ^{2}\left (f x +e \right )\right )-7\right ) \left (a \sin \left (f x +e \right )\right )^{\frac {7}{2}} \cos \left (f x +e \right )}{21 f \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(f*x+e))^(7/2)/(b*tan(f*x+e))^(1/2),x)

[Out]

2/21/f*(3*cos(f*x+e)^2-7)*(a*sin(f*x+e))^(7/2)*cos(f*x+e)/(b*sin(f*x+e)/cos(f*x+e))^(1/2)/sin(f*x+e)^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {7}{2}}}{\sqrt {b \tan \left (f x + e\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(7/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(7/2)/sqrt(b*tan(f*x + e)), x)

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mupad [B]  time = 4.82, size = 88, normalized size = 1.29 \[ -\frac {a^3\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\sqrt {-\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{2\,{\sin \left (e+f\,x\right )}^2-2}}\,\left (22\,\sin \left (e+f\,x\right )+19\,\sin \left (3\,e+3\,f\,x\right )-3\,\sin \left (5\,e+5\,f\,x\right )\right )}{168\,b\,f\,{\sin \left (e+f\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(e + f*x))^(7/2)/(b*tan(e + f*x))^(1/2),x)

[Out]

-(a^3*(a*sin(e + f*x))^(1/2)*(-(b*sin(2*e + 2*f*x))/(2*sin(e + f*x)^2 - 2))^(1/2)*(22*sin(e + f*x) + 19*sin(3*
e + 3*f*x) - 3*sin(5*e + 5*f*x)))/(168*b*f*sin(e + f*x)^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))**(7/2)/(b*tan(f*x+e))**(1/2),x)

[Out]

Timed out

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