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]

(e*x+d)^(1+m)*(a+b*arcsinh(c*x))/e/(1+m)-b*c*(e*x+d)^(2+m)*AppellF1(2+m,1/2,1/2,3+m,(e*x+d)/(d-e/(-c^2)^(1/2))
,(e*x+d)/(d+e/(-c^2)^(1/2)))*(1+(-e*x-d)/(d-e/(-c^2)^(1/2)))^(1/2)*(1+(-e*x-d)/(d+e/(-c^2)^(1/2)))^(1/2)/e^2/(
1+m)/(2+m)/(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 179, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int (d+e x)^m \left (a+b \sinh ^{-1}(c x)\right ) \, dx \]

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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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} {\left (e x + d\right )}^{m}, x\right ) \]

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

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)

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maple [F]  time = 3.60, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{m} \left (a +b \arcsinh \left (c x \right )\right )\, dx \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ b {\left (\frac {{\left (e x + d\right )} {\left (e x + d\right )}^{m} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{e {\left (m + 1\right )}} - \int \frac {{\left (c^{2} e x^{2} + c^{2} d x\right )} {\left (e x + d\right )}^{m}}{c^{2} e {\left (m + 1\right )} x^{2} + e {\left (m + 1\right )}}\,{d x} - \int \frac {{\left (c e x + c d\right )} {\left (e x + d\right )}^{m}}{c^{3} e {\left (m + 1\right )} x^{3} + c e {\left (m + 1\right )} x + {\left (c^{2} e {\left (m + 1\right )} x^{2} + e {\left (m + 1\right )}\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x}\right )} + \frac {{\left (e x + d\right )}^{m + 1} a}{e {\left (m + 1\right )}} \]

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]

b*((e*x + d)*(e*x + d)^m*log(c*x + sqrt(c^2*x^2 + 1))/(e*(m + 1)) - integrate((c^2*e*x^2 + c^2*d*x)*(e*x + d)^
m/(c^2*e*(m + 1)*x^2 + e*(m + 1)), x) - integrate((c*e*x + c*d)*(e*x + d)^m/(c^3*e*(m + 1)*x^3 + c*e*(m + 1)*x
+ (c^2*e*(m + 1)*x^2 + e*(m + 1))*sqrt(c^2*x^2 + 1)), x)) + (e*x + d)^(m + 1)*a/(e*(m + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))*(d + e*x)^m,x)

[Out]

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

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

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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