3.538 \(\int \frac {a+b \sin ^{-1}(c x)}{(d+c d x)^{3/2} (f-c f x)^{5/2}} \, dx\)

Optimal. Leaf size=255 \[ \frac {2 d x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {d (c x+1) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b d \left (1-c^2 x^2\right )^{5/2}}{6 c (1-c x) (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {b d \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b d \left (1-c^2 x^2\right )^{5/2} \tanh ^{-1}(c x)}{6 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]

[Out]

-1/6*b*d*(-c^2*x^2+1)^(5/2)/c/(-c*x+1)/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)+1/3*d*(c*x+1)*(-c^2*x^2+1)*(a+b*arcsin
(c*x))/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)+2/3*d*x*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*f*x+f)^
(5/2)-1/6*b*d*(-c^2*x^2+1)^(5/2)*arctanh(c*x)/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)+1/3*b*d*(-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 = 0.25, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4673, 639, 191, 4761, 627, 44, 207, 260} \[ \frac {2 d x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {d (c x+1) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b d \left (1-c^2 x^2\right )^{5/2}}{6 c (1-c x) (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {b d \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b d \left (1-c^2 x^2\right )^{5/2} \tanh ^{-1}(c x)}{6 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/((d + c*d*x)^(3/2)*(f - c*f*x)^(5/2)),x]

[Out]

-(b*d*(1 - c^2*x^2)^(5/2))/(6*c*(1 - c*x)*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) + (d*(1 + c*x)*(1 - c^2*x^2)*(a
+ b*ArcSin[c*x]))/(3*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) + (2*d*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*d*(1 - c^2*x^2)^(5/2)*ArcTanh[c*x])/(6*c*(d + c*d*x)^(5/2)*(f - c*f*
x)^(5/2)) + (b*d*(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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 207

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

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 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]

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])

Rubi steps

\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{(d+c d x)^{3/2} (f-c f x)^{5/2}} \, dx &=\frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {(d+c d x) \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=\frac {d (1+c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d 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 (b c \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (\frac {d (1+c x)}{3 c \left (1-c^2 x^2\right )^2}+\frac {2 d x}{3 \left (1-c^2 x^2\right )}\right ) \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=\frac {d (1+c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d 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 (b d \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1+c x}{\left (1-c^2 x^2\right )^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (2 b c d \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 {d (1+c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d 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 d \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}}-\frac {\left (b d \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1}{(1-c x)^2 (1+c x)} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=\frac {d (1+c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d 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 d \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}}-\frac {\left (b d \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (\frac {1}{2 (-1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {b d \left (1-c^2 x^2\right )^{5/2}}{6 c (1-c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {d (1+c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d 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 d \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}}+\frac {\left (b d \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1}{-1+c^2 x^2} \, dx}{6 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {b d \left (1-c^2 x^2\right )^{5/2}}{6 c (1-c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {d (1+c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d 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 d \left (1-c^2 x^2\right )^{5/2} \tanh ^{-1}(c x)}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b d \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 = 0.69, size = 184, normalized size = 0.72 \[ \frac {\sqrt {c d x+d} \left (8 a c^2 x^2-8 a c x-4 a+5 b c x \sqrt {1-c^2 x^2} \log (f-c f x)+3 b (c x-1) \sqrt {1-c^2 x^2} \log (-f (c x+1))-5 b \sqrt {1-c^2 x^2} \log (f-c f x)+2 b \sqrt {1-c^2 x^2}+4 b \left (2 c^2 x^2-2 c x-1\right ) \sin ^{-1}(c x)\right )}{12 c d^2 f^2 \left (c^2 x^2-1\right ) \sqrt {f-c f x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])/((d + c*d*x)^(3/2)*(f - c*f*x)^(5/2)),x]

[Out]

(Sqrt[d + c*d*x]*(-4*a - 8*a*c*x + 8*a*c^2*x^2 + 2*b*Sqrt[1 - c^2*x^2] + 4*b*(-1 - 2*c*x + 2*c^2*x^2)*ArcSin[c
*x] + 3*b*(-1 + c*x)*Sqrt[1 - c^2*x^2]*Log[-(f*(1 + c*x))] - 5*b*Sqrt[1 - c^2*x^2]*Log[f - c*f*x] + 5*b*c*x*Sq
rt[1 - c^2*x^2]*Log[f - c*f*x]))/(12*c*d^2*f^2*Sqrt[f - c*f*x]*(-1 + c^2*x^2))

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{5} d^{2} f^{3} x^{5} - c^{4} d^{2} f^{3} x^{4} - 2 \, c^{3} d^{2} f^{3} x^{3} + 2 \, c^{2} d^{2} f^{3} x^{2} + c d^{2} f^{3} x - d^{2} f^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-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^5*d^2*f^3*x^5 - c^4*d^2*f^3*x^4 - 2*c^3*d^2*
f^3*x^3 + 2*c^2*d^2*f^3*x^2 + c*d^2*f^3*x - d^2*f^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c f x + f\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-c*f*x+f)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)/((c*d*x + d)^(3/2)*(-c*f*x + f)^(5/2)), x)

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {a +b \arcsin \left (c x \right )}{\left (c d x +d \right )^{\frac {3}{2}} \left (-c f x +f \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-c*f*x+f)^(5/2),x)

[Out]

int((a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-c*f*x+f)^(5/2),x)

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maxima [A]  time = 0.55, size = 237, normalized size = 0.93 \[ \frac {1}{12} \, b c {\left (\frac {2 \, \sqrt {d} \sqrt {f}}{c^{3} d^{2} f^{3} x - c^{2} d^{2} f^{3}} + \frac {3 \, \log \left (c x + 1\right )}{c^{2} d^{\frac {3}{2}} f^{\frac {5}{2}}} + \frac {5 \, \log \left (c x - 1\right )}{c^{2} d^{\frac {3}{2}} f^{\frac {5}{2}}}\right )} - \frac {1}{3} \, b {\left (\frac {1}{\sqrt {-c^{2} d f x^{2} + d f} c^{2} d f^{2} x - \sqrt {-c^{2} d f x^{2} + d f} c d f^{2}} - \frac {2 \, x}{\sqrt {-c^{2} d f x^{2} + d f} d f^{2}}\right )} \arcsin \left (c x\right ) - \frac {1}{3} \, a {\left (\frac {1}{\sqrt {-c^{2} d f x^{2} + d f} c^{2} d f^{2} x - \sqrt {-c^{2} d f x^{2} + d f} c d f^{2}} - \frac {2 \, x}{\sqrt {-c^{2} d f x^{2} + d f} d f^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-c*f*x+f)^(5/2),x, algorithm="maxima")

[Out]

1/12*b*c*(2*sqrt(d)*sqrt(f)/(c^3*d^2*f^3*x - c^2*d^2*f^3) + 3*log(c*x + 1)/(c^2*d^(3/2)*f^(5/2)) + 5*log(c*x -
1)/(c^2*d^(3/2)*f^(5/2))) - 1/3*b*(1/(sqrt(-c^2*d*f*x^2 + d*f)*c^2*d*f^2*x - sqrt(-c^2*d*f*x^2 + d*f)*c*d*f^2
) - 2*x/(sqrt(-c^2*d*f*x^2 + d*f)*d*f^2))*arcsin(c*x) - 1/3*a*(1/(sqrt(-c^2*d*f*x^2 + d*f)*c^2*d*f^2*x - sqrt(
-c^2*d*f*x^2 + d*f)*c*d*f^2) - 2*x/(sqrt(-c^2*d*f*x^2 + d*f)*d*f^2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (d+c\,d\,x\right )}^{3/2}\,{\left (f-c\,f\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))/((d + c*d*x)^(3/2)*(f - c*f*x)^(5/2)),x)

[Out]

int((a + b*asin(c*x))/((d + c*d*x)^(3/2)*(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)**(3/2)/(-c*f*x+f)**(5/2),x)

[Out]

Timed out

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