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

Optimal. Leaf size=187 \[ \frac {2 b^2 (e (c+d x))^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};-(c+d x)^2\right )}{d e^3 (m+1) (m+2) (m+3)}-\frac {2 b (e (c+d x))^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2 (m+1) (m+2)}+\frac {(e (c+d x))^{m+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e (m+1)} \]

[Out]

(e*(d*x+c))^(1+m)*(a+b*arcsinh(d*x+c))^2/d/e/(1+m)-2*b*(e*(d*x+c))^(2+m)*(a+b*arcsinh(d*x+c))*hypergeom([1/2,
1+1/2*m],[2+1/2*m],-(d*x+c)^2)/d/e^2/(1+m)/(2+m)+2*b^2*(e*(d*x+c))^(3+m)*HypergeometricPFQ([1, 3/2+1/2*m, 3/2+
1/2*m],[2+1/2*m, 5/2+1/2*m],-(d*x+c)^2)/d/e^3/(3+m)/(m^2+3*m+2)

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Rubi [A]  time = 0.21, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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]

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
]

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*}

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Mathematica [A]  time = 0.12, size = 155, normalized size = 0.83 \[ \frac {(c+d x) (e (c+d x))^m \left (\frac {2 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};-(c+d x)^2\right )}{(m+2) (m+3)}-\frac {2 b (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{m+2}+\left (a+b \sinh ^{-1}(c+d x)\right )^2\right )}{d (m+1)} \]

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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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arsinh}\left (d x + c\right ) + a^{2}\right )} {\left (d e x + c e\right )}^{m}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2} {\left (d e x + c e\right )}^{m}\,{d x} \]

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)

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maple [F]  time = 1.95, size = 0, normalized size = 0.00 \[ \int \left (d e x +c e \right )^{m} \left (a +b \arcsinh \left (d x +c \right )\right )^{2}\, dx \]

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

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,e+d\,e\,x\right )}^m\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \left (c + d x\right )\right )^{m} \left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{2}\, dx \]

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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