\(\int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} (f-c f x)^{3/2}} \, dx\) [532]

Optimal result

Integrand size = 30, antiderivative size = 96 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} (f-c f x)^{3/2}} \, dx=\frac {x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{(d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {b \left (1-c^2 x^2\right )^{3/2} \log \left (1-c^2 x^2\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4763, 4745, 266} \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} (f-c f x)^{3/2}} \, dx=\frac {x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{(c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {b \left (1-c^2 x^2\right )^{3/2} \log \left (1-c^2 x^2\right )}{2 c (c d x+d)^{3/2} (f-c f x)^{3/2}} \]

[In]

Int[(a + b*ArcSin[c*x])/((d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)),x]

[Out]

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

Rule 266

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 4745

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/d)*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]], Int[x*((a + b*ArcSin
[c*x])^(n - 1)/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 4763

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

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} (f-c f x)^{3/2}} \, dx=\frac {\sqrt {d+c d x} \left (2 a c x+2 b c x \arcsin (c x)+b \sqrt {1-c^2 x^2} \log (-f (1+c x))+b \sqrt {1-c^2 x^2} \log (f-c f x)\right )}{2 c d^2 f (1+c x) \sqrt {f-c f x}} \]

[In]

Integrate[(a + b*ArcSin[c*x])/((d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)),x]

[Out]

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

Maple [F]

\[\int \frac {a +b \arcsin \left (c x \right )}{\left (c d x +d \right )^{\frac {3}{2}} \left (-c f x +f \right )^{\frac {3}{2}}}d x\]

[In]

int((a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2),x)

[Out]

int((a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2),x)

Fricas [F]

\[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} (f-c f x)^{3/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c f x + f\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} (f-c f x)^{3/2}} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\left (d \left (c x + 1\right )\right )^{\frac {3}{2}} \left (- f \left (c x - 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((a+b*asin(c*x))/(c*d*x+d)**(3/2)/(-c*f*x+f)**(3/2),x)

[Out]

Integral((a + b*asin(c*x))/((d*(c*x + 1))**(3/2)*(-f*(c*x - 1))**(3/2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.90 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} (f-c f x)^{3/2}} \, dx=\frac {b x \arcsin \left (c x\right )}{\sqrt {-c^{2} d f x^{2} + d f} d f} + \frac {a x}{\sqrt {-c^{2} d f x^{2} + d f} d f} - \frac {b \sqrt {\frac {1}{d f}} \log \left (x^{2} - \frac {1}{c^{2}}\right )}{2 \, c d f} \]

[In]

integrate((a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} (f-c f x)^{3/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c f x + f\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)/((c*d*x + d)^(3/2)*(-c*f*x + f)^(3/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} (f-c f x)^{3/2}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (d+c\,d\,x\right )}^{3/2}\,{\left (f-c\,f\,x\right )}^{3/2}} \,d x \]

[In]

int((a + b*asin(c*x))/((d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)),x)

[Out]

int((a + b*asin(c*x))/((d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)), x)