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

Optimal. Leaf size=187 \[ \frac{2 b^2 (e (c+d x))^{m+3} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2}\right \},\left \{\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2}\right \},-(c+d x)^2\right )}{d e^3 (m+1) (m+2) (m+3)}-\frac{2 b (e (c+d x))^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2 (m+1) (m+2)}+\frac{(e (c+d x))^{m+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e (m+1)} \]

[Out]

((e*(c + d*x))^(1 + m)*(a + b*ArcSinh[c + d*x])^2)/(d*e*(1 + m)) - (2*b*(e*(c + d*x))^(2 + m)*(a + b*ArcSinh[c
+ d*x])*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -(c + d*x)^2])/(d*e^2*(1 + m)*(2 + m)) + (2*b^2*(e*(c +
d*x))^(3 + m)*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5/2 + m/2}, -(c + d*x)^2])/(d*e^3*(1 + m)
*(2 + m)*(3 + m))

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Rubi [A]  time = 0.206553, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {5865, 5661, 5762} \[ \frac{2 b^2 (e (c+d x))^{m+3} \, _3F_2\left (1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2};\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2};-(c+d x)^2\right )}{d e^3 (m+1) (m+2) (m+3)}-\frac{2 b (e (c+d x))^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2 (m+1) (m+2)}+\frac{(e (c+d x))^{m+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*(c + d*x))^(1 + m)*(a + b*ArcSinh[c + d*x])^2)/(d*e*(1 + m)) - (2*b*(e*(c + d*x))^(2 + m)*(a + b*ArcSinh[c
+ d*x])*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -(c + d*x)^2])/(d*e^2*(1 + m)*(2 + m)) + (2*b^2*(e*(c +
d*x))^(3 + m)*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5/2 + m/2}, -(c + d*x)^2])/(d*e^3*(1 + m)
*(2 + m)*(3 + m))

Rule 5865

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

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[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 5762

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x
)^(m + 1)*(a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)])/(Sqrt[d]*f*(m + 1)),
x] - Simp[(b*c*(f*x)^(m + 2)*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, -(c^2*x^2)])/(Sqrt
[d]*f^2*(m + 1)*(m + 2)), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[d, 0] &&  !IntegerQ[m]

Rubi steps

\begin{align*} \int (c e+d e x)^m \left (a+b \sinh ^{-1}(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int (e x)^m \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{(e (c+d x))^{1+m} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e (1+m)}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{(e x)^{1+m} \left (a+b \sinh ^{-1}(x)\right )}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{d e (1+m)}\\ &=\frac{(e (c+d x))^{1+m} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e (1+m)}-\frac{2 b (e (c+d x))^{2+m} \left (a+b \sinh ^{-1}(c+d x)\right ) \, _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)}+\frac{2 b^2 (e (c+d x))^{3+m} \, _3F_2\left (1,\frac{3}{2}+\frac{m}{2},\frac{3}{2}+\frac{m}{2};2+\frac{m}{2},\frac{5}{2}+\frac{m}{2};-(c+d x)^2\right )}{d e^3 (1+m) (2+m) (3+m)}\\ \end{align*}

Mathematica [A]  time = 0.129712, size = 155, normalized size = 0.83 \[ \frac{(c+d x) (e (c+d x))^m \left (\frac{2 b^2 (c+d x)^2 \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2}\right \},\left \{\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2}\right \},-(c+d x)^2\right )}{(m+2) (m+3)}-\frac{2 b (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{m+2}+\left (a+b \sinh ^{-1}(c+d x)\right )^2\right )}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c + d*x)*(e*(c + d*x))^m*((a + b*ArcSinh[c + d*x])^2 - (2*b*(c + d*x)*(a + b*ArcSinh[c + d*x])*Hypergeometri
c2F1[1/2, (2 + m)/2, (4 + m)/2, -(c + d*x)^2])/(2 + m) + (2*b^2*(c + d*x)^2*HypergeometricPFQ[{1, 3/2 + m/2, 3
/2 + m/2}, {2 + m/2, 5/2 + m/2}, -(c + d*x)^2])/((2 + m)*(3 + m))))/(d*(1 + m))

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Maple [F]  time = 1.458, size = 0, normalized size = 0. \begin{align*} \int \left ( dex+ce \right ) ^{m} \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((d*e*x+c*e)^m*(a+b*arcsinh(d*x+c))^2,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*arcsinh(d*x+c))^2,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^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 2 \, a b \operatorname{arsinh}\left (d x + c\right ) + a^{2}\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*arcsinh(d*x+c))^2,x, algorithm="fricas")

[Out]

integral((b^2*arcsinh(d*x + c)^2 + 2*a*b*arcsinh(d*x + c) + a^2)*(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{asinh}{\left (c + d x \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{2}{\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*arcsinh(d*x+c))^2,x, algorithm="giac")

[Out]

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