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

Optimal. Leaf size=161 \[ -\frac{b (f x)^{m+2} \left (c^2 d (m+3)^2+e (m+1) (m+2)\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{c f^2 (m+1) (m+2) (m+3)^2}+\frac{d (f x)^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{f (m+1)}+\frac{e (f x)^{m+3} \left (a+b \sin ^{-1}(c x)\right )}{f^3 (m+3)}+\frac{b e \sqrt{1-c^2 x^2} (f x)^{m+2}}{c f^2 (m+3)^2} \]

[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*(e*(1 + m)*(2 + m) + c^2*d*(3 + m)^2)*(f*x)^(2 + m)*H
ypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/(c*f^2*(1 + m)*(2 + m)*(3 + m)^2)

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Rubi [A]  time = 0.166187, 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 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 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]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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])

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*}

Mathematica [A]  time = 0.180262, size = 122, normalized size = 0.76 \[ x (f x)^m \left (\frac{\frac{\left (d (m+3)+e (m+1) x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{m+1}-\frac{b c e x^3 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m}{2}+2,\frac{m}{2}+3,c^2 x^2\right )}{m+4}}{m+3}-\frac{b c d x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m}{2}+1,\frac{m}{2}+2,c^2 x^2\right )}{m^2+3 m+2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 3.367, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m} \left ( e{x}^{2}+d \right ) \left ( a+b\arcsin \left ( cx \right ) \right ) \, dx \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (f x\right )^{m} \left (a + b \operatorname{asin}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}{\left (b \arcsin \left (c x\right ) + a\right )} \left (f x\right )^{m}\,{d x} \end{align*}

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)