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

Optimal. Leaf size=399 \[ -\frac {3 b c d x^{m+2} \sqrt {d-c^2 d x^2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{(m+1) (m+2)^2 (m+4) \sqrt {1-c^2 x^2}}+\frac {3 d x^{m+1} \sqrt {d-c^2 d x^2} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (m^3+7 m^2+14 m+8\right ) \sqrt {1-c^2 x^2}}+\frac {3 d x^{m+1} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{m^2+6 m+8}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{m+4}-\frac {b c d x^{m+2} \sqrt {d-c^2 d x^2}}{\left (m^2+6 m+8\right ) \sqrt {1-c^2 x^2}}-\frac {3 b c d x^{m+2} \sqrt {d-c^2 d x^2}}{(m+2)^2 (m+4) \sqrt {1-c^2 x^2}}+\frac {b c^3 d x^{m+4} \sqrt {d-c^2 d x^2}}{(m+4)^2 \sqrt {1-c^2 x^2}} \]

[Out]

x^(1+m)*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/(4+m)+3*d*x^(1+m)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/(m^2+6
*m+8)-3*b*c*d*x^(2+m)*(-c^2*d*x^2+d)^(1/2)/(2+m)^2/(4+m)/(-c^2*x^2+1)^(1/2)-b*c*d*x^(2+m)*(-c^2*d*x^2+d)^(1/2)
/(m^2+6*m+8)/(-c^2*x^2+1)^(1/2)+b*c^3*d*x^(4+m)*(-c^2*d*x^2+d)^(1/2)/(4+m)^2/(-c^2*x^2+1)^(1/2)+3*d*x^(1+m)*(a
+b*arcsin(c*x))*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],c^2*x^2)*(-c^2*d*x^2+d)^(1/2)/(m^3+7*m^2+14*m+8)/(-c^2*
x^2+1)^(1/2)-3*b*c*d*x^(2+m)*HypergeometricPFQ([1, 1+1/2*m, 1+1/2*m],[3/2+1/2*m, 2+1/2*m],c^2*x^2)*(-c^2*d*x^2
+d)^(1/2)/(2+m)^2/(m^2+5*m+4)/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 399, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4699, 4697, 4711, 30, 14} \[ -\frac {3 b c d x^{m+2} \sqrt {d-c^2 d x^2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{(m+1) (m+2)^2 (m+4) \sqrt {1-c^2 x^2}}+\frac {3 d x^{m+1} \sqrt {d-c^2 d x^2} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (m^3+7 m^2+14 m+8\right ) \sqrt {1-c^2 x^2}}+\frac {3 d x^{m+1} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{m^2+6 m+8}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{m+4}-\frac {b c d x^{m+2} \sqrt {d-c^2 d x^2}}{\left (m^2+6 m+8\right ) \sqrt {1-c^2 x^2}}-\frac {3 b c d x^{m+2} \sqrt {d-c^2 d x^2}}{(m+2)^2 (m+4) \sqrt {1-c^2 x^2}}+\frac {b c^3 d x^{m+4} \sqrt {d-c^2 d x^2}}{(m+4)^2 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*c*d*x^(2 + m)*Sqrt[d - c^2*d*x^2])/((2 + m)^2*(4 + m)*Sqrt[1 - c^2*x^2]) - (b*c*d*x^(2 + m)*Sqrt[d - c^2
*d*x^2])/((8 + 6*m + m^2)*Sqrt[1 - c^2*x^2]) + (b*c^3*d*x^(4 + m)*Sqrt[d - c^2*d*x^2])/((4 + m)^2*Sqrt[1 - c^2
*x^2]) + (3*d*x^(1 + m)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(8 + 6*m + m^2) + (x^(1 + m)*(d - c^2*d*x^2)^
(3/2)*(a + b*ArcSin[c*x]))/(4 + m) + (3*d*x^(1 + m)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])*Hypergeometric2F1[
1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/((8 + 14*m + 7*m^2 + m^3)*Sqrt[1 - c^2*x^2]) - (3*b*c*d*x^(2 + m)*Sqrt[d
- c^2*d*x^2]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/((1 + m)*(2 + m)^2*(4 +
m)*Sqrt[1 - c^2*x^2])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

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

Rule 4697

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

Rule 4699

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n)/(f*(m + 2*p + 1)), x] + (Dist[(2*d*p)/(m + 2*p + 1), Int[(
f*x)^m*(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(
f*(m + 2*p + 1)*(1 - c^2*x^2)^FracPart[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, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]
&& (RationalQ[m] || EqQ[n, 1])

Rule 4711

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)
^(m + 1)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(Sqrt[d]*f*(m + 1)), x] -
Simp[(b*c*(f*x)^(m + 2)*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(Sqrt[d]*f^2*
(m + 1)*(m + 2)), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int x^m \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {x^{1+m} \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{4+m}+\frac {(3 d) \int x^m \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{4+m}-\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int x^{1+m} \left (1-c^2 x^2\right ) \, dx}{(4+m) \sqrt {1-c^2 x^2}}\\ &=\frac {3 d x^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8+6 m+m^2}+\frac {x^{1+m} \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{4+m}-\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int \left (x^{1+m}-c^2 x^{3+m}\right ) \, dx}{(4+m) \sqrt {1-c^2 x^2}}+\frac {\left (3 d \sqrt {d-c^2 d x^2}\right ) \int \frac {x^m \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{(2+m) (4+m) \sqrt {1-c^2 x^2}}-\frac {\left (3 b c d \sqrt {d-c^2 d x^2}\right ) \int x^{1+m} \, dx}{(2+m) (4+m) \sqrt {1-c^2 x^2}}\\ &=-\frac {3 b c d x^{2+m} \sqrt {d-c^2 d x^2}}{(2+m)^2 (4+m) \sqrt {1-c^2 x^2}}-\frac {b c d x^{2+m} \sqrt {d-c^2 d x^2}}{\left (8+6 m+m^2\right ) \sqrt {1-c^2 x^2}}+\frac {b c^3 d x^{4+m} \sqrt {d-c^2 d x^2}}{(4+m)^2 \sqrt {1-c^2 x^2}}+\frac {3 d x^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8+6 m+m^2}+\frac {x^{1+m} \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{4+m}+\frac {3 d x^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{\left (8+14 m+7 m^2+m^3\right ) \sqrt {1-c^2 x^2}}-\frac {3 b c d x^{2+m} \sqrt {d-c^2 d x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )}{(1+m) (2+m)^2 (4+m) \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 237, normalized size = 0.59 \[ \frac {d x^{m+1} \sqrt {d-c^2 d x^2} \left (-\frac {3 \left (b c x \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )-(m+2) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-(m+1) (m+2) \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+b c (m+1) x\right )}{(m+1) (m+2)^2 \sqrt {1-c^2 x^2}}+\left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {b c x \left (-c^2 (m+2) x^2+m+4\right )}{(m+2) (m+4) \sqrt {1-c^2 x^2}}\right )}{m+4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x^(1 + m)*Sqrt[d - c^2*d*x^2]*(-((b*c*x*(4 + m - c^2*(2 + m)*x^2))/((2 + m)*(4 + m)*Sqrt[1 - c^2*x^2])) + (
1 - c^2*x^2)*(a + b*ArcSin[c*x]) - (3*(b*c*(1 + m)*x - (1 + m)*(2 + m)*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]) -
(2 + m)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2] + b*c*x*HypergeometricPFQ[{
1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2]))/((1 + m)*(2 + m)^2*Sqrt[1 - c^2*x^2])))/(4 + m)

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fricas [F]  time = 1.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a c^{2} d x^{2} - a d + {\left (b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a*c^2*d*x^2 - a*d + (b*c^2*d*x^2 - b*d)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d)*x^m, x)

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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^m*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),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 [F]  time = 3.92, size = 0, normalized size = 0.00 \[ \int x^{m} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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