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

Optimal. Leaf size=164 \[ \frac {d^3 (c x+1)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {2 b d^3 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b d^3 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4673, 651, 4761, 12, 627, 43} \[ \frac {d^3 (c x+1)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {2 b d^3 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b d^3 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 651

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

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

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Mathematica [A]  time = 0.54, size = 126, normalized size = 0.77 \[ \frac {\sqrt {c d x+d} \sqrt {f-c f x} \left ((c x+1) \left (a \sqrt {1-c^2 x^2}+b c x-b\right )+b (c x+1) \sqrt {1-c^2 x^2} \sin ^{-1}(c x)-b (c x-1)^2 \log (f-c f x)\right )}{3 c f^3 (c x-1)^2 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 520, normalized size = 3.17 \[ \left [\frac {{\left (b c^{3} f x^{3} - b c^{2} f x^{2} - b c f x + b f\right )} \sqrt {\frac {d}{f}} \log \left (\frac {c^{6} d x^{6} - 4 \, c^{5} d x^{5} + 5 \, c^{4} d x^{4} - 4 \, c^{2} d x^{2} + 4 \, c d x + {\left (c^{4} x^{4} - 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} - 4 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {\frac {d}{f}} - 2 \, d}{c^{4} x^{4} - 2 \, c^{3} x^{3} + 2 \, c x - 1}\right ) + 2 \, {\left (a c^{2} x^{2} - 2 \, \sqrt {-c^{2} x^{2} + 1} b c x + 2 \, a c x + {\left (b c^{2} x^{2} + 2 \, b c x + b\right )} \arcsin \left (c x\right ) + a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{6 \, {\left (c^{4} f^{3} x^{3} - c^{3} f^{3} x^{2} - c^{2} f^{3} x + c f^{3}\right )}}, -\frac {{\left (b c^{3} f x^{3} - b c^{2} f x^{2} - b c f x + b f\right )} \sqrt {-\frac {d}{f}} \arctan \left (\frac {{\left (c^{2} x^{2} - 2 \, c x + 2\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {-\frac {d}{f}}}{c^{4} d x^{4} - 2 \, c^{3} d x^{3} - c^{2} d x^{2} + 2 \, c d x}\right ) - {\left (a c^{2} x^{2} - 2 \, \sqrt {-c^{2} x^{2} + 1} b c x + 2 \, a c x + {\left (b c^{2} x^{2} + 2 \, b c x + b\right )} \arcsin \left (c x\right ) + a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{3 \, {\left (c^{4} f^{3} x^{3} - c^{3} f^{3} x^{2} - c^{2} f^{3} x + c f^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(1/2)*(a+b*arcsin(c*x))/(-c*f*x+f)^(5/2),x, algorithm="fricas")

[Out]

[1/6*((b*c^3*f*x^3 - b*c^2*f*x^2 - b*c*f*x + b*f)*sqrt(d/f)*log((c^6*d*x^6 - 4*c^5*d*x^5 + 5*c^4*d*x^4 - 4*c^2
*d*x^2 + 4*c*d*x + (c^4*x^4 - 4*c^3*x^3 + 6*c^2*x^2 - 4*c*x)*sqrt(-c^2*x^2 + 1)*sqrt(c*d*x + d)*sqrt(-c*f*x +
f)*sqrt(d/f) - 2*d)/(c^4*x^4 - 2*c^3*x^3 + 2*c*x - 1)) + 2*(a*c^2*x^2 - 2*sqrt(-c^2*x^2 + 1)*b*c*x + 2*a*c*x +
(b*c^2*x^2 + 2*b*c*x + b)*arcsin(c*x) + a)*sqrt(c*d*x + d)*sqrt(-c*f*x + f))/(c^4*f^3*x^3 - c^3*f^3*x^2 - c^2
*f^3*x + c*f^3), -1/3*((b*c^3*f*x^3 - b*c^2*f*x^2 - b*c*f*x + b*f)*sqrt(-d/f)*arctan((c^2*x^2 - 2*c*x + 2)*sqr
t(-c^2*x^2 + 1)*sqrt(c*d*x + d)*sqrt(-c*f*x + f)*sqrt(-d/f)/(c^4*d*x^4 - 2*c^3*d*x^3 - c^2*d*x^2 + 2*c*d*x)) -
(a*c^2*x^2 - 2*sqrt(-c^2*x^2 + 1)*b*c*x + 2*a*c*x + (b*c^2*x^2 + 2*b*c*x + b)*arcsin(c*x) + a)*sqrt(c*d*x + d
)*sqrt(-c*f*x + f))/(c^4*f^3*x^3 - c^3*f^3*x^2 - c^2*f^3*x + c*f^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(1/2)*(a+b*arcsin(c*x))/(-c*f*x+f)^(5/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c d x +d}\, \left (a +b \arcsin \left (c x \right )\right )}{\left (-c f x +f \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^(1/2)*(a+b*arcsin(c*x))/(-c*f*x+f)^(5/2),x)

[Out]

int((c*d*x+d)^(1/2)*(a+b*arcsin(c*x))/(-c*f*x+f)^(5/2),x)

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maxima [A]  time = 0.50, size = 217, normalized size = 1.32 \[ \frac {1}{3} \, b c {\left (\frac {2 \, \sqrt {d}}{c^{3} f^{\frac {5}{2}} x - c^{2} f^{\frac {5}{2}}} - \frac {\sqrt {d} \log \left (c x - 1\right )}{c^{2} f^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {2 \, \sqrt {-c^{2} d f x^{2} + d f}}{c^{3} f^{3} x^{2} - 2 \, c^{2} f^{3} x + c f^{3}} + \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} f^{3} x - c f^{3}}\right )} \arcsin \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {2 \, \sqrt {-c^{2} d f x^{2} + d f}}{c^{3} f^{3} x^{2} - 2 \, c^{2} f^{3} x + c f^{3}} + \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} f^{3} x - c f^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(1/2)*(a+b*arcsin(c*x))/(-c*f*x+f)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d+c\,d\,x}}{{\left (f-c\,f\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asin(c*x))*(d + c*d*x)^(1/2))/(f - c*f*x)^(5/2),x)

[Out]

int(((a + b*asin(c*x))*(d + c*d*x)^(1/2))/(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((c*d*x+d)**(1/2)*(a+b*asin(c*x))/(-c*f*x+f)**(5/2),x)

[Out]

Timed out

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