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

Optimal. Leaf size=83 \[ \frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{2 b^{3/2} f}-\frac{\cos (e+f x) \sqrt{a-b \cos ^2(e+f x)+b}}{2 b f} \]

[Out]

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

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Rubi [A]  time = 0.0955315, antiderivative size = 83, 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, 388, 217, 203} \[ \frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{2 b^{3/2} f}-\frac{\cos (e+f x) \sqrt{a-b \cos ^2(e+f x)+b}}{2 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b)*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(2*b^(3/2)*f) - (Cos[e + f*x]*Sqrt[a +
b - b*Cos[e + f*x]^2])/(2*b*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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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)}{\sqrt{a+b \sin ^2(e+f x)}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{\sqrt{a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sqrt{a+b-b \cos ^2(e+f x)}}{2 b f}+\frac{(a-b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{2 b f}\\ &=-\frac{\cos (e+f x) \sqrt{a+b-b \cos ^2(e+f x)}}{2 b f}+\frac{(a-b) \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 )}{2 b f}\\ &=\frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a+b-b \cos ^2(e+f x)}}\right )}{2 b^{3/2} f}-\frac{\cos (e+f x) \sqrt{a+b-b \cos ^2(e+f x)}}{2 b f}\\ \end{align*}

Mathematica [A]  time = 0.278719, size = 105, normalized size = 1.27 \[ \frac{(a-b) \log \left (\sqrt{2 a-b \cos (2 (e+f x))+b}+\sqrt{2} \sqrt{-b} \cos (e+f x)\right )}{2 \sqrt{-b} b f}-\frac{\cos (e+f x) \sqrt{2 a-b \cos (2 (e+f x))+b}}{2 \sqrt{2} b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 1.256, size = 186, normalized size = 2.2 \begin{align*} -{\frac{1}{4\,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 ( 2\,{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}}+ba\arctan \left ({\frac{-2\,b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+a+b}{2}{\frac{1}{\sqrt{b}}}{\frac{1}{\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \right ) -{b}^{2}\arctan \left ({\frac{-2\,b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+a+b}{2}{\frac{1}{\sqrt{b}}}{\frac{1}{\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \right ) \right ){b}^{-{\frac{5}{2}}}{\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)^(1/2),x)

[Out]

-1/4*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(2*b^(3/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)+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))-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^(5/2)/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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.32528, size = 1076, normalized size = 12.96 \begin{align*} \left [-\frac{8 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right ) -{\left (a - b\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 )}{16 \, b^{2} f}, -\frac{{\left (a - b\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 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right )}{8 \, b^{2} 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)^(1/2),x, algorithm="fricas")

[Out]

[-1/16*(8*sqrt(-b*cos(f*x + e)^2 + a + b)*b*cos(f*x + e) - (a - b)*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))*sq
rt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)))/(b^2*f), -1/8*((a - b)*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*sqrt(-b*cos(f*x + e)^2 + a + b)*b
*cos(f*x + e))/(b^2*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)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.83821, size = 126, normalized size = 1.52 \begin{align*} -\frac{\sqrt{-{\left (\cos \left (f x + e\right )^{2} - 1\right )} b + a} \cos \left (f x + e\right )}{2 \, b f} + \frac{{\left (a - b\right )} \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 )}{2 \, \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)^(1/2),x, algorithm="giac")

[Out]

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