3.6.34 \(\int \frac {(d+c d x)^{5/2} (a+b \text {ArcSin}(c x))}{(f-c f x)^{5/2}} \, dx\) [534]

Optimal. Leaf size=419 \[ \frac {b d^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {8 b d^5 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 b d^5 \left (1-c^2 x^2\right )^{5/2} \text {ArcSin}(c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^5 (1+c x)^4 \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {10 d^5 (1+c x)^2 \left (1-c^2 x^2\right )^2 (a+b \text {ArcSin}(c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 d^5 \left (1-c^2 x^2\right )^3 (a+b \text {ArcSin}(c x))}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 d^5 \left (1-c^2 x^2\right )^{5/2} \text {ArcSin}(c x) (a+b \text {ArcSin}(c x))}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {28 b d^5 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}} \]

[Out]

b*d^5*x*(-c^2*x^2+1)^(5/2)/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)-8/3*b*d^5*(-c^2*x^2+1)^(5/2)/c/(-c*x+1)/(c*d*x+d)^
(5/2)/(-c*f*x+f)^(5/2)-5/2*b*d^5*(-c^2*x^2+1)^(5/2)*arcsin(c*x)^2/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)+2/3*d^5*(
c*x+1)^4*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)-10/3*d^5*(c*x+1)^2*(-c^2*x^2+1)^2*(
a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)-5*d^5*(-c^2*x^2+1)^3*(a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(
-c*f*x+f)^(5/2)+5*d^5*(-c^2*x^2+1)^(5/2)*arcsin(c*x)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)-28/3
*b*d^5*(-c^2*x^2+1)^(5/2)*ln(-c*x+1)/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)

________________________________________________________________________________________

Rubi [A]
time = 0.27, antiderivative size = 419, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4763, 683, 655, 222, 4845, 641, 45, 4737} \begin {gather*} -\frac {5 d^5 \left (1-c^2 x^2\right )^3 (a+b \text {ArcSin}(c x))}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {10 d^5 (c x+1)^2 \left (1-c^2 x^2\right )^2 (a+b \text {ArcSin}(c x))}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^5 (c x+1)^4 \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {5 d^5 \left (1-c^2 x^2\right )^{5/2} \text {ArcSin}(c x) (a+b \text {ArcSin}(c x))}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {5 b d^5 \left (1-c^2 x^2\right )^{5/2} \text {ArcSin}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {b d^5 x \left (1-c^2 x^2\right )^{5/2}}{(c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {8 b d^5 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {28 b d^5 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d^5*x*(1 - c^2*x^2)^(5/2))/((d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) - (8*b*d^5*(1 - c^2*x^2)^(5/2))/(3*c*(1 -
c*x)*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) - (5*b*d^5*(1 - c^2*x^2)^(5/2)*ArcSin[c*x]^2)/(2*c*(d + c*d*x)^(5/2)
*(f - c*f*x)^(5/2)) + (2*d^5*(1 + c*x)^4*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(3*c*(d + c*d*x)^(5/2)*(f - c*f*x)
^(5/2)) - (10*d^5*(1 + c*x)^2*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/(3*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) -
(5*d^5*(1 - c^2*x^2)^3*(a + b*ArcSin[c*x]))/(c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) + (5*d^5*(1 - c^2*x^2)^(5
/2)*ArcSin[c*x]*(a + b*ArcSin[c*x]))/(c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) - (28*b*d^5*(1 - c^2*x^2)^(5/2)*L
og[1 - c*x])/(3*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))

Rule 45

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 222

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 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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 655

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 683

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 4737

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

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]

Rule 4845

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 {(d+c d x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{(f-c f x)^{5/2}} \, dx &=\frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {(d+c d x)^5 \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 {2 d^5 (1+c x)^4 \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 {10 d^5 (1+c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 d^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (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 {5 d^5}{c}+\frac {2 d^5 (1+c x)^4}{3 c \left (1-c^2 x^2\right )^2}-\frac {10 d^5 (1+c x)^2}{3 c \left (1-c^2 x^2\right )}+\frac {5 d^5 \sin ^{-1}(c x)}{c \sqrt {1-c^2 x^2}}\right ) \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=\frac {5 b d^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^5 (1+c x)^4 \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 {10 d^5 (1+c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 d^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (2 b d^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {(1+c x)^4}{\left (1-c^2 x^2\right )^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {\left (10 b d^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {(1+c x)^2}{1-c^2 x^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (5 b d^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {\sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=\frac {5 b d^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 b d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^5 (1+c x)^4 \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 {10 d^5 (1+c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 d^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (2 b d^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {(1+c x)^2}{(1-c x)^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {\left (10 b d^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1+c x}{1-c x} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=\frac {5 b d^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 b d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^5 (1+c x)^4 \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 {10 d^5 (1+c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 d^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (2 b d^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (1+\frac {4}{(-1+c x)^2}+\frac {4}{-1+c x}\right ) \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {\left (10 b d^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (-1-\frac {2}{-1+c x}\right ) \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=\frac {b d^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {8 b d^5 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 b d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^5 (1+c x)^4 \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 {10 d^5 (1+c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 d^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 d^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {28 b d^5 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(850\) vs. \(2(419)=838\).
time = 3.52, size = 850, normalized size = 2.03 \begin {gather*} \frac {d^2 \left (-\frac {4 a \sqrt {d+c d x} \sqrt {f-c f x} \left (23-34 c x+3 c^2 x^2\right )}{(-1+c x)^2}-60 a \sqrt {d} \sqrt {f} \text {ArcTan}\left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+\frac {2 b \sqrt {d+c d x} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right ) \left (-4+3 \text {ArcSin}(c x)-6 \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )\right )-\cos \left (\frac {3}{2} \text {ArcSin}(c x)\right ) \left (\text {ArcSin}(c x)-2 \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )\right )+2 \left (2+\left (2+\sqrt {1-c^2 x^2}\right ) \text {ArcSin}(c x)+2 \left (2+\sqrt {1-c^2 x^2}\right ) \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )\right ) \sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}{\left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )^4 \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}+\frac {2 b \sqrt {d+c d x} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right ) \left (-8-6 \text {ArcSin}(c x)+9 \text {ArcSin}(c x)^2-84 \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )\right )+\cos \left (\frac {3}{2} \text {ArcSin}(c x)\right ) \left (-\text {ArcSin}(c x) (14+3 \text {ArcSin}(c x))+28 \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )\right )+2 \left (4+2 \left (2+7 \sqrt {1-c^2 x^2}\right ) \text {ArcSin}(c x)-3 \left (2+\sqrt {1-c^2 x^2}\right ) \text {ArcSin}(c x)^2+28 \left (2+\sqrt {1-c^2 x^2}\right ) \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )\right ) \sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}{\left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )^4 \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}+\frac {b \sqrt {d+c d x} \sqrt {f-c f x} \left (2 \left (-7+6 c x+3 \cos (2 \text {ArcSin}(c x))+52 (-1+c x) \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )+18 \text {ArcSin}(c x)^2 \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )^3+\text {ArcSin}(c x) \left (-24 \cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-35 \cos \left (\frac {3}{2} \text {ArcSin}(c x)\right )+3 \cos \left (\frac {5}{2} \text {ArcSin}(c x)\right )-24 \sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )+35 \sin \left (\frac {3}{2} \text {ArcSin}(c x)\right )+3 \sin \left (\frac {5}{2} \text {ArcSin}(c x)\right )\right )\right )}{\left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )^4 \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}\right )}{12 c f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*((-4*a*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(23 - 34*c*x + 3*c^2*x^2))/(-1 + c*x)^2 - 60*a*Sqrt[d]*Sqrt[f]*Arc
Tan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + (2*b*Sqrt[d + c*d*x]*Sqrt[f - c*
f*x]*(Cos[ArcSin[c*x]/2]*(-4 + 3*ArcSin[c*x] - 6*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]]) - Cos[(3*ArcSin
[c*x])/2]*(ArcSin[c*x] - 2*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]]) + 2*(2 + (2 + Sqrt[1 - c^2*x^2])*ArcS
in[c*x] + 2*(2 + Sqrt[1 - c^2*x^2])*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/((Cos[A
rcSin[c*x]/2] - Sin[ArcSin[c*x]/2])^4*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])) + (2*b*Sqrt[d + c*d*x]*Sqrt[f
- c*f*x]*(Cos[ArcSin[c*x]/2]*(-8 - 6*ArcSin[c*x] + 9*ArcSin[c*x]^2 - 84*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c
*x]/2]]) + Cos[(3*ArcSin[c*x])/2]*(-(ArcSin[c*x]*(14 + 3*ArcSin[c*x])) + 28*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSi
n[c*x]/2]]) + 2*(4 + 2*(2 + 7*Sqrt[1 - c^2*x^2])*ArcSin[c*x] - 3*(2 + Sqrt[1 - c^2*x^2])*ArcSin[c*x]^2 + 28*(2
+ Sqrt[1 - c^2*x^2])*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/((Cos[ArcSin[c*x]/2]
- Sin[ArcSin[c*x]/2])^4*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])) + (b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(2*(-7
+ 6*c*x + 3*Cos[2*ArcSin[c*x]] + 52*(-1 + c*x)*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*(Cos[ArcSin[c*x]
/2] - Sin[ArcSin[c*x]/2]) + 18*ArcSin[c*x]^2*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])^3 + ArcSin[c*x]*(-24*Co
s[ArcSin[c*x]/2] - 35*Cos[(3*ArcSin[c*x])/2] + 3*Cos[(5*ArcSin[c*x])/2] - 24*Sin[ArcSin[c*x]/2] + 35*Sin[(3*Ar
cSin[c*x])/2] + 3*Sin[(5*ArcSin[c*x])/2])))/((Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])^4*(Cos[ArcSin[c*x]/2] +
Sin[ArcSin[c*x]/2]))))/(12*c*f^3)

________________________________________________________________________________________

Maple [F]
time = 0.30, size = 0, normalized size = 0.00 \[\int \frac {\left (c d x +d \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\left (-c f x +f \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*(-c^2*d*f*x^2 + d*f)^(5/2)/(c^5*f^5*x^4 - 4*c^4*f^5*x^3 + 6*c^3*f^5*x^2 - 4*c^2*f^5*x + c*f^5) + 5*(-c
^2*d*f*x^2 + d*f)^(3/2)*d/(c^4*f^4*x^3 - 3*c^3*f^4*x^2 + 3*c^2*f^4*x - c*f^4) - 10*sqrt(-c^2*d*f*x^2 + d*f)*d^
2/(c^3*f^3*x^2 - 2*c^2*f^3*x + c*f^3) - 35*sqrt(-c^2*d*f*x^2 + d*f)*d^2/(c^2*f^3*x - c*f^3) - 15*d^3*arcsin(c*
x)/(c*f^3*sqrt(d/f)))*a + b*sqrt(d)*integrate((c^2*d^2*x^2 + 2*c*d^2*x + d^2)*sqrt(c*x + 1)*arctan2(c*x, sqrt(
c*x + 1)*sqrt(-c*x + 1))/((c^2*f^2*x^2 - 2*c*f^2*x + f^2)*sqrt(-c*x + 1)), x)/sqrt(f)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________