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

Optimal. Leaf size=114 \[ -\frac{3 (a+b) \cos (e+f x) \sqrt{a-b \cos ^2(e+f x)+b}}{8 f}-\frac{\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 f}-\frac{3 (a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{8 \sqrt{b} f} \]

[Out]

(-3*(a + b)^2*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(8*Sqrt[b]*f) - (3*(a + b)*Cos[e
+ f*x]*Sqrt[a + b - b*Cos[e + f*x]^2])/(8*f) - (Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^(3/2))/(4*f)

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Rubi [A]  time = 0.075981, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3186, 195, 217, 203} \[ -\frac{3 (a+b) \cos (e+f x) \sqrt{a-b \cos ^2(e+f x)+b}}{8 f}-\frac{\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 f}-\frac{3 (a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{8 \sqrt{b} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(a + b)^2*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(8*Sqrt[b]*f) - (3*(a + b)*Cos[e
+ f*x]*Sqrt[a + b - b*Cos[e + f*x]^2])/(8*f) - (Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^(3/2))/(4*f)

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [A]  time = 0.400842, size = 113, normalized size = 0.99 \[ -\frac{\frac{\cos (e+f x) \sqrt{2 a-b \cos (2 (e+f x))+b} (5 a-b \cos (2 (e+f x))+4 b)}{\sqrt{2}}+\frac{3 (a+b)^2 \log \left (\sqrt{2 a-b \cos (2 (e+f x))+b}+\sqrt{2} \sqrt{-b} \cos (e+f x)\right )}{\sqrt{-b}}}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[e + f*x]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]*(5*a + 4*b - b*Cos[2*(e + f*x)]))/Sqrt[2] + (3*(a + b)^2*Lo
g[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/Sqrt[-b])/(8*f)

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Maple [B]  time = 1.507, size = 309, normalized size = 2.7 \begin{align*}{\frac{1}{16\,f\cos \left ( fx+e \right ) }\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) } \left ( 4\,{b}^{3/2}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( \cos \left ( fx+e \right ) \right ) ^{2}-10\,{b}^{3/2}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-10\,a\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{b}+3\,\arctan \left ( 1/2\,{\frac{-2\,b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+a+b}{\sqrt{b}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}} \right ){a}^{2}+6\,ba\arctan \left ( 1/2\,{\frac{-2\,b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+a+b}{\sqrt{b}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}} \right ) +3\,{b}^{2}\arctan \left ( 1/2\,{\frac{-2\,b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+a+b}{\sqrt{b}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}} \right ) \right ){\frac{1}{\sqrt{b}}}{\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)*(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

1/16*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(4*b^(3/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*cos(f*x+e)^
2-10*b^(3/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)-10*a*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^(1/2
)+3*arctan(1/2*(-2*b*cos(f*x+e)^2+a+b)/b^(1/2)/(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2))*a^2+6*b*a*arctan(1/
2*(-2*b*cos(f*x+e)^2+a+b)/b^(1/2)/(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2))+3*b^2*arctan(1/2*(-2*b*cos(f*x+e
)^2+a+b)/b^(1/2)/(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)))/b^(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)*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.64765, size = 1208, normalized size = 10.6 \begin{align*} \left [-\frac{3 \,{\left (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 (2 \, b^{2} \cos \left (f x + e\right )^{3} - 5 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{64 \, b f}, \frac{3 \,{\left (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 (2 \, b^{2} \cos \left (f x + e\right )^{3} - 5 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{32 \, b f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/64*(3*(a^2 + 2*a*b + b^2)*sqrt(-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
)) - 8*(2*b^2*cos(f*x + e)^3 - 5*(a*b + b^2)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/(b*f), 1/32*(3*(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))) + 4*(2*b^2*cos(f*x + e)^3 - 5*(a*b + b^2)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a +
b))/(b*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)*(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.71035, size = 173, normalized size = 1.52 \begin{align*} \frac{{\left (2 \, b \cos \left (f x + e\right )^{2} - \frac{5 \,{\left (a b^{2} f^{4} + b^{3} f^{4}\right )}}{b^{2} f^{4}}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )}{8 \, f} - \frac{3 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left ({\left | \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} + \frac{\sqrt{-b f^{2}} \cos \left (f x + e\right )}{f} \right |}\right )}{8 \, \sqrt{-b}{\left | f \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*b*cos(f*x + e)^2 - 5*(a*b^2*f^4 + b^3*f^4)/(b^2*f^4))*sqrt(-b*cos(f*x + e)^2 + a + b)*cos(f*x + e)/f -
3/8*(a^2 + 2*a*b + b^2)*log(abs(sqrt(-b*cos(f*x + e)^2 + a + b) + sqrt(-b*f^2)*cos(f*x + e)/f))/(sqrt(-b)*abs(
f))