3.2.32 \(\int \sin ^3(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [132]

Optimal. Leaf size=169 \[ \frac {(a-5 b) (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac {(a-5 b) (a+b) \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{16 b f}+\frac {(a-5 b) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}{24 b f}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{5/2}}{6 b f} \]

[Out]

1/16*(a-5*b)*(a+b)^2*arctan(cos(f*x+e)*b^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/b^(3/2)/f+1/24*(a-5*b)*cos(f*x+e)*(
a+b-b*cos(f*x+e)^2)^(3/2)/b/f-1/6*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(5/2)/b/f+1/16*(a-5*b)*(a+b)*cos(f*x+e)*(a+b
-b*cos(f*x+e)^2)^(1/2)/b/f

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Rubi [A]
time = 0.10, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3265, 396, 201, 223, 209} \begin {gather*} \frac {(a-5 b) (a+b)^2 \text {ArcTan}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{16 b^{3/2} f}-\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{5/2}}{6 b f}+\frac {(a-5 b) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{24 b f}+\frac {(a-5 b) (a+b) \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{16 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a - 5*b)*(a + b)^2*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(16*b^(3/2)*f) + ((a - 5*b
)*(a + b)*Cos[e + f*x]*Sqrt[a + b - b*Cos[e + f*x]^2])/(16*b*f) + ((a - 5*b)*Cos[e + f*x]*(a + b - b*Cos[e + f
*x]^2)^(3/2))/(24*b*f) - (Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^(5/2))/(6*b*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 396

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 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 ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=-\frac {\text {Subst}\left (\int \left (1-x^2\right ) \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 )^{5/2}}{6 b f}+\frac {(a-5 b) \text {Subst}\left (\int \left (a+b-b x^2\right )^{3/2} \, dx,x,\cos (e+f x)\right )}{6 b f}\\ &=\frac {(a-5 b) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}{24 b f}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{5/2}}{6 b f}+\frac {((a-5 b) (a+b)) \text {Subst}\left (\int \sqrt {a+b-b x^2} \, dx,x,\cos (e+f x)\right )}{8 b f}\\ &=\frac {(a-5 b) (a+b) \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{16 b f}+\frac {(a-5 b) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}{24 b f}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{5/2}}{6 b f}+\frac {\left ((a-5 b) (a+b)^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{16 b f}\\ &=\frac {(a-5 b) (a+b) \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{16 b f}+\frac {(a-5 b) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}{24 b f}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{5/2}}{6 b f}+\frac {\left ((a-5 b) (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 )}{16 b f}\\ &=\frac {(a-5 b) (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac {(a-5 b) (a+b) \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{16 b f}+\frac {(a-5 b) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}{24 b f}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{5/2}}{6 b f}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 152, normalized size = 0.90 \begin {gather*} \frac {-\frac {\cos (e+f x) \sqrt {2 a+b-b \cos (2 (e+f x))} \left (3 a^2+29 a b+23 b^2-b (7 a+9 b) \cos (2 (e+f x))+b^2 \cos (4 (e+f x))\right )}{3 \sqrt {2} b}+\frac {(a+b)^2 (-a+5 b) \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{(-b)^{3/2}}}{16 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1/3*(Cos[e + f*x]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]*(3*a^2 + 29*a*b + 23*b^2 - b*(7*a + 9*b)*Cos[2*(e + f*x
)] + b^2*Cos[4*(e + f*x)]))/(Sqrt[2]*b) + ((a + b)^2*(-a + 5*b)*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a +
b - b*Cos[2*(e + f*x)]]])/(-b)^(3/2))/(16*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(445\) vs. \(2(149)=298\).
time = 9.45, size = 446, normalized size = 2.64

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 (16 \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 {7}{2}} \left (\cos ^{4}\left (f x +e \right )\right )-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 {5}{2}} \left (13 b +7 a \right ) \left (\cos ^{2}\left (f x +e \right )\right )+66 b^{\frac {7}{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 )}+72 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 )}\, b^{\frac {5}{2}}+6 a^{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 )}\, b^{\frac {3}{2}}+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^{3} b -9 \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} b^{2}-27 b^{3} 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 )-15 b^{4} \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 )\right )}{96 b^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(446\)

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,method=_RETURNVERBOSE)

[Out]

-1/96*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(16*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^(7/2)*cos(f*x+e
)^4-4*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^(5/2)*(13*b+7*a)*cos(f*x+e)^2+66*b^(7/2)*(-b*cos(f*x+e)^4+(
a+b)*cos(f*x+e)^2)^(1/2)+72*a*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^(5/2)+6*a^2*(-b*cos(f*x+e)^4+(a+b)*
cos(f*x+e)^2)^(1/2)*b^(3/2)+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^3*b-9*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*b^2-
27*b^3*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))-15*b^4*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 [A]
time = 0.50, size = 262, normalized size = 1.55 \begin {gather*} \frac {\frac {3 \, {\left (a + b\right )}^{2} a \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} + \frac {3 \, {\left (a + b\right )}^{2} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - \frac {18 \, {\left (a + b\right )} a \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - 18 \, {\left (a + b\right )} \sqrt {b} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) - 12 \, {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \cos \left (f x + e\right ) - 18 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \cos \left (f x + e\right ) - \frac {8 \, {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}} \cos \left (f x + e\right )}{b} + \frac {2 \, {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \cos \left (f x + e\right )}{b} + \frac {3 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \cos \left (f x + e\right )}{b}}{48 \, f} \end {gather*}

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]

1/48*(3*(a + b)^2*a*arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/b^(3/2) + 3*(a + b)^2*arcsin(b*cos(f*x + e)/sqrt((a
+ b)*b))/sqrt(b) - 18*(a + b)*a*arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/sqrt(b) - 18*(a + b)*sqrt(b)*arcsin(b*
cos(f*x + e)/sqrt((a + b)*b)) - 12*(-b*cos(f*x + e)^2 + a + b)^(3/2)*cos(f*x + e) - 18*sqrt(-b*cos(f*x + e)^2
+ a + b)*(a + b)*cos(f*x + e) - 8*(-b*cos(f*x + e)^2 + a + b)^(5/2)*cos(f*x + e)/b + 2*(-b*cos(f*x + e)^2 + a
+ b)^(3/2)*(a + b)*cos(f*x + e)/b + 3*sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)^2*cos(f*x + e)/b)/f

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Fricas [A]
time = 1.43, size = 579, normalized size = 3.43 \begin {gather*} \left [\frac {3 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\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 (8 \, b^{3} \cos \left (f x + e\right )^{5} - 2 \, {\left (7 \, a b^{2} + 13 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a^{2} b + 12 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{384 \, b^{2} f}, -\frac {3 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\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 (8 \, b^{3} \cos \left (f x + e\right )^{5} - 2 \, {\left (7 \, a b^{2} + 13 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a^{2} b + 12 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{192 \, b^{2} f}\right ] \end {gather*}

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]

[1/384*(3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*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*(8*b^3*cos(f*x + e)^5 - 2*(7*a*b^2 + 13*b^3)*cos(f*x + e)^3 + 3*(a^2*b + 12*a*b^2 + 11*b^
3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/(b^2*f), -1/192*(3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*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*(8*b^3*cos(f*x + e)^5 - 2*(7*a*b^2 + 13*b^3)*cos(f*x + e)^3 + 3*(a^2*b + 12*a*b^2 + 11*b^3)*cos(f*x + e)
)*sqrt(-b*cos(f*x + e)^2 + a + b))/(b^2*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)**3*(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [A]
time = 0.86, size = 197, normalized size = 1.17 \begin {gather*} -\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (\frac {2 \, {\left (4 \, b f^{2} \cos \left (f x + e\right )^{2} - \frac {7 \, a b^{4} f^{10} + 13 \, b^{5} f^{10}}{b^{4} f^{8}}\right )} \cos \left (f x + e\right )^{2}}{f^{2}} + \frac {3 \, {\left (a^{2} b^{3} f^{8} + 12 \, a b^{4} f^{8} + 11 \, b^{5} f^{8}\right )}}{b^{4} f^{8}}\right )} \cos \left (f x + e\right )}{48 \, f} + \frac {{\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\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 )}{16 \, \sqrt {-b} b {\left | f \right |}} \end {gather*}

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]

-1/48*sqrt(-b*cos(f*x + e)^2 + a + b)*(2*(4*b*f^2*cos(f*x + e)^2 - (7*a*b^4*f^10 + 13*b^5*f^10)/(b^4*f^8))*cos
(f*x + e)^2/f^2 + 3*(a^2*b^3*f^8 + 12*a*b^4*f^8 + 11*b^5*f^8)/(b^4*f^8))*cos(f*x + e)/f + 1/16*(a^3 - 3*a^2*b
- 9*a*b^2 - 5*b^3)*log(abs(sqrt(-b*cos(f*x + e)^2 + a + b) + sqrt(-b*f^2)*cos(f*x + e)/f))/(sqrt(-b)*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 )}^3\,{\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)^3*(a + b*sin(e + f*x)^2)^(3/2),x)

[Out]

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

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