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

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

[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)

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Rubi [A]  time = 0.148433, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3186, 388, 195, 217, 203} \[ \frac{(a-5 b) (a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{16 b^{3/2} f}-\frac{\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{5/2}}{6 b f}+\frac{(a-5 b) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{24 b f}+\frac{(a-5 b) (a+b) \cos (e+f x) \sqrt{a-b \cos ^2(e+f x)+b}}{16 b f} \]

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

Mathematica [A]  time = 0.753845, size = 152, normalized size = 0.9 \[ \frac{\frac{(a+b)^2 (5 b-a) \log \left (\sqrt{2 a-b \cos (2 (e+f x))+b}+\sqrt{2} \sqrt{-b} \cos (e+f x)\right )}{(-b)^{3/2}}-\frac{\cos (e+f x) \sqrt{2 a-b \cos (2 (e+f x))+b} \left (3 a^2-b (7 a+9 b) \cos (2 (e+f x))+29 a b+b^2 \cos (4 (e+f x))+23 b^2\right )}{3 \sqrt{2} b}}{16 f} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[e + f*x]^3*(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)]]*(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)]))/(3*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]  time = 1.565, size = 446, normalized size = 2.6 \begin{align*} \text{result too large to display} \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]

-1/96*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(16*b^(7/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*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 [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 [A]  time = 12.4483, size = 1400, normalized size = 8.28 \begin{align*} \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{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]

[1/384*(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)]  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 = 1.7358, size = 266, normalized size = 1.57 \begin{align*} -\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{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]

-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))