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

Optimal. Leaf size=171 \[ \frac{a \sin (e+f x)}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{a}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}+\frac{a x \tan (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{a x \sec (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}} \]

[Out]

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

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Rubi [A]  time = 0.972936, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 12, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.387, Rules used = {4604, 6741, 12, 6742, 4185, 4181, 2279, 2391, 3757, 3767, 8, 4413} \[ \frac{a \sin (e+f x)}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{a}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}+\frac{a x \tan (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{a x \sec (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3757

Int[(x_)^(m_.)*Sec[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tan[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> Simp[(x^(
m - n + 1)*Sec[a + b*x^n]^p)/(b*n*p), x] - Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sec[a + b*x^n]^p, x], x] /;
FreeQ[{a, b, p}, x] && IntegerQ[n] && GeQ[m, n] && EqQ[q, 1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4413

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]*Tan[(a_.) + (b_.)*(x_)]^(p_), x_Symbol] :> -Int[(c + d*
x)^m*Sec[a + b*x]*Tan[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Sec[a + b*x]^3*Tan[a + b*x]^(p - 2), x] /; FreeQ[
{a, b, c, d, m}, x] && IGtQ[p/2, 0]

Rubi steps

\begin{align*} \int \frac{x \sqrt{a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx &=\frac{\cos (e+f x) \int x \sec ^3(e+f x) (a-a \sin (e+f x))^2 \, dx}{a c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=\frac{\cos (e+f x) \int a^2 x \sec ^3(e+f x) (1-\sin (e+f x))^2 \, dx}{a c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=\frac{(a \cos (e+f x)) \int x \sec ^3(e+f x) (1-\sin (e+f x))^2 \, dx}{c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=\frac{(a \cos (e+f x)) \int \left (x \sec ^3(e+f x)-2 x \sec ^2(e+f x) \tan (e+f x)+x \sec (e+f x) \tan ^2(e+f x)\right ) \, dx}{c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=\frac{(a \cos (e+f x)) \int x \sec ^3(e+f x) \, dx}{c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(a \cos (e+f x)) \int x \sec (e+f x) \tan ^2(e+f x) \, dx}{c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(2 a \cos (e+f x)) \int x \sec ^2(e+f x) \tan (e+f x) \, dx}{c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\frac{a}{2 c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{a x \sec (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{a x \tan (e+f x)}{2 c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(a \cos (e+f x)) \int x \sec (e+f x) \, dx}{2 c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(a \cos (e+f x)) \int x \sec (e+f x) \, dx}{c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(a \cos (e+f x)) \int x \sec ^3(e+f x) \, dx}{c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(a \cos (e+f x)) \int \sec ^2(e+f x) \, dx}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\frac{a}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{i a x \tan ^{-1}\left (e^{i (e+f x)}\right ) \cos (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{a x \sec (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{a x \tan (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(a \cos (e+f x)) \int x \sec (e+f x) \, dx}{2 c \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(a \cos (e+f x)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(a \cos (e+f x)) \int \log \left (1-i e^{i (e+f x)}\right ) \, dx}{2 c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(a \cos (e+f x)) \int \log \left (1+i e^{i (e+f x)}\right ) \, dx}{2 c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(a \cos (e+f x)) \int \log \left (1-i e^{i (e+f x)}\right ) \, dx}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(a \cos (e+f x)) \int \log \left (1+i e^{i (e+f x)}\right ) \, dx}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\frac{a}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{a x \sec (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{a \sin (e+f x)}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{a x \tan (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(i a \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{2 c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(i a \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{2 c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(i a \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(i a \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(a \cos (e+f x)) \int \log \left (1-i e^{i (e+f x)}\right ) \, dx}{2 c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(a \cos (e+f x)) \int \log \left (1+i e^{i (e+f x)}\right ) \, dx}{2 c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\frac{a}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{i a \cos (e+f x) \text{Li}_2\left (-i e^{i (e+f x)}\right )}{2 c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{i a \cos (e+f x) \text{Li}_2\left (i e^{i (e+f x)}\right )}{2 c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{a x \sec (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{a \sin (e+f x)}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{a x \tan (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(i a \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{2 c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(i a \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{2 c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\frac{a}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{a x \sec (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{a \sin (e+f x)}{c f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{a x \tan (e+f x)}{c f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.621979, size = 150, normalized size = 0.88 \[ -\frac{\sqrt{a-a \sin (e+f x)} \sqrt{c (\sin (e+f x)+1)} \left (f x \sin \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}+f x\right )+\cos \left (\frac{e}{2}\right ) (f x-1)+\cos \left (\frac{e}{2}+f x\right )+\sin \left (\frac{e}{2}\right )\right )}{c^2 f^2 \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.079, size = 0, normalized size = 0. \begin{align*} \int{x\sqrt{a-a\sin \left ( fx+e \right ) } \left ( c+c\sin \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.3572, size = 180, normalized size = 1.05 \begin{align*} -\frac{{\left (f x + \cos \left (f x + e\right )\right )} \sqrt{-a \sin \left (f x + e\right ) + a} \sqrt{c \sin \left (f x + e\right ) + c}}{c^{2} f^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + c^{2} f^{2} \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(f*x + cos(f*x + e))*sqrt(-a*sin(f*x + e) + a)*sqrt(c*sin(f*x + e) + c)/(c^2*f^2*cos(f*x + e)*sin(f*x + e) +
c^2*f^2*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sqrt{- a \left (\sin{\left (e + f x \right )} - 1\right )}}{\left (c \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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