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

Optimal. Leaf size=221 \[ -\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^6 d^3}+\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^6 d^2}+\frac {a+b \sin ^{-1}(c x)}{c^6 d \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {d-c^2 d x^2} \tanh ^{-1}(c x)}{c^6 d^2 \sqrt {1-c^2 x^2}}-\frac {5 b x \sqrt {d-c^2 d x^2}}{3 c^5 d^2 \sqrt {1-c^2 x^2}}-\frac {b x^3 \sqrt {d-c^2 d x^2}}{9 c^3 d^2 \sqrt {1-c^2 x^2}} \]

[Out]

-1/3*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/c^6/d^3+(a+b*arcsin(c*x))/c^6/d/(-c^2*d*x^2+d)^(1/2)+2*(a+b*arcsin
(c*x))*(-c^2*d*x^2+d)^(1/2)/c^6/d^2-5/3*b*x*(-c^2*d*x^2+d)^(1/2)/c^5/d^2/(-c^2*x^2+1)^(1/2)-1/9*b*x^3*(-c^2*d*
x^2+d)^(1/2)/c^3/d^2/(-c^2*x^2+1)^(1/2)-b*arctanh(c*x)*(-c^2*d*x^2+d)^(1/2)/c^6/d^2/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 229, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4703, 4707, 4677, 8, 30, 302, 206} \[ \frac {4 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2}+\frac {8 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^6 d^2}+\frac {x^4 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {b x^3 \sqrt {1-c^2 x^2}}{9 c^3 d \sqrt {d-c^2 d x^2}}-\frac {5 b x \sqrt {1-c^2 x^2}}{3 c^5 d \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{c^6 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*x*Sqrt[1 - c^2*x^2])/(3*c^5*d*Sqrt[d - c^2*d*x^2]) - (b*x^3*Sqrt[1 - c^2*x^2])/(9*c^3*d*Sqrt[d - c^2*d*x
^2]) + (x^4*(a + b*ArcSin[c*x]))/(c^2*d*Sqrt[d - c^2*d*x^2]) + (8*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(3*
c^6*d^2) + (4*x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(3*c^4*d^2) - (b*Sqrt[1 - c^2*x^2]*ArcTanh[c*x])/(c
^6*d*Sqrt[d - c^2*d*x^2])

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 206

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

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 4677

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[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 4703

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)/(2*e*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e*(p +
1)), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Dist[(b*f*n*d^IntPart[p]*(d + e*x^2
)^FracPart[p])/(2*c*(p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSi
n[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && Gt
Q[m, 1]

Rule 4707

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

Rubi steps

\begin {align*} \int \frac {x^5 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx &=\frac {x^4 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {4 \int \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \, dx}{c^2 d}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {x^4}{1-c^2 x^2} \, dx}{c d \sqrt {d-c^2 d x^2}}\\ &=\frac {x^4 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {4 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2}-\frac {8 \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \, dx}{3 c^4 d}-\frac {\left (4 b \sqrt {1-c^2 x^2}\right ) \int x^2 \, dx}{3 c^3 d \sqrt {d-c^2 d x^2}}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \left (-\frac {1}{c^4}-\frac {x^2}{c^2}+\frac {1}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx}{c d \sqrt {d-c^2 d x^2}}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{c^5 d \sqrt {d-c^2 d x^2}}-\frac {b x^3 \sqrt {1-c^2 x^2}}{9 c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^4 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {8 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^6 d^2}+\frac {4 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {1}{1-c^2 x^2} \, dx}{c^5 d \sqrt {d-c^2 d x^2}}-\frac {\left (8 b \sqrt {1-c^2 x^2}\right ) \int 1 \, dx}{3 c^5 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {5 b x \sqrt {1-c^2 x^2}}{3 c^5 d \sqrt {d-c^2 d x^2}}-\frac {b x^3 \sqrt {1-c^2 x^2}}{9 c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^4 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {8 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^6 d^2}+\frac {4 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2}-\frac {b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{c^6 d \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.31, size = 166, normalized size = 0.75 \[ \frac {\sqrt {d-c^2 d x^2} \left (\sqrt {-c^2} \left (3 a \left (c^4 x^4+4 c^2 x^2-8\right )+b c x \sqrt {1-c^2 x^2} \left (c^2 x^2+15\right )+3 b \left (c^4 x^4+4 c^2 x^2-8\right ) \sin ^{-1}(c x)\right )-9 i b c \sqrt {1-c^2 x^2} F\left (\left .i \sinh ^{-1}\left (\sqrt {-c^2} x\right )\right |1\right )\right )}{9 c^6 \sqrt {-c^2} d^2 \left (c^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d - c^2*d*x^2]*(Sqrt[-c^2]*(b*c*x*Sqrt[1 - c^2*x^2]*(15 + c^2*x^2) + 3*a*(-8 + 4*c^2*x^2 + c^4*x^4) + 3*
b*(-8 + 4*c^2*x^2 + c^4*x^4)*ArcSin[c*x]) - (9*I)*b*c*Sqrt[1 - c^2*x^2]*EllipticF[I*ArcSinh[Sqrt[-c^2]*x], 1])
)/(9*c^6*Sqrt[-c^2]*d^2*(-1 + c^2*x^2))

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fricas [A]  time = 1.01, size = 441, normalized size = 2.00 \[ \left [\frac {9 \, {\left (b c^{2} x^{2} - b\right )} \sqrt {d} \log \left (-\frac {c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} + 4 \, {\left (c^{3} x^{3} + c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} \sqrt {d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right ) + 4 \, {\left (b c^{3} x^{3} + 15 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + 12 \, {\left (a c^{4} x^{4} + 4 \, a c^{2} x^{2} + {\left (b c^{4} x^{4} + 4 \, b c^{2} x^{2} - 8 \, b\right )} \arcsin \left (c x\right ) - 8 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{36 \, {\left (c^{8} d^{2} x^{2} - c^{6} d^{2}\right )}}, -\frac {9 \, {\left (b c^{2} x^{2} - b\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} c \sqrt {-d} x}{c^{4} d x^{4} - d}\right ) - 2 \, {\left (b c^{3} x^{3} + 15 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} - 6 \, {\left (a c^{4} x^{4} + 4 \, a c^{2} x^{2} + {\left (b c^{4} x^{4} + 4 \, b c^{2} x^{2} - 8 \, b\right )} \arcsin \left (c x\right ) - 8 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{18 \, {\left (c^{8} d^{2} x^{2} - c^{6} d^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

[1/36*(9*(b*c^2*x^2 - b)*sqrt(d)*log(-(c^6*d*x^6 + 5*c^4*d*x^4 - 5*c^2*d*x^2 + 4*(c^3*x^3 + c*x)*sqrt(-c^2*d*x
^2 + d)*sqrt(-c^2*x^2 + 1)*sqrt(d) - d)/(c^6*x^6 - 3*c^4*x^4 + 3*c^2*x^2 - 1)) + 4*(b*c^3*x^3 + 15*b*c*x)*sqrt
(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) + 12*(a*c^4*x^4 + 4*a*c^2*x^2 + (b*c^4*x^4 + 4*b*c^2*x^2 - 8*b)*arcsin(c*x
) - 8*a)*sqrt(-c^2*d*x^2 + d))/(c^8*d^2*x^2 - c^6*d^2), -1/18*(9*(b*c^2*x^2 - b)*sqrt(-d)*arctan(2*sqrt(-c^2*d
*x^2 + d)*sqrt(-c^2*x^2 + 1)*c*sqrt(-d)*x/(c^4*d*x^4 - d)) - 2*(b*c^3*x^3 + 15*b*c*x)*sqrt(-c^2*d*x^2 + d)*sqr
t(-c^2*x^2 + 1) - 6*(a*c^4*x^4 + 4*a*c^2*x^2 + (b*c^4*x^4 + 4*b*c^2*x^2 - 8*b)*arcsin(c*x) - 8*a)*sqrt(-c^2*d*
x^2 + d))/(c^8*d^2*x^2 - c^6*d^2)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.50, size = 419, normalized size = 1.90 \[ -\frac {a \,x^{4}}{3 c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {4 a \,x^{2}}{3 c^{4} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {8 a}{3 c^{6} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )}{d^{2} c^{6} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \cos \left (4 \arcsin \left (c x \right )\right )}{24 d^{2} c^{6} \left (c^{2} x^{2}-1\right )}+\frac {5 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x^{2}}{3 d^{2} c^{4} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )}{d^{2} c^{6} \left (c^{2} x^{2}-1\right )}+\frac {31 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, x}{18 d^{2} c^{5} \left (c^{2} x^{2}-1\right )}-\frac {65 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{24 d^{2} c^{6} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sin \left (4 \arcsin \left (c x \right )\right )}{72 d^{2} c^{6} \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*x^4/c^2/d/(-c^2*d*x^2+d)^(1/2)-4/3*a/c^4*x^2/d/(-c^2*d*x^2+d)^(1/2)+8/3*a/c^6/d/(-c^2*d*x^2+d)^(1/2)+b*
(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^6/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)+1/24*b*(-d*(c^2*x
^2-1))^(1/2)/d^2/c^6/(c^2*x^2-1)*arcsin(c*x)*cos(4*arcsin(c*x))+5/3*b*(-d*(c^2*x^2-1))^(1/2)/d^2/c^4/(c^2*x^2-
1)*arcsin(c*x)*x^2-b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^6/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)
-I)+31/18*b*(-d*(c^2*x^2-1))^(1/2)/d^2/c^5/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x-65/24*b*(-d*(c^2*x^2-1))^(1/2)/d^2
/c^6/(c^2*x^2-1)*arcsin(c*x)-1/72*b*(-d*(c^2*x^2-1))^(1/2)/d^2/c^6/(c^2*x^2-1)*sin(4*arcsin(c*x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{3} \, a {\left (\frac {x^{4}}{\sqrt {-c^{2} d x^{2} + d} c^{2} d} + \frac {4 \, x^{2}}{\sqrt {-c^{2} d x^{2} + d} c^{4} d} - \frac {8}{\sqrt {-c^{2} d x^{2} + d} c^{6} d}\right )} - \frac {\frac {1}{30} \, {\left (\sqrt {c x + 1} \sqrt {-c x + 1} c^{6} d^{2} {\left (\frac {2 \, {\left (3 \, c^{4} x^{5} + 25 \, c^{2} x^{3} - 45 \, x\right )}}{c^{5} d^{2}} + \frac {45 \, \log \left (c x + 1\right )}{c^{6} d^{2}} - \frac {45 \, \log \left (c x - 1\right )}{c^{6} d^{2}}\right )} + 30 \, {\left (c^{4} x^{4} + 4 \, c^{2} x^{2} - 8\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} b}{3 \, \sqrt {c x + 1} \sqrt {-c x + 1} c^{6} d^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

-1/3*a*(x^4/(sqrt(-c^2*d*x^2 + d)*c^2*d) + 4*x^2/(sqrt(-c^2*d*x^2 + d)*c^4*d) - 8/(sqrt(-c^2*d*x^2 + d)*c^6*d)
) - 1/3*(3*sqrt(c*x + 1)*sqrt(-c*x + 1)*c^6*d^2*integrate(1/3*(c^4*x^6 + 4*c^2*x^4 - 8*x^2)/(c^7*d^2*x^4 - c^5
*d^2*x^2 + (c^5*d^2*x^2 - c^3*d^2)*e^(log(c*x + 1) + log(-c*x + 1))), x) + (c^4*x^4 + 4*c^2*x^2 - 8)*arctan2(c
*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*b/(sqrt(c*x + 1)*sqrt(-c*x + 1)*c^6*d^(3/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**5*(a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(3/2), x)

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