3.656 \(\int (f x)^m (d+e x^2) (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=161 \[ \frac {d (f x)^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \sin ^{-1}(c x)\right )}{f^3 (m+3)}-\frac {b (f x)^{m+2} \left (c^2 d (m+3)^2+e (m+1) (m+2)\right ) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{c f^2 (m+1) (m+2) (m+3)^2}+\frac {b e \sqrt {1-c^2 x^2} (f x)^{m+2}}{c f^2 (m+3)^2} \]

[Out]

d*(f*x)^(1+m)*(a+b*arcsin(c*x))/f/(1+m)+e*(f*x)^(3+m)*(a+b*arcsin(c*x))/f^3/(3+m)-b*(e*(1+m)*(2+m)+c^2*d*(3+m)
^2)*(f*x)^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],c^2*x^2)/c/f^2/(1+m)/(2+m)/(3+m)^2+b*e*(f*x)^(2+m)*(-c^2*x^
2+1)^(1/2)/c/f^2/(3+m)^2

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Rubi [A]  time = 0.17, antiderivative size = 148, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {14, 4731, 12, 459, 364} \[ \frac {d (f x)^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \sin ^{-1}(c x)\right )}{f^3 (m+3)}-\frac {b c (f x)^{m+2} \left (\frac {e}{c^2 (m+3)^2}+\frac {d}{m^2+3 m+2}\right ) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{f^2}+\frac {b e \sqrt {1-c^2 x^2} (f x)^{m+2}}{c f^2 (m+3)^2} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^m*(d + e*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

(b*e*(f*x)^(2 + m)*Sqrt[1 - c^2*x^2])/(c*f^2*(3 + m)^2) + (d*(f*x)^(1 + m)*(a + b*ArcSin[c*x]))/(f*(1 + m)) +
(e*(f*x)^(3 + m)*(a + b*ArcSin[c*x]))/(f^3*(3 + m)) - (b*c*(e/(c^2*(3 + m)^2) + d/(2 + 3*m + m^2))*(f*x)^(2 +
m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/f^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 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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 459

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

Rule 4731

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 -
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] ||
(IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

Rubi steps

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

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Mathematica [C]  time = 2.22, size = 508, normalized size = 3.16 \[ \frac {x (f x)^m \left (2 e x^2 \left (b c x \Gamma \left (\frac {m}{2}+2\right ) \, _2\tilde {F}_1\left (\frac {1}{2},\frac {m}{2}+1;\frac {m}{2}+2;c^2 x^2\right )-\frac {b c x \Gamma \left (\frac {m}{2}+2\right ) \Gamma \left (\frac {m+3}{2}\right ) \, _2\tilde {F}_1\left (\frac {1}{2},\frac {m+3}{2};\frac {m+5}{2};c^2 x^2\right )}{\Gamma \left (\frac {m}{2}+1\right )}-a+\frac {b c x \, _2F_1\left (\frac {1}{2},\frac {m+3}{2};\frac {m+5}{2};c^2 x^2\right )}{m+3}-b \sin ^{-1}(c x)\right )-\frac {e x^2 \left (b c (m+3) x \Gamma \left (\frac {m}{2}+1\right ) \Gamma \left (\frac {m}{2}+2\right ) \, _2\tilde {F}_1\left (\frac {1}{2},\frac {m}{2}+1;\frac {m}{2}+3;c^2 x^2\right )+\frac {2 b c x \Gamma \left (\frac {m}{2}+2\right ) \left ((m+3) \Gamma \left (\frac {m}{2}+2\right ) \Gamma \left (\frac {m+3}{2}\right )-\Gamma \left (\frac {m}{2}+1\right ) \Gamma \left (\frac {m+5}{2}\right )\right ) \, _2\tilde {F}_1\left (\frac {1}{2},\frac {m}{2}+2;\frac {m}{2}+3;c^2 x^2\right )}{\Gamma \left (\frac {m+3}{2}\right )}-2 b c x \left ((m+3) \Gamma \left (\frac {m}{2}+2\right ) \Gamma \left (\frac {m+3}{2}\right )-\Gamma \left (\frac {m}{2}+1\right ) \Gamma \left (\frac {m+5}{2}\right )\right ) \, _2\tilde {F}_1\left (\frac {1}{2},\frac {m+3}{2};\frac {m+5}{2};c^2 x^2\right )-\frac {2 \Gamma \left (\frac {m}{2}+1\right ) \left ((m+4) \left (a+b \sin ^{-1}(c x)\right )-b c x \, _2F_1\left (\frac {1}{2},\frac {m}{2}+2;\frac {m}{2}+3;c^2 x^2\right )\right )}{m+4}\right )}{(m+3) \Gamma \left (\frac {m}{2}+1\right )}-\left (d+e x^2\right ) \left (b c x \, _2F_1\left (\frac {1}{2},\frac {m}{2}+1;\frac {m}{2}+2;c^2 x^2\right )-(m+2) \left (a+b \sin ^{-1}(c x)\right )\right )\right )}{(m+1) (m+2)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(f*x)^m*(d + e*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

(x*(f*x)^m*(-((d + e*x^2)*(-((2 + m)*(a + b*ArcSin[c*x])) + b*c*x*Hypergeometric2F1[1/2, 1 + m/2, 2 + m/2, c^2
*x^2])) + 2*e*x^2*(-a - b*ArcSin[c*x] + (b*c*x*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, c^2*x^2])/(3 + m)
+ b*c*x*Gamma[2 + m/2]*HypergeometricPFQRegularized[{1/2, 1 + m/2}, {2 + m/2}, c^2*x^2] - (b*c*x*Gamma[2 + m/2
]*Gamma[(3 + m)/2]*HypergeometricPFQRegularized[{1/2, (3 + m)/2}, {(5 + m)/2}, c^2*x^2])/Gamma[1 + m/2]) - (e*
x^2*((-2*Gamma[1 + m/2]*((4 + m)*(a + b*ArcSin[c*x]) - b*c*x*Hypergeometric2F1[1/2, 2 + m/2, 3 + m/2, c^2*x^2]
))/(4 + m) + b*c*(3 + m)*x*Gamma[1 + m/2]*Gamma[2 + m/2]*HypergeometricPFQRegularized[{1/2, 1 + m/2}, {3 + m/2
}, c^2*x^2] + (2*b*c*x*Gamma[2 + m/2]*((3 + m)*Gamma[2 + m/2]*Gamma[(3 + m)/2] - Gamma[1 + m/2]*Gamma[(5 + m)/
2])*HypergeometricPFQRegularized[{1/2, 2 + m/2}, {3 + m/2}, c^2*x^2])/Gamma[(3 + m)/2] - 2*b*c*x*((3 + m)*Gamm
a[2 + m/2]*Gamma[(3 + m)/2] - Gamma[1 + m/2]*Gamma[(5 + m)/2])*HypergeometricPFQRegularized[{1/2, (3 + m)/2},
{(5 + m)/2}, c^2*x^2]))/((3 + m)*Gamma[1 + m/2])))/((1 + m)*(2 + m))

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a e x^{2} + a d + {\left (b e x^{2} + b d\right )} \arcsin \left (c x\right )\right )} \left (f x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral((a*e*x^2 + a*d + (b*e*x^2 + b*d)*arcsin(c*x))*(f*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arcsin(c*x) + a)*(f*x)^m, x)

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maple [F]  time = 4.86, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{m} \left (e \,x^{2}+d \right ) \left (a +b \arcsin \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x)

[Out]

int((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x)

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maxima [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((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))*(f*x)^m*(d + e*x^2),x)

[Out]

int((a + b*asin(c*x))*(f*x)^m*(d + e*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{m} \left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*(e*x**2+d)*(a+b*asin(c*x)),x)

[Out]

Integral((f*x)**m*(a + b*asin(c*x))*(d + e*x**2), x)

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