3.56 \(\int \frac{(f+g x) (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=228 \[ \frac{\left (c^2 f x+g\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}+\frac{2 f x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}-\frac{b (f+g x)}{6 c d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{b f \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt{d-c^2 d x^2}}-\frac{b g \sqrt{1-c^2 x^2} \tanh ^{-1}(c x)}{6 c^2 d^2 \sqrt{d-c^2 d x^2}} \]

[Out]

-(b*(f + g*x))/(6*c*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (2*f*x*(a + b*ArcSin[c*x]))/(3*d^2*Sqrt[d - c
^2*d*x^2]) + ((g + c^2*f*x)*(a + b*ArcSin[c*x]))/(3*c^2*d^2*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]) - (b*g*Sqrt[1 -
c^2*x^2]*ArcTanh[c*x])/(6*c^2*d^2*Sqrt[d - c^2*d*x^2]) + (b*f*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(3*c*d^2*Sq
rt[d - c^2*d*x^2])

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Rubi [A]  time = 0.194799, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4777, 639, 191, 4761, 206, 260} \[ \frac{\left (c^2 f x+g\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}+\frac{2 f x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}-\frac{b (f+g x)}{6 c d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{b f \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt{d-c^2 d x^2}}-\frac{b g \sqrt{1-c^2 x^2} \tanh ^{-1}(c x)}{6 c^2 d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x)*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

-(b*(f + g*x))/(6*c*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (2*f*x*(a + b*ArcSin[c*x]))/(3*d^2*Sqrt[d - c
^2*d*x^2]) + ((g + c^2*f*x)*(a + b*ArcSin[c*x]))/(3*c^2*d^2*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]) - (b*g*Sqrt[1 -
c^2*x^2]*ArcTanh[c*x])/(6*c^2*d^2*Sqrt[d - c^2*d*x^2]) + (b*f*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(3*c*d^2*Sq
rt[d - c^2*d*x^2])

Rule 4777

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Dist[(d^IntPart[p]*(d + e*x^2)^FracPart[p])/(1 - c^2*x^2)^FracPart[p], Int[(f + g*x)^m*(1 - c^2*x^2)^p*(a +
b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && IntegerQ
[p - 1/2] &&  !GtQ[d, 0]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 4761

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With
[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[Dist[1/Sqrt[1 - c^
2*x^2], u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && ILtQ[p + 1/2,
0] && GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{(f+g x) \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac{\sqrt{1-c^2 x^2} \int \frac{(f+g x) \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{2 f x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}-\frac{\left (b c \sqrt{1-c^2 x^2}\right ) \int \left (\frac{g+c^2 f x}{3 c^2 \left (1-c^2 x^2\right )^2}+\frac{2 f x}{3 \left (1-c^2 x^2\right )}\right ) \, dx}{d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{2 f x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}-\frac{\left (b \sqrt{1-c^2 x^2}\right ) \int \frac{g+c^2 f x}{\left (1-c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (2 b c f \sqrt{1-c^2 x^2}\right ) \int \frac{x}{1-c^2 x^2} \, dx}{3 d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b (f+g x)}{6 c d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{2 f x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}+\frac{b f \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (b g \sqrt{1-c^2 x^2}\right ) \int \frac{1}{1-c^2 x^2} \, dx}{6 c d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b (f+g x)}{6 c d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{2 f x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}-\frac{b g \sqrt{1-c^2 x^2} \tanh ^{-1}(c x)}{6 c^2 d^2 \sqrt{d-c^2 d x^2}}+\frac{b f \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.808149, size = 208, normalized size = 0.91 \[ -\frac{\sqrt{d-c^2 d x^2} \left (\sqrt{-c^2} \left (-4 a c^4 f x^3+6 a c^2 f x+2 a g+2 b \sin ^{-1}(c x) \left (c^2 f x \left (3-2 c^2 x^2\right )+g\right )-b c f \sqrt{1-c^2 x^2}+2 b c f \left (1-c^2 x^2\right )^{3/2} \log \left (c^2 x^2-1\right )-b c g x \sqrt{1-c^2 x^2}\right )+i b c g \left (1-c^2 x^2\right )^{3/2} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-c^2} x\right ),1\right )\right )}{6 \left (-c^2\right )^{3/2} d^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

-(Sqrt[d - c^2*d*x^2]*(I*b*c*g*(1 - c^2*x^2)^(3/2)*EllipticF[I*ArcSinh[Sqrt[-c^2]*x], 1] + Sqrt[-c^2]*(2*a*g +
6*a*c^2*f*x - 4*a*c^4*f*x^3 - b*c*f*Sqrt[1 - c^2*x^2] - b*c*g*x*Sqrt[1 - c^2*x^2] + 2*b*(g + c^2*f*x*(3 - 2*c
^2*x^2))*ArcSin[c*x] + 2*b*c*f*(1 - c^2*x^2)^(3/2)*Log[-1 + c^2*x^2])))/(6*(-c^2)^(3/2)*d^3*(-1 + c^2*x^2)^2)

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Maple [C]  time = 0.309, size = 2236, normalized size = 9.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x)

[Out]

-4/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*arcsin(c*x)*(-c^2*x^2+1)*x^4*g-8/3*I
*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/c*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*f+2/3*I*b*(
-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4*(-c^2*x^2+1)*x^6*g+2/3*I*b*(-d*(c^2*x^2-1))^
(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4*(-c^2*x^2+1)*x^5*f-5/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6
*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*(-c^2*x^2+1)*x^4*g-5/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+
11*c^2*x^2-4)*c^2*(-c^2*x^2+1)*x^3*f+I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*(-c^2*
x^2+1)*x*f-4/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4*arcsin(c*x)*x^6*g-1/6*b*(-
c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^2/d^3/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)-I)*g-2/3*b*(-c^2*x^2+1
)^(1/2)*(-d*(c^2*x^2-1))^(1/2)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)/c/d^3/(c^2*x^2-1)*f+1/6*b*(-c^2*x^2+1)^(1/2)*(-d
*(c^2*x^2-1))^(1/2)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)/c^2/d^3/(c^2*x^2-1)*g-1/2*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c
^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c*(-c^2*x^2+1)^(1/2)*x^2*f-2/3*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c/d
^3/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)-I)*f-1/2*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*
x^2-4)*c*(-c^2*x^2+1)^(1/2)*x^3*g+17/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*ar
csin(c*x)*x^3*f-7/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4*x^5*f+8/3*I*b*(-d*(
c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*x^4*g+2/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*
x^6-10*c^4*x^4+11*c^2*x^2-4)*c^6*x^8*g+2/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*
c^6*x^7*f+8/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*arcsin(c*x)*(-c^2*x^2+1)*x^2*g-
4/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/c^2*arcsin(c*x)*(-c^2*x^2+1)*g+2/3*b*(-d*
(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/c*(-c^2*x^2+1)^(1/2)*x*g+8/3*I*b*(-d*(c^2*x^2-1))^(
1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*x^3*f-7/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^
4+11*c^2*x^2-4)*c^4*x^6*g-2*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4*arcsin(c*x)*x
^5*f+4*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*arcsin(c*x)*x^4*g+I*b*(-d*(c^2*x^2
-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*(-c^2*x^2+1)*x^2*g+2/3*a*f/d^2*x/(-c^2*d*x^2+d)^(1/2)+1/3*a
*f/d*x/(-c^2*d*x^2+d)^(3/2)+1/3*a*g/c^2/d/(-c^2*d*x^2+d)^(3/2)+2/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*
c^4*x^4+11*c^2*x^2-4)/c*(-c^2*x^2+1)^(1/2)*f-3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4
)*arcsin(c*x)*x^2*g-4*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*arcsin(c*x)*x*f-I*b*(-d
*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*x^2*g-I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10
*c^4*x^4+11*c^2*x^2-4)*x*f-2*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^3*arcsin(c*x
)*(-c^2*x^2+1)^(1/2)*x^4*f+14/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c*arcsin(c*
x)*(-c^2*x^2+1)^(1/2)*x^2*f+4/3*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/d^3/(c^2*x^2-1)*f*arcsin(c*x)

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Maxima [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((g*x+f)*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-c^{2} d x^{2} + d}{\left (a g x + a f +{\left (b g x + b f\right )} \arcsin \left (c x\right )\right )}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(a*g*x + a*f + (b*g*x + b*f)*arcsin(c*x))/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*
d^3*x^2 - d^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asin}{\left (c x \right )}\right ) \left (f + g x\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Integral((a + b*asin(c*x))*(f + g*x)/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x + f\right )}{\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((g*x + f)*(b*arcsin(c*x) + a)/(-c^2*d*x^2 + d)^(5/2), x)