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

Optimal. Leaf size=168 \[ \frac{c \sin (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{4 f^2}+\frac{c \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f^2}+\frac{x \sec (e+f x) \sqrt{a-a \sin (e+f x)} (c \sin (e+f x)+c)^{5/2}}{2 c f}-\frac{3 c x \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{4 f} \]

[Out]

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

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Rubi [A]  time = 0.143336, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {4604, 4422, 2644} \[ \frac{c \sin (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{4 f^2}+\frac{c \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f^2}+\frac{x \sec (e+f x) \sqrt{a-a \sin (e+f x)} (c \sin (e+f x)+c)^{5/2}}{2 c f}-\frac{3 c x \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2644

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[((2*a^2 + b^2)*x)/2, x] + (-Simp[(2*a*b*Cos[c
+ d*x])/d, x] - Simp[(b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d), x]) /; FreeQ[{a, b, c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.620265, size = 73, normalized size = 0.43 \[ \frac{c \sqrt{a-a \sin (e+f x)} \sqrt{c (\sin (e+f x)+1)} (\sin (e+f x)+4 f x \tan (e+f x)-f x \cos (2 (e+f x)) \sec (e+f x)+4)}{4 f^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*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]]*(4 - f*x*Cos[2*(e + f*x)]*Sec[e + f*x] + Sin[e + f*x] +
4*f*x*Tan[e + f*x]))/(4*f^2)

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Maple [F]  time = 0.076, size = 0, normalized size = 0. \begin{align*} \int x \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*(c+c*sin(f*x+e))^(3/2)*(a-a*sin(f*x+e))^(1/2),x)

[Out]

int(x*(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\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(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, 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*(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*(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*} \int \sqrt{-a \sin \left (f x + e\right ) + a}{\left (c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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