3.2.33 \(\int \sin (e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [133]

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

[Out]

-1/4*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(3/2)/f-3/8*(a+b)^2*arctan(cos(f*x+e)*b^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))
/f/b^(1/2)-3/8*(a+b)*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)/f

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Rubi [A]
time = 0.05, antiderivative size = 114, normalized size of antiderivative = 1.00, 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 = {3265, 201, 223, 209} \begin {gather*} -\frac {3 (a+b)^2 \text {ArcTan}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{8 \sqrt {b} f}-\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} \end {gather*}

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 201

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 209

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

Rule 223

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 3265

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]

Rubi steps

\begin {align*} \int \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=-\frac {\text {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)) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.27, size = 113, normalized size = 0.99 \begin {gather*} -\frac {\frac {\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}}+\frac {3 (a+b)^2 \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{\sqrt {-b}}}{8 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((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*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/Sqrt[-b])/f

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(308\) vs. \(2(98)=196\).
time = 8.80, size = 309, normalized size = 2.71

method	result	size
default	\(\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (4 \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b^{\frac {3}{2}} \left (\cos ^{2}\left (f x +e \right )\right )-10 b^{\frac {3}{2}} \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+3 \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right ) a^{2}+6 b a \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )+3 b^{2} \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )-10 a \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {b}\right )}{16 \sqrt {b}\, \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(309\)

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,method=_RETURNVERBOSE)

[Out]

1/16*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(4*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^(3/2)*cos(f*x+e)^
2-10*b^(3/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)+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))-10*a*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^(1/2))/b^(1/2)/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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Maxima [A]
time = 0.50, size = 112, normalized size = 0.98 \begin {gather*} -\frac {\frac {3 \, {\left (a + b\right )} a \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + 3 \, {\left (a + b\right )} \sqrt {b} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 2 \, {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \cos \left (f x + e\right ) + 3 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \cos \left (f x + e\right )}{8 \, f} \end {gather*}

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]

-1/8*(3*(a + b)*a*arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/sqrt(b) + 3*(a + b)*sqrt(b)*arcsin(b*cos(f*x + e)/sqr
t((a + b)*b)) + 2*(-b*cos(f*x + e)^2 + a + b)^(3/2)*cos(f*x + e) + 3*sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)*c
os(f*x + e))/f

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Fricas [A]
time = 0.65, size = 495, normalized size = 4.34 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.79, size = 128, normalized size = 1.12 \begin {gather*} \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 {gather*}

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))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sin \left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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