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

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

[Out]

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

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Rubi [A]  time = 0.261779, antiderivative size = 255, normalized size of antiderivative = 1., 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} \[ \frac{2 f x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{f (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{b f \left (1-c^2 x^2\right )^{5/2}}{6 c (c x+1) (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{b f \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{b f \left (1-c^2 x^2\right )^{5/2} \tanh ^{-1}(c x)}{6 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*f*(1 - c^2*x^2)^(5/2))/(6*c*(1 + c*x)*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) - (f*(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*f*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*f*(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*f*(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 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 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 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 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])

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.598244, size = 180, normalized size = 0.71 \[ \frac{\sqrt{c d x+d} \left (8 a c^2 x^2+8 a c x-4 a+3 b c x \sqrt{1-c^2 x^2} \log (f-c f x)+5 b (c x+1) \sqrt{1-c^2 x^2} \log (-f (c x+1))+3 b \sqrt{1-c^2 x^2} \log (f-c f x)-2 b \sqrt{1-c^2 x^2}+4 b \left (2 c^2 x^2+2 c x-1\right ) \sin ^{-1}(c x)\right )}{12 c d^3 f (c x+1)^2 \sqrt{f-c f x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])/((d + c*d*x)^(5/2)*(f - c*f*x)^(3/2)),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] + 5*b*(1 + c*x)*Sqrt[1 - c^2*x^2]*Log[-(f*(1 + c*x))] + 3*b*Sqrt[1 - c^2*x^2]*Log[f - c*f*x] + 3*b*c*x*Sqr
t[1 - c^2*x^2]*Log[f - c*f*x]))/(12*c*d^3*f*(1 + c*x)^2*Sqrt[f - c*f*x])

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Maple [F]  time = 0.227, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arcsin \left ( cx \right ) ) \left ( cdx+d \right ) ^{-{\frac{5}{2}}} \left ( -cfx+f \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

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)^(3/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{3} f^{2} x^{5} + c^{4} d^{3} f^{2} x^{4} - 2 \, c^{3} d^{3} f^{2} x^{3} - 2 \, c^{2} d^{3} f^{2} x^{2} + c d^{3} f^{2} x + d^{3} f^{2}}, x\right ) \end{align*}

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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{3}{2}}}\,{d x} \end{align*}

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)^(3/2),x, algorithm="giac")

[Out]

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