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

Optimal. Leaf size=89 \[ \frac{(e (c+d x))^{m+1} \left (a+b \sin ^{-1}(c+d x)\right )}{d e (m+1)}-\frac{b (e (c+d x))^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},(c+d x)^2\right )}{d e^2 (m+1) (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.0621221, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4805, 4627, 364} \[ \frac{(e (c+d x))^{m+1} \left (a+b \sin ^{-1}(c+d x)\right )}{d e (m+1)}-\frac{b (e (c+d x))^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};(c+d x)^2\right )}{d e^2 (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4805

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

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

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

Mathematica [A]  time = 0.0418228, size = 77, normalized size = 0.87 \[ -\frac{(c+d x) (e (c+d x))^m \left (b (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},(c+d x)^2\right )-(m+2) \left (a+b \sin ^{-1}(c+d x)\right )\right )}{d (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.263, size = 0, normalized size = 0. \begin{align*} \int \left ( dex+ce \right ) ^{m} \left ( a+b\arcsin \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((d*e*x+c*e)^m*(a+b*arcsin(d*x+c)),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((d*e*x+c*e)^m*(a+b*arcsin(d*x+c)),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 (b \arcsin \left (d x + c\right ) + a\right )}{\left (d e x + c e\right )}^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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