3.172 \(\int x^2 \sqrt{a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=265 \[ \frac{2 c x \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f^2}+\frac{c x \sin (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{2 f^2}-\frac{c \sin (e+f x) \tan (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{4 f^3}-\frac{2 c \tan (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f^3}+\frac{x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} (c \sin (e+f x)+c)^{5/2}}{2 c f}-\frac{3 c x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{4 f} \]

[Out]

(2*c*x*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])/f^2 - (3*c*x^2*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]
*Sqrt[c + c*Sin[e + f*x]])/(4*f) + (c*x*Sin[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])/(2*f^2
) + (x^2*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*(c + c*Sin[e + f*x])^(5/2))/(2*c*f) - (2*c*Sqrt[a - a*Sin[e + f
*x]]*Sqrt[c + c*Sin[e + f*x]]*Tan[e + f*x])/f^3 - (c*Sin[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e +
f*x]]*Tan[e + f*x])/(4*f^3)

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Rubi [A]  time = 0.273629, antiderivative size = 265, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4604, 4422, 3317, 3296, 2637, 3310, 30} \[ \frac{2 c x \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f^2}+\frac{c x \sin (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{2 f^2}-\frac{c \sin (e+f x) \tan (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{4 f^3}-\frac{2 c \tan (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f^3}+\frac{x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} (c \sin (e+f x)+c)^{5/2}}{2 c f}-\frac{3 c x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*x*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])/f^2 - (3*c*x^2*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]
*Sqrt[c + c*Sin[e + f*x]])/(4*f) + (c*x*Sin[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])/(2*f^2
) + (x^2*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*(c + c*Sin[e + f*x])^(5/2))/(2*c*f) - (2*c*Sqrt[a - a*Sin[e + f
*x]]*Sqrt[c + c*Sin[e + f*x]]*Tan[e + f*x])/f^3 - (c*Sin[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e +
f*x]]*Tan[e + f*x])/(4*f^3)

Rule 4604

Int[((g_.) + (h_.)*(x_))^(p_.)*((a_) + (b_.)*Sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*Sin[(e_.) + (f_.)*(x_
)])^(n_), x_Symbol] :> Dist[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e + f*x])^F
racPart[m])/Cos[e + f*x]^(2*FracPart[m]), Int[(g + h*x)^p*Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x],
x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p] && IntegerQ
[2*m] && IGeQ[n - m, 0]

Rule 4422

Int[Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol]
:> Simp[((e + f*x)^m*(a + b*Sin[c + d*x])^(n + 1))/(b*d*(n + 1)), x] - Dist[(f*m)/(b*d*(n + 1)), Int[(e + f*x
)^(m - 1)*(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x^2 \sqrt{a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx &=\left (\sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}\right ) \int x^2 \cos (e+f x) (c+c \sin (e+f x)) \, dx\\ &=\frac{x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac{\left (\sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}\right ) \int x (c+c \sin (e+f x))^2 \, dx}{c f}\\ &=\frac{x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac{\left (\sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}\right ) \int \left (c^2 x+2 c^2 x \sin (e+f x)+c^2 x \sin ^2(e+f x)\right ) \, dx}{c f}\\ &=-\frac{c x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}{2 f}+\frac{x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac{\left (c \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}\right ) \int x \sin ^2(e+f x) \, dx}{f}-\frac{\left (2 c \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}\right ) \int x \sin (e+f x) \, dx}{f}\\ &=\frac{2 c x \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}{f^2}-\frac{c x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}{2 f}+\frac{c x \sin (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}{2 f^2}+\frac{x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac{c \sin (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)} \tan (e+f x)}{4 f^3}-\frac{\left (2 c \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}\right ) \int \cos (e+f x) \, dx}{f^2}-\frac{\left (c \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}\right ) \int x \, dx}{2 f}\\ &=\frac{2 c x \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}{f^2}-\frac{3 c x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}{4 f}+\frac{c x \sin (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}{2 f^2}+\frac{x^2 \sec (e+f x) \sqrt{a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac{2 c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)} \tan (e+f x)}{f^3}-\frac{c \sin (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)} \tan (e+f x)}{4 f^3}\\ \end{align*}

Mathematica [A]  time = 0.823209, size = 95, normalized size = 0.36 \[ \frac{c \sqrt{a-a \sin (e+f x)} \sqrt{c (\sin (e+f x)+1)} \left (8 \left (f^2 x^2-2\right ) \tan (e+f x)-\left (2 f^2 x^2-1\right ) \cos (2 (e+f x)) \sec (e+f x)+4 f x \sin (e+f x)+16 f x\right )}{8 f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(16*f*x - (-1 + 2*f^2*x^2)*Cos[2*(e + f*x)]*Sec[e + f*x
] + 4*f*x*Sin[e + f*x] + 8*(-2 + f^2*x^2)*Tan[e + f*x]))/(8*f^3)

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Maple [F]  time = 0.08, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( c+c\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\sqrt{a-a\sin \left ( fx+e \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c+c*sin(f*x+e))^(3/2)*(a-a*sin(f*x+e))^(1/2),x)

[Out]

int(x^2*(c+c*sin(f*x+e))^(3/2)*(a-a*sin(f*x+e))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-a \sin \left (f x + e\right ) + a}{\left (c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c+c*sin(f*x+e))^(3/2)*(a-a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-a*sin(f*x + e) + a)*(c*sin(f*x + e) + c)^(3/2)*x^2, x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c+c*sin(f*x+e))^(3/2)*(a-a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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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(x**2*(c+c*sin(f*x+e))**(3/2)*(a-a*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c+c*sin(f*x+e))^(3/2)*(a-a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

sage2