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

Optimal. Leaf size=188 \[ \frac {2 x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b \left (1-c^2 x^2\right )^{3/2}}{6 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {b \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]

[Out]

-1/6*b*(-c^2*x^2+1)^(3/2)/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)+1/3*x*(-c^2*x^2+1)*(a+b*arcsin(c*x))/(c*d*x+d)^(5
/2)/(-c*f*x+f)^(5/2)+2/3*x*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)+1/3*b*(-c^2*x^2+1
)^(5/2)*ln(-c^2*x^2+1)/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)

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Rubi [A]  time = 0.20, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4673, 4655, 4651, 260, 261} \[ \frac {2 x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b \left (1-c^2 x^2\right )^{3/2}}{6 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {b \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(1 - c^2*x^2)^(3/2))/(6*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) + (x*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(3*
(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) + (2*x*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/(3*(d + c*d*x)^(5/2)*(f - c*f
*x)^(5/2)) + (b*(1 - c^2*x^2)^(5/2)*Log[1 - c^2*x^2])/(3*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))

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]

Rule 261

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

Rule 4651

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n)/Sqrt[d], Int[(x*(a + b*ArcSin[c*x])^(n - 1))/(d + e*x^2), x], x
] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[d, 0]

Rule 4655

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSin[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a + b*
ArcSin[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 - c^2*x^2)^FracPart[p
]), Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 4673

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> D
ist[((d + e*x)^q*(f + g*x)^q)/(1 - c^2*x^2)^q, Int[(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x]
, x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q]
&& GeQ[p - q, 0]

Rubi steps

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

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Mathematica [A]  time = 0.64, size = 178, normalized size = 0.95 \[ \frac {\sqrt {c d x+d} \left (4 a c^3 x^3-6 a c x+2 b c^2 x^2 \sqrt {1-c^2 x^2} \log (f-c f x)-2 b \left (1-c^2 x^2\right )^{3/2} \log (-f (c x+1))-2 b \sqrt {1-c^2 x^2} \log (f-c f x)+b \sqrt {1-c^2 x^2}+2 b c x \left (2 c^2 x^2-3\right ) \sin ^{-1}(c x)\right )}{6 c d^3 (c x-1) \sqrt {f-c f x} (c f x+f)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + c*d*x]*(-6*a*c*x + 4*a*c^3*x^3 + b*Sqrt[1 - c^2*x^2] + 2*b*c*x*(-3 + 2*c^2*x^2)*ArcSin[c*x] - 2*b*(1
- c^2*x^2)^(3/2)*Log[-(f*(1 + c*x))] - 2*b*Sqrt[1 - c^2*x^2]*Log[f - c*f*x] + 2*b*c^2*x^2*Sqrt[1 - c^2*x^2]*L
og[f - c*f*x]))/(6*c*d^3*(-1 + c*x)*Sqrt[f - c*f*x]*(f + c*f*x)^2)

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{6} d^{3} f^{3} x^{6} - 3 \, c^{4} d^{3} f^{3} x^{4} + 3 \, c^{2} d^{3} f^{3} x^{2} - d^{3} f^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c f x + f\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {a +b \arcsin \left (c x \right )}{\left (c d x +d \right )^{\frac {5}{2}} \left (-c f x +f \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.52, size = 177, normalized size = 0.94 \[ \frac {1}{6} \, b c {\left (\frac {1}{c^{4} d^{\frac {5}{2}} f^{\frac {5}{2}} x^{2} - c^{2} d^{\frac {5}{2}} f^{\frac {5}{2}}} + \frac {2 \, \log \left (c x + 1\right )}{c^{2} d^{\frac {5}{2}} f^{\frac {5}{2}}} + \frac {2 \, \log \left (c x - 1\right )}{c^{2} d^{\frac {5}{2}} f^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {x}{{\left (-c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} d f} + \frac {2 \, x}{\sqrt {-c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} \arcsin \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {x}{{\left (-c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} d f} + \frac {2 \, x}{\sqrt {-c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b*c*(1/(c^4*d^(5/2)*f^(5/2)*x^2 - c^2*d^(5/2)*f^(5/2)) + 2*log(c*x + 1)/(c^2*d^(5/2)*f^(5/2)) + 2*log(c*x
- 1)/(c^2*d^(5/2)*f^(5/2))) + 1/3*b*(x/((-c^2*d*f*x^2 + d*f)^(3/2)*d*f) + 2*x/(sqrt(-c^2*d*f*x^2 + d*f)*d^2*f^
2))*arcsin(c*x) + 1/3*a*(x/((-c^2*d*f*x^2 + d*f)^(3/2)*d*f) + 2*x/(sqrt(-c^2*d*f*x^2 + d*f)*d^2*f^2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (d+c\,d\,x\right )}^{5/2}\,{\left (f-c\,f\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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