3.6.37 \(\int \frac {a+b \text {ArcSin}(c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx\) [537]

Optimal. Leaf size=265 \[ -\frac {b d^2 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^2 (1+c x) \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {d^2 x \left (1-c^2 x^2\right )^2 (a+b \text {ArcSin}(c x))}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \tanh ^{-1}(c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}} \]

[Out]

-1/3*b*d^2*(-c^2*x^2+1)^(5/2)/c/(-c*x+1)/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)+2/3*d^2*(c*x+1)*(-c^2*x^2+1)*(a+b*ar
csin(c*x))/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)+1/3*d^2*x*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*f
*x+f)^(5/2)-1/3*b*d^2*(-c^2*x^2+1)^(5/2)*arctanh(c*x)/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)+1/6*b*d^2*(-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 = 0.21, antiderivative size = 265, 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 = {4763, 667, 197, 4845, 641, 46, 213, 266} \begin {gather*} \frac {d^2 x \left (1-c^2 x^2\right )^2 (a+b \text {ArcSin}(c x))}{3 (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^2 (c x+1) \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 {b d^2 \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^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \tanh ^{-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[(a + b*ArcSin[c*x])/(Sqrt[d + c*d*x]*(f - c*f*x)^(5/2)),x]

[Out]

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 197

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 213

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

Rule 266

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 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 667

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)*((a + c*x^2)^(p + 1)/(c*(p
+ 1))), x] - Dist[e^2*((p + 2)/(c*(p + 1))), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -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 {a+b \sin ^{-1}(c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx &=\frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {(d+c d x)^2 \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^2 (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 {d^2 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 {2 d^2 (1+c x)}{3 c \left (1-c^2 x^2\right )^2}+\frac {d^2 x}{3 \left (1-c^2 x^2\right )}\right ) \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=\frac {2 d^2 (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 {d^2 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 (2 b d^2 \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 (b c d^2 \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 {2 d^2 (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 {d^2 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^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (2 b d^2 \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 {2 d^2 (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 {d^2 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^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {\left (2 b d^2 \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^2 \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 {2 d^2 (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 {d^2 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^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {\left (b d^2 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac {1}{-1+c^2 x^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac {b d^2 \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 {2 d^2 (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 {d^2 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^2 \left (1-c^2 x^2\right )^{5/2} \tanh ^{-1}(c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 130, normalized size = 0.49 \begin {gather*} \frac {\sqrt {d+c d x} \sqrt {f-c f x} \left (-\left ((-2+c x) \left (-b+b c x+a \sqrt {1-c^2 x^2}\right )\right )-b (-2+c x) \sqrt {1-c^2 x^2} \text {ArcSin}(c x)+b (-1+c x)^2 \log (f-c f x)\right )}{3 c d f^3 (-1+c x)^2 \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {a +b \arcsin \left (c x \right )}{\sqrt {c d x +d}\, \left (-c f x +f \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 227, normalized size = 0.86 \begin {gather*} \frac {1}{3} \, b c {\left (\frac {1}{c^{3} \sqrt {d} f^{\frac {5}{2}} x - c^{2} \sqrt {d} f^{\frac {5}{2}}} + \frac {\log \left (c x - 1\right )}{c^{2} \sqrt {d} f^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d f^{3} x^{2} - 2 \, c^{2} d f^{3} x + c d f^{3}} - \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d f^{3} x - c d f^{3}}\right )} \arcsin \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d f^{3} x^{2} - 2 \, c^{2} d f^{3} x + c d f^{3}} - \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d f^{3} x - c d f^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.55, size = 527, normalized size = 1.99 \begin {gather*} \left [\frac {{\left (b c^{3} x^{3} - b c^{2} x^{2} - b c x + b\right )} \sqrt {d f} \log \left (\frac {c^{6} d f x^{6} - 4 \, c^{5} d f x^{5} + 5 \, c^{4} d f x^{4} - 4 \, c^{2} d f x^{2} + 4 \, c d f 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 {d f} - 2 \, d f}{c^{4} x^{4} - 2 \, c^{3} x^{3} + 2 \, c x - 1}\right ) - 2 \, {\left (a c^{2} x^{2} + \sqrt {-c^{2} x^{2} + 1} b c x - a c x + {\left (b c^{2} x^{2} - b c x - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{6 \, {\left (c^{4} d f^{3} x^{3} - c^{3} d f^{3} x^{2} - c^{2} d f^{3} x + c d f^{3}\right )}}, \frac {{\left (b c^{3} x^{3} - b c^{2} x^{2} - b c x + b\right )} \sqrt {-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 {-d f}}{c^{4} d f x^{4} - 2 \, c^{3} d f x^{3} - c^{2} d f x^{2} + 2 \, c d f x}\right ) - {\left (a c^{2} x^{2} + \sqrt {-c^{2} x^{2} + 1} b c x - a c x + {\left (b c^{2} x^{2} - b c x - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{3 \, {\left (c^{4} d f^{3} x^{3} - c^{3} d f^{3} x^{2} - c^{2} d f^{3} x + c d f^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\sqrt {d \left (c x + 1\right )} \left (- f \left (c x - 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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