3.153 \(\int \frac{\sin ^3(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=79 \[ \frac{a \cos (e+f x)}{b f (a+b) \sqrt{a-b \cos ^2(e+f x)+b}}-\frac{\tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{b^{3/2} f} \]

[Out]

-(ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]]/(b^(3/2)*f)) + (a*Cos[e + f*x])/(b*(a + b)*f*S
qrt[a + b - b*Cos[e + f*x]^2])

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Rubi [A]  time = 0.0988078, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3186, 385, 217, 203} \[ \frac{a \cos (e+f x)}{b f (a+b) \sqrt{a-b \cos ^2(e+f x)+b}}-\frac{\tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{b^{3/2} f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

-(ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]]/(b^(3/2)*f)) + (a*Cos[e + f*x])/(b*(a + b)*f*S
qrt[a + b - b*Cos[e + f*x]^2])

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.4146, size = 96, normalized size = 1.22 \[ \frac{\frac{\sqrt{2} a b \cos (e+f x)}{(a+b) \sqrt{2 a-b \cos (2 (e+f x))+b}}+\sqrt{-b} \log \left (\sqrt{2 a-b \cos (2 (e+f x))+b}+\sqrt{2} \sqrt{-b} \cos (e+f x)\right )}{b^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

((Sqrt[2]*a*b*Cos[e + f*x])/((a + b)*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]) + Sqrt[-b]*Log[Sqrt[2]*Sqrt[-b]*Cos[e
+ f*x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/(b^2*f)

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Maple [B]  time = 1.646, size = 156, normalized size = 2. \begin{align*}{\frac{1}{f\cos \left ( fx+e \right ) }\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ({\frac{1}{2}\arctan \left ({\sqrt{b} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{2}-{\frac{-a+b}{2\,b}} \right ){\frac{1}{\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \right ){b}^{-{\frac{3}{2}}}}+{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( a+b \right ) b}{\frac{1}{\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \right ){\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.62754, size = 1339, normalized size = 16.95 \begin{align*} \left [-\frac{8 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} a b \cos \left (f x + e\right ) +{\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sqrt{-b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \,{\left (a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{6} + 160 \,{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 32 \,{\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \,{\left (16 \, b^{3} \cos \left (f x + e\right )^{7} - 24 \,{\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \,{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} -{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-b}\right )}{8 \,{\left ({\left (a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f\right )}}, -\frac{4 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} a b \cos \left (f x + e\right ) -{\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sqrt{b} \arctan \left (\frac{{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{b}}{4 \,{\left (2 \, b^{3} \cos \left (f x + e\right )^{5} - 3 \,{\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} +{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )}}\right )}{4 \,{\left ({\left (a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(8*sqrt(-b*cos(f*x + e)^2 + a + b)*a*b*cos(f*x + e) + ((a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*s
qrt(-b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + b^4)*cos(f*x + e)^6 + 160*(a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x
+ e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2 + 8
*(16*b^3*cos(f*x + e)^7 - 24*(a*b^2 + b^3)*cos(f*x + e)^5 + 10*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^3 - (a^3 +
3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)))/((a*b^3 + b^4)*f*cos(f*x +
e)^2 - (a^2*b^2 + 2*a*b^3 + b^4)*f), -1/4*(4*sqrt(-b*cos(f*x + e)^2 + a + b)*a*b*cos(f*x + e) - ((a*b + b^2)*c
os(f*x + e)^2 - a^2 - 2*a*b - b^2)*sqrt(b)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + b^2)*cos(f*x + e)^2 + a
^2 + 2*a*b + b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b)/(2*b^3*cos(f*x + e)^5 - 3*(a*b^2 + b^3)*cos(f*x + e)
^3 + (a^2*b + 2*a*b^2 + b^3)*cos(f*x + e))))/((a*b^3 + b^4)*f*cos(f*x + e)^2 - (a^2*b^2 + 2*a*b^3 + b^4)*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**3/(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.92863, size = 154, normalized size = 1.95 \begin{align*} -\frac{\sqrt{-{\left (\cos \left (f x + e\right )^{2} - 1\right )} b + a} a \cos \left (f x + e\right )}{{\left ({\left (\cos \left (f x + e\right )^{2} - 1\right )} b - a\right )}{\left (a b + b^{2}\right )} f} - \frac{\log \left ({\left | \sqrt{-{\left (\cos \left (f x + e\right )^{2} - 1\right )} b + a} + \frac{\sqrt{-b f^{2}} \cos \left (f x + e\right )}{f} \right |}\right )}{\sqrt{-b} b{\left | f \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

-sqrt(-(cos(f*x + e)^2 - 1)*b + a)*a*cos(f*x + e)/(((cos(f*x + e)^2 - 1)*b - a)*(a*b + b^2)*f) - log(abs(sqrt(
-(cos(f*x + e)^2 - 1)*b + a) + sqrt(-b*f^2)*cos(f*x + e)/f))/(sqrt(-b)*b*abs(f))