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

Optimal. Leaf size=179 \[ \frac{(d+e x)^{m+1} \left (a+b \sinh ^{-1}(c x)\right )}{e (m+1)}-\frac{b c \sqrt{1-\frac{d+e x}{d-\frac{e}{\sqrt{-c^2}}}} \sqrt{1-\frac{d+e x}{\frac{e}{\sqrt{-c^2}}+d}} (d+e x)^{m+2} F_1\left (m+2;\frac{1}{2},\frac{1}{2};m+3;\frac{d+e x}{d-\frac{e}{\sqrt{-c^2}}},\frac{d+e x}{d+\frac{e}{\sqrt{-c^2}}}\right )}{e^2 (m+1) (m+2) \sqrt{c^2 x^2+1}} \]

[Out]

-((b*c*(d + e*x)^(2 + m)*Sqrt[1 - (d + e*x)/(d - e/Sqrt[-c^2])]*Sqrt[1 - (d + e*x)/(d + e/Sqrt[-c^2])]*AppellF
1[2 + m, 1/2, 1/2, 3 + m, (d + e*x)/(d - e/Sqrt[-c^2]), (d + e*x)/(d + e/Sqrt[-c^2])])/(e^2*(1 + m)*(2 + m)*Sq
rt[1 + c^2*x^2])) + ((d + e*x)^(1 + m)*(a + b*ArcSinh[c*x]))/(e*(1 + m))

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Rubi [A]  time = 0.100638, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {5801, 760, 133} \[ \frac{(d+e x)^{m+1} \left (a+b \sinh ^{-1}(c x)\right )}{e (m+1)}-\frac{b c \sqrt{1-\frac{d+e x}{d-\frac{e}{\sqrt{-c^2}}}} \sqrt{1-\frac{d+e x}{\frac{e}{\sqrt{-c^2}}+d}} (d+e x)^{m+2} F_1\left (m+2;\frac{1}{2},\frac{1}{2};m+3;\frac{d+e x}{d-\frac{e}{\sqrt{-c^2}}},\frac{d+e x}{d+\frac{e}{\sqrt{-c^2}}}\right )}{e^2 (m+1) (m+2) \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*(d + e*x)^(2 + m)*Sqrt[1 - (d + e*x)/(d - e/Sqrt[-c^2])]*Sqrt[1 - (d + e*x)/(d + e/Sqrt[-c^2])]*AppellF
1[2 + m, 1/2, 1/2, 3 + m, (d + e*x)/(d - e/Sqrt[-c^2]), (d + e*x)/(d + e/Sqrt[-c^2])])/(e^2*(1 + m)*(2 + m)*Sq
rt[1 + c^2*x^2])) + ((d + e*x)^(1 + m)*(a + b*ArcSinh[c*x]))/(e*(1 + m))

Rule 5801

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

Rule 760

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[(a + c*x^
2)^p/(e*(1 - (d + e*x)/(d + (e*q)/c))^p*(1 - (d + e*x)/(d - (e*q)/c))^p), Subst[Int[x^m*Simp[1 - x/(d + (e*q)/
c), x]^p*Simp[1 - x/(d - (e*q)/c), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a
*e^2, 0] &&  !IntegerQ[p]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int (d+e x)^m \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{(d+e x)^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{e (1+m)}-\frac{(b c) \int \frac{(d+e x)^{1+m}}{\sqrt{1+c^2 x^2}} \, dx}{e (1+m)}\\ &=\frac{(d+e x)^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{e (1+m)}-\frac{\left (b c \sqrt{1-\frac{d+e x}{d-\frac{\sqrt{-c^2} e}{c^2}}} \sqrt{1-\frac{d+e x}{d+\frac{\sqrt{-c^2} e}{c^2}}}\right ) \operatorname{Subst}\left (\int \frac{x^{1+m}}{\sqrt{1-\frac{x}{d-\frac{e}{\sqrt{-c^2}}}} \sqrt{1-\frac{x}{d+\frac{e}{\sqrt{-c^2}}}}} \, dx,x,d+e x\right )}{e^2 (1+m) \sqrt{1+c^2 x^2}}\\ &=-\frac{b c (d+e x)^{2+m} \sqrt{1-\frac{d+e x}{d-\frac{e}{\sqrt{-c^2}}}} \sqrt{1-\frac{d+e x}{d+\frac{e}{\sqrt{-c^2}}}} F_1\left (2+m;\frac{1}{2},\frac{1}{2};3+m;\frac{d+e x}{d-\frac{e}{\sqrt{-c^2}}},\frac{d+e x}{d+\frac{e}{\sqrt{-c^2}}}\right )}{e^2 (1+m) (2+m) \sqrt{1+c^2 x^2}}+\frac{(d+e x)^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{e (1+m)}\\ \end{align*}

Mathematica [F]  time = 0.0432109, size = 0, normalized size = 0. \[ \int (d+e x)^m \left (a+b \sinh ^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 2.781, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{m} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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