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

Optimal. Leaf size=465 \[ -\frac {5 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{2 c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {b c f^4 x^2 \left (1-c^2 x^2\right )^{3/2}}{(c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {5 b f^4 (1-c x)^2 \left (1-c^2 x^2\right )^{3/2}}{4 c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {3 b f^4 x \left (1-c^2 x^2\right )^{3/2}}{2 (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {8 b f^4 \left (1-c^2 x^2\right )^{3/2} \log (c x+1)}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {15 b f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x)^2}{4 c (c d x+d)^{3/2} (f-c f x)^{3/2}} \]

[Out]

3/2*b*f^4*x*(-c^2*x^2+1)^(3/2)/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)+b*c*f^4*x^2*(-c^2*x^2+1)^(3/2)/(c*d*x+d)^(3/2)
/(-c*f*x+f)^(3/2)-5/4*b*f^4*(-c*x+1)^2*(-c^2*x^2+1)^(3/2)/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)+15/4*b*f^4*(-c^2*
x^2+1)^(3/2)*arcsin(c*x)^2/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)-2*f^4*(-c*x+1)^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))/
c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)-15/2*f^4*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2
)-5/2*f^4*(-c*x+1)*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)-15/2*f^4*(-c^2*x^2+1)^(
3/2)*arcsin(c*x)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)+8*b*f^4*(-c^2*x^2+1)^(3/2)*ln(c*x+1)/c/(
c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 0.38, antiderivative size = 465, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4673, 669, 671, 641, 216, 4761, 627, 43, 4641} \[ -\frac {5 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{2 c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {b c f^4 x^2 \left (1-c^2 x^2\right )^{3/2}}{(c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {5 b f^4 (1-c x)^2 \left (1-c^2 x^2\right )^{3/2}}{4 c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {3 b f^4 x \left (1-c^2 x^2\right )^{3/2}}{2 (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {8 b f^4 \left (1-c^2 x^2\right )^{3/2} \log (c x+1)}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {15 b f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x)^2}{4 c (c d x+d)^{3/2} (f-c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*f^4*x*(1 - c^2*x^2)^(3/2))/(2*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) + (b*c*f^4*x^2*(1 - c^2*x^2)^(3/2))/((
d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) - (5*b*f^4*(1 - c*x)^2*(1 - c^2*x^2)^(3/2))/(4*c*(d + c*d*x)^(3/2)*(f - c*
f*x)^(3/2)) + (15*b*f^4*(1 - c^2*x^2)^(3/2)*ArcSin[c*x]^2)/(4*c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) - (2*f^4*
(1 - c*x)^3*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) - (15*f^4*(1 - c^2*x^2)
^2*(a + b*ArcSin[c*x]))/(2*c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) - (5*f^4*(1 - c*x)*(1 - c^2*x^2)^2*(a + b*Ar
cSin[c*x]))/(2*c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) - (15*f^4*(1 - c^2*x^2)^(3/2)*ArcSin[c*x]*(a + b*ArcSin[
c*x]))/(2*c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) + (8*b*f^4*(1 - c^2*x^2)^(3/2)*Log[1 + c*x])/(c*(d + c*d*x)^(
3/2)*(f - c*f*x)^(3/2))

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 216

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

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 641

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

Rule 669

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[(2*c*d*(m + p))/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rule 4641

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

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 {(f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{(d+c d x)^{3/2}} \, dx &=\frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {(f-c f x)^4 \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{(d+c d x)^{3/2} (f-c f x)^{3/2}}\\ &=-\frac {2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {5 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{2 c (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 \left (-\frac {15 f^4}{2 c}-\frac {5 f^4 (1-c x)}{2 c}-\frac {2 f^4 (1-c x)^3}{c \left (1-c^2 x^2\right )}-\frac {15 f^4 \sin ^{-1}(c x)}{2 c \sqrt {1-c^2 x^2}}\right ) \, dx}{(d+c d x)^{3/2} (f-c f x)^{3/2}}\\ &=\frac {15 b f^4 x \left (1-c^2 x^2\right )^{3/2}}{2 (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {5 b f^4 (1-c x)^2 \left (1-c^2 x^2\right )^{3/2}}{4 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {5 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {\left (2 b f^4 \left (1-c^2 x^2\right )^{3/2}\right ) \int \frac {(1-c x)^3}{1-c^2 x^2} \, dx}{(d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {\left (15 b f^4 \left (1-c^2 x^2\right )^{3/2}\right ) \int \frac {\sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{2 (d+c d x)^{3/2} (f-c f x)^{3/2}}\\ &=\frac {15 b f^4 x \left (1-c^2 x^2\right )^{3/2}}{2 (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {5 b f^4 (1-c x)^2 \left (1-c^2 x^2\right )^{3/2}}{4 c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {15 b f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x)^2}{4 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {5 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {\left (2 b f^4 \left (1-c^2 x^2\right )^{3/2}\right ) \int \frac {(1-c x)^2}{1+c x} \, dx}{(d+c d x)^{3/2} (f-c f x)^{3/2}}\\ &=\frac {15 b f^4 x \left (1-c^2 x^2\right )^{3/2}}{2 (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {5 b f^4 (1-c x)^2 \left (1-c^2 x^2\right )^{3/2}}{4 c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {15 b f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x)^2}{4 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {5 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {\left (2 b f^4 \left (1-c^2 x^2\right )^{3/2}\right ) \int \left (-3+c x+\frac {4}{1+c x}\right ) \, dx}{(d+c d x)^{3/2} (f-c f x)^{3/2}}\\ &=\frac {3 b f^4 x \left (1-c^2 x^2\right )^{3/2}}{2 (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {b c f^4 x^2 \left (1-c^2 x^2\right )^{3/2}}{(d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {5 b f^4 (1-c x)^2 \left (1-c^2 x^2\right )^{3/2}}{4 c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {15 b f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x)^2}{4 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {5 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {15 f^4 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {8 b f^4 \left (1-c^2 x^2\right )^{3/2} \log (1+c x)}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 3.77, size = 685, normalized size = 1.47 \[ \frac {f^2 \left (8 a \sqrt {1-c^2 x^2} \left (c^2 x^2-7 c x-24\right ) \sqrt {c d x+d} \sqrt {f-c f x} \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+120 a \sqrt {d} \sqrt {f} (c x+1) \sqrt {1-c^2 x^2} \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right ) \tan ^{-1}\left (\frac {c x \sqrt {c d x+d} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (c^2 x^2-1\right )}\right )-32 b (c x+1) \sqrt {c d x+d} \sqrt {f-c f x} \left (\sin ^{-1}(c x) \left (\left (\sqrt {1-c^2 x^2}-2\right ) \sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\left (\sqrt {1-c^2 x^2}+2\right ) \cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+\sin ^{-1}(c x)^2 \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )-\left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right ) \left (c x+4 \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )\right )-8 b (c x+1) \sqrt {c d x+d} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right ) \left (\sin ^{-1}(c x) \left (\sin ^{-1}(c x)+4\right )-8 \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )+\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right ) \left (\left (\sin ^{-1}(c x)-4\right ) \sin ^{-1}(c x)-8 \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )\right )-b (c x+1) \sqrt {c d x+d} \sqrt {f-c f x} \left (20 \sin ^{-1}(c x)^2 \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+2 \sin ^{-1}(c x) \left (-24 \sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+7 \sin \left (\frac {3}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {5}{2} \sin ^{-1}(c x)\right )+24 \cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )+7 \cos \left (\frac {3}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {5}{2} \sin ^{-1}(c x)\right )\right )-2 \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right ) \left (16 c x+\cos \left (2 \sin ^{-1}(c x)\right )+32 \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )\right )\right )}{16 c d^2 (c x+1) \sqrt {1-c^2 x^2} \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^2*(8*a*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*Sqrt[1 - c^2*x^2]*(-24 - 7*c*x + c^2*x^2)*(Cos[ArcSin[c*x]/2] + Sin[
ArcSin[c*x]/2]) + 120*a*Sqrt[d]*Sqrt[f]*(1 + c*x)*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x
])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))]*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) - 8*b*(1 + c*x)*Sqrt[d + c*d*x]
*Sqrt[f - c*f*x]*(Cos[ArcSin[c*x]/2]*(ArcSin[c*x]*(4 + ArcSin[c*x]) - 8*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*
x]/2]]) + ((-4 + ArcSin[c*x])*ArcSin[c*x] - 8*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]
) - 32*b*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(ArcSin[c*x]^2*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) -
(c*x + 4*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) + ArcSin[c*x]
*((2 + Sqrt[1 - c^2*x^2])*Cos[ArcSin[c*x]/2] + (-2 + Sqrt[1 - c^2*x^2])*Sin[ArcSin[c*x]/2])) - b*(1 + c*x)*Sqr
t[d + c*d*x]*Sqrt[f - c*f*x]*(20*ArcSin[c*x]^2*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) - 2*(16*c*x + Cos[2*A
rcSin[c*x]] + 32*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) + 2*A
rcSin[c*x]*(24*Cos[ArcSin[c*x]/2] + 7*Cos[(3*ArcSin[c*x])/2] + Cos[(5*ArcSin[c*x])/2] - 24*Sin[ArcSin[c*x]/2]
+ 7*Sin[(3*ArcSin[c*x])/2] - Sin[(5*ArcSin[c*x])/2]))))/(16*c*d^2*(1 + c*x)*Sqrt[1 - c^2*x^2]*(Cos[ArcSin[c*x]
/2] + Sin[ArcSin[c*x]/2]))

________________________________________________________________________________________

fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a c^{2} f^{2} x^{2} - 2 \, a c f^{2} x + a f^{2} + {\left (b c^{2} f^{2} x^{2} - 2 \, b c f^{2} x + b f^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{c^{2} d^{2} x^{2} + 2 \, c d^{2} x + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {\left (-c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\left (c d x +d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, {\left (\frac {c^{2} f^{3} x^{3}}{\sqrt {-c^{2} d f x^{2} + d f} d} - \frac {8 \, c f^{3} x^{2}}{\sqrt {-c^{2} d f x^{2} + d f} d} - \frac {17 \, f^{3} x}{\sqrt {-c^{2} d f x^{2} + d f} d} + \frac {15 \, f^{3} \arcsin \left (c x\right )}{\sqrt {d f} c d} + \frac {24 \, f^{3}}{\sqrt {-c^{2} d f x^{2} + d f} c d}\right )} a + \frac {\frac {{\left (c^{2} \int \frac {\sqrt {-c x + 1} x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{\sqrt {c x + 1} c x + \sqrt {c x + 1}}\,{d x} - 2 \, c \int \frac {\sqrt {-c x + 1} x \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{\sqrt {c x + 1} c x + \sqrt {c x + 1}}\,{d x} + \int \frac {\sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{{\left (c x + 1\right )}^{\frac {3}{2}}}\,{d x}\right )} b f^{\frac {5}{2}}}{d}}{\sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c^2*f^3*x^3/(sqrt(-c^2*d*f*x^2 + d*f)*d) - 8*c*f^3*x^2/(sqrt(-c^2*d*f*x^2 + d*f)*d) - 17*f^3*x/(sqrt(-c^
2*d*f*x^2 + d*f)*d) + 15*f^3*arcsin(c*x)/(sqrt(d*f)*c*d) + 24*f^3/(sqrt(-c^2*d*f*x^2 + d*f)*c*d))*a + b*sqrt(f
)*integrate((c^2*f^2*x^2 - 2*c*f^2*x + f^2)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/((c*d*x
+ d)*sqrt(c*x + 1)), x)/sqrt(d)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________