\(\int \frac {(a \sin (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx\) [127]

Optimal result

Integrand size = 25, antiderivative size = 68 \[ \int \frac {(a \sin (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=-\frac {8 a^2 b (a \sin (e+f x))^{3/2}}{21 f (b \tan (e+f x))^{3/2}}-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f (b \tan (e+f x))^{3/2}} \]

[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)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2678, 2669} \[ \int \frac {(a \sin (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=-\frac {8 a^2 b (a \sin (e+f x))^{3/2}}{21 f (b \tan (e+f x))^{3/2}}-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f (b \tan (e+f x))^{3/2}} \]

[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 2669

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 2678

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*
Tan[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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {(a \sin (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\frac {a^3 \cos (e+f x) (-11+3 \cos (2 (e+f x))) \sqrt {a \sin (e+f x)}}{21 f \sqrt {b \tan (e+f x)}} \]

[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]])

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {2 a^{3} \sqrt {\sin \left (f x +e \right ) a}\, \left (3 \left (\cos ^{3}\left (f x +e \right )\right )-7 \cos \left (f x +e \right )\right )}{21 f \sqrt {b \tan \left (f x +e \right )}}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.04 \[ \int \frac {(a \sin (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\frac {2 \, {\left (3 \, a^{3} \cos \left (f x + e\right )^{4} - 7 \, a^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{21 \, b f \sin \left (f x + e\right )} \]

[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))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a \sin (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a \sin (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {7}{2}}}{\sqrt {b \tan \left (f x + e\right )}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {(a \sin (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {7}{2}}}{\sqrt {b \tan \left (f x + e\right )}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 5.39 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.29 \[ \int \frac {(a \sin (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=-\frac {a^3\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\sqrt {-\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{2\,{\sin \left (e+f\,x\right )}^2-2}}\,\left (22\,\sin \left (e+f\,x\right )+19\,\sin \left (3\,e+3\,f\,x\right )-3\,\sin \left (5\,e+5\,f\,x\right )\right )}{168\,b\,f\,{\sin \left (e+f\,x\right )}^2} \]

[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)