3.114 \(\int (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx\)

Optimal. Leaf size=68 \[ -\frac {8 a^2 b \sqrt {a \sin (e+f x)}}{5 f \sqrt {b \tan (e+f x)}}-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt {b \tan (e+f x)}} \]

[Out]

-2/5*b*(a*sin(f*x+e))^(5/2)/f/(b*tan(f*x+e))^(1/2)-8/5*a^2*b*(a*sin(f*x+e))^(1/2)/f/(b*tan(f*x+e))^(1/2)

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Rubi [A]  time = 0.09, 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} \[ -\frac {8 a^2 b \sqrt {a \sin (e+f x)}}{5 f \sqrt {b \tan (e+f x)}}-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt {b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]],x]

[Out]

(-8*a^2*b*Sqrt[a*Sin[e + f*x]])/(5*f*Sqrt[b*Tan[e + f*x]]) - (2*b*(a*Sin[e + f*x])^(5/2))/(5*f*Sqrt[b*Tan[e +
f*x]])

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 (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx &=-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt {b \tan (e+f x)}}+\frac {1}{5} \left (4 a^2\right ) \int \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)} \, dx\\ &=-\frac {8 a^2 b \sqrt {a \sin (e+f x)}}{5 f \sqrt {b \tan (e+f x)}}-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt {b \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 51, normalized size = 0.75 \[ -\frac {a^2 \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)} (\sin (2 (e+f x))+8 \cot (e+f x))}{5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]],x]

[Out]

-1/5*(a^2*Sqrt[a*Sin[e + f*x]]*(8*Cot[e + f*x] + Sin[2*(e + f*x)])*Sqrt[b*Tan[e + f*x]])/f

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fricas [A]  time = 0.43, size = 65, normalized size = 0.96 \[ \frac {2 \, {\left (a^{2} \cos \left (f x + e\right )^{3} - 5 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{5 \, f \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/5*(a^2*cos(f*x + e)^3 - 5*a^2*cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))/(f*sin(f*
x + e))

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giac [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))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.64, size = 493, normalized size = 7.25 \[ -\frac {\left (a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \left (5 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (-\frac {2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+2 \cos \left (f x +e \right )-1}{\sin \left (f x +e \right )^{2}}\right )-5 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (-\frac {2 \left (2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+2 \cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )^{2}}\right )-4 \left (\cos ^{3}\left (f x +e \right )\right )+5 \ln \left (-\frac {2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+2 \cos \left (f x +e \right )-1}{\sin \left (f x +e \right )^{2}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-5 \ln \left (-\frac {2 \left (2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+2 \cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )^{2}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+20 \cos \left (f x +e \right )\right ) \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}}{10 f \sin \left (f x +e \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x)

[Out]

-1/10/f*(a*sin(f*x+e))^(5/2)*(5*cos(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*ln(-(2*(-cos(f*x+e)/(1+cos(f*x
+e))^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)-5
*cos(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2-co
s(f*x+e)^2-2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)-4*cos(f*x+e)^3+5*ln(-(2*(-cos(
f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*cos(f*x+e)-1
)/sin(f*x+e)^2)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)-5*ln(-2*(2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e
)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)*(-cos(f*x+e)/(1+cos(f*x+
e))^2)^(1/2)+20*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 \[ \int \left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}} \sqrt {b \tan \left (f x + e\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(5/2)*sqrt(b*tan(f*x + e)), x)

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mupad [B]  time = 4.33, size = 80, normalized size = 1.18 \[ \frac {a^2\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\left (18\,\sin \left (2\,e+2\,f\,x\right )-\sin \left (4\,e+4\,f\,x\right )\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}}{10\,f\,\left (\cos \left (2\,e+2\,f\,x\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(e + f*x))^(5/2)*(b*tan(e + f*x))^(1/2),x)

[Out]

(a^2*(a*sin(e + f*x))^(1/2)*(18*sin(2*e + 2*f*x) - sin(4*e + 4*f*x))*((b*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) +
1))^(1/2))/(10*f*(cos(2*e + 2*f*x) - 1))

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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))**(5/2)*(b*tan(f*x+e))**(1/2),x)

[Out]

Timed out

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