3.6.22 \(\int \frac {(d+c d x)^{5/2} (a+b \text {ArcSin}(c x))}{\sqrt {f-c f x}} \, dx\) [522]

Optimal. Leaf size=345 \[ \frac {11 b d^3 x \sqrt {1-c^2 x^2}}{3 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {3 b c d^3 x^2 \sqrt {1-c^2 x^2}}{4 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {b c^2 d^3 x^3 \sqrt {1-c^2 x^2}}{9 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {11 d^3 \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{3 c \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {3 d^3 x \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{2 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {c d^3 x^2 \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{3 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {5 d^3 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{4 b c \sqrt {d+c d x} \sqrt {f-c f x}} \]

[Out]

-11/3*d^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-3/2*d^3*x*(-c^2*x^2+1)*(a+b*arcsin
(c*x))/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-1/3*c*d^3*x^2*(-c^2*x^2+1)*(a+b*arcsin(c*x))/(c*d*x+d)^(1/2)/(-c*f*x+f
)^(1/2)+11/3*b*d^3*x*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+3/4*b*c*d^3*x^2*(-c^2*x^2+1)^(1/2)/(c
*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+1/9*b*c^2*d^3*x^3*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+5/4*d^3*(
a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/b/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)

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Rubi [A]
time = 0.40, antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {4763, 4847, 4737, 4767, 8, 4795, 30} \begin {gather*} \frac {5 d^3 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{4 b c \sqrt {c d x+d} \sqrt {f-c f x}}-\frac {c d^3 x^2 \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{3 \sqrt {c d x+d} \sqrt {f-c f x}}-\frac {3 d^3 x \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{2 \sqrt {c d x+d} \sqrt {f-c f x}}-\frac {11 d^3 \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{3 c \sqrt {c d x+d} \sqrt {f-c f x}}+\frac {3 b c d^3 x^2 \sqrt {1-c^2 x^2}}{4 \sqrt {c d x+d} \sqrt {f-c f x}}+\frac {11 b d^3 x \sqrt {1-c^2 x^2}}{3 \sqrt {c d x+d} \sqrt {f-c f x}}+\frac {b c^2 d^3 x^3 \sqrt {1-c^2 x^2}}{9 \sqrt {c d x+d} \sqrt {f-c f x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(11*b*d^3*x*Sqrt[1 - c^2*x^2])/(3*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) + (3*b*c*d^3*x^2*Sqrt[1 - c^2*x^2])/(4*Sqrt
[d + c*d*x]*Sqrt[f - c*f*x]) + (b*c^2*d^3*x^3*Sqrt[1 - c^2*x^2])/(9*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) - (11*d^3
*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(3*c*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) - (3*d^3*x*(1 - c^2*x^2)*(a + b*ArcS
in[c*x]))/(2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) - (c*d^3*x^2*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(3*Sqrt[d + c*d*
x]*Sqrt[f - c*f*x]) + (5*d^3*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(4*b*c*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 4767

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

Rule 4795

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[f
*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(e*(m + 2*p + 1))), x] + (Dist[f^2*((m - 1)/(c^2*(m
+ 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + Dist[b*f*(n/(c*(m + 2*p + 1)))*S
imp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x],
x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0]

Rule 4847

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g},
x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && (m == 1 || p > 0 ||
(n == 1 && p > -1) || (m == 2 && p < -2))

Rubi steps

\begin {align*} \int \frac {(d+c d x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {f-c f x}} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {(d+c d x)^3 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}\\ &=\frac {\sqrt {1-c^2 x^2} \int \left (\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {3 c d^3 x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {3 c^2 d^3 x^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {c^3 d^3 x^3 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}\right ) \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}\\ &=\frac {\left (d^3 \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\left (3 c d^3 \sqrt {1-c^2 x^2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\left (3 c^2 d^3 \sqrt {1-c^2 x^2}\right ) \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\left (c^3 d^3 \sqrt {1-c^2 x^2}\right ) \int \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}\\ &=-\frac {3 d^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {3 d^3 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {c d^3 x^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {d^3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\left (3 d^3 \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{2 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\left (3 b d^3 \sqrt {1-c^2 x^2}\right ) \int 1 \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\left (2 c d^3 \sqrt {1-c^2 x^2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{3 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\left (3 b c d^3 \sqrt {1-c^2 x^2}\right ) \int x \, dx}{2 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\left (b c^2 d^3 \sqrt {1-c^2 x^2}\right ) \int x^2 \, dx}{3 \sqrt {d+c d x} \sqrt {f-c f x}}\\ &=\frac {3 b d^3 x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {3 b c d^3 x^2 \sqrt {1-c^2 x^2}}{4 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {b c^2 d^3 x^3 \sqrt {1-c^2 x^2}}{9 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {11 d^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {3 d^3 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {c d^3 x^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {5 d^3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\left (2 b d^3 \sqrt {1-c^2 x^2}\right ) \int 1 \, dx}{3 \sqrt {d+c d x} \sqrt {f-c f x}}\\ &=\frac {11 b d^3 x \sqrt {1-c^2 x^2}}{3 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {3 b c d^3 x^2 \sqrt {1-c^2 x^2}}{4 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {b c^2 d^3 x^3 \sqrt {1-c^2 x^2}}{9 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {11 d^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {3 d^3 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {c d^3 x^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {5 d^3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt {d+c d x} \sqrt {f-c f x}}\\ \end {align*}

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Mathematica [A]
time = 1.27, size = 270, normalized size = 0.78 \begin {gather*} -\frac {d^2 \left (-90 b \sqrt {d+c d x} \sqrt {f-c f x} \text {ArcSin}(c x)^2+180 a \sqrt {d} \sqrt {f} \sqrt {1-c^2 x^2} \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 )+6 b \sqrt {d+c d x} \sqrt {f-c f x} \text {ArcSin}(c x) \left (9 (5+2 c x) \sqrt {1-c^2 x^2}-\cos (3 \text {ArcSin}(c x))\right )+\sqrt {d+c d x} \sqrt {f-c f x} \left (-270 b c x+12 a \sqrt {1-c^2 x^2} \left (22+9 c x+2 c^2 x^2\right )+27 b \cos (2 \text {ArcSin}(c x))+2 b \sin (3 \text {ArcSin}(c x))\right )\right )}{72 c f \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/72*(d^2*(-90*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 + 180*a*Sqrt[d]*Sqrt[f]*Sqrt[1 - c^2*x^2]*ArcT
an[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + 6*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*
x]*ArcSin[c*x]*(9*(5 + 2*c*x)*Sqrt[1 - c^2*x^2] - Cos[3*ArcSin[c*x]]) + Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-270*
b*c*x + 12*a*Sqrt[1 - c^2*x^2]*(22 + 9*c*x + 2*c^2*x^2) + 27*b*Cos[2*ArcSin[c*x]] + 2*b*Sin[3*ArcSin[c*x]])))/
(c*f*Sqrt[1 - c^2*x^2])

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

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)^(1/2),x)

[Out]

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

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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)^(1/2),x, algorithm="maxima")

[Out]

-1/6*(2*sqrt(-c^2*d*f*x^2 + d*f)*c*d^2*x^2/f + 9*sqrt(-c^2*d*f*x^2 + d*f)*d^2*x/f - 15*d^3*arcsin(c*x)/(sqrt(d
*f)*c) + 22*sqrt(-c^2*d*f*x^2 + d*f)*d^2/(c*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))/sqrt(-c*x + 1), x)/sqrt(f)

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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)^(1/2),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*f*x - f), x)

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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)**(1/2),x)

[Out]

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

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

[Out]

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

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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}}{\sqrt {f-c\,f\,x}} \,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)^(1/2),x)

[Out]

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

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