3.3.37 \(\int (c e+d e x)^{5/2} (a+b \sinh ^{-1}(c+d x))^2 \, dx\) [237]

Optimal. Leaf size=134 \[ \frac {2 (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{7 d e}-\frac {8 b (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};-(c+d x)^2\right )}{63 d e^2}+\frac {16 b^2 (e (c+d x))^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};-(c+d x)^2\right )}{693 d e^3} \]

[Out]

2/7*(e*(d*x+c))^(7/2)*(a+b*arcsinh(d*x+c))^2/d/e-8/63*b*(e*(d*x+c))^(9/2)*(a+b*arcsinh(d*x+c))*hypergeom([1/2,
9/4],[13/4],-(d*x+c)^2)/d/e^2+16/693*b^2*(e*(d*x+c))^(11/2)*hypergeom([1, 11/4, 11/4],[13/4, 15/4],-(d*x+c)^2
)/d/e^3

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Rubi [A]
time = 0.16, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5859, 5776, 5817} \begin {gather*} \frac {16 b^2 (e (c+d x))^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};-(c+d x)^2\right )}{693 d e^3}-\frac {8 b (e (c+d x))^{9/2} \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{63 d e^2}+\frac {2 (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{7 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^(5/2)*(a + b*ArcSinh[c + d*x])^2,x]

[Out]

(2*(e*(c + d*x))^(7/2)*(a + b*ArcSinh[c + d*x])^2)/(7*d*e) - (8*b*(e*(c + d*x))^(9/2)*(a + b*ArcSinh[c + d*x])
*Hypergeometric2F1[1/2, 9/4, 13/4, -(c + d*x)^2])/(63*d*e^2) + (16*b^2*(e*(c + d*x))^(11/2)*HypergeometricPFQ[
{1, 11/4, 11/4}, {13/4, 15/4}, -(c + d*x)^2])/(693*d*e^3)

Rule 5776

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 5817

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

Rule 5859

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)^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int (e x)^{5/2} \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{7 d e}-\frac {(4 b) \text {Subst}\left (\int \frac {(e x)^{7/2} \left (a+b \sinh ^{-1}(x)\right )}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{7 d e}\\ &=\frac {2 (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{7 d e}-\frac {8 b (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};-(c+d x)^2\right )}{63 d e^2}+\frac {16 b^2 (e (c+d x))^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};-(c+d x)^2\right )}{693 d e^3}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 110, normalized size = 0.82 \begin {gather*} \frac {2 (e (c+d x))^{7/2} \left (99 \left (a+b \sinh ^{-1}(c+d x)\right )^2-44 b (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};-(c+d x)^2\right )+8 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};-(c+d x)^2\right )\right )}{693 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^(5/2)*(a + b*ArcSinh[c + d*x])^2,x]

[Out]

(2*(e*(c + d*x))^(7/2)*(99*(a + b*ArcSinh[c + d*x])^2 - 44*b*(c + d*x)*(a + b*ArcSinh[c + d*x])*Hypergeometric
2F1[1/2, 9/4, 13/4, -(c + d*x)^2] + 8*b^2*(c + d*x)^2*HypergeometricPFQ[{1, 11/4, 11/4}, {13/4, 15/4}, -(c + d
*x)^2]))/(693*d*e)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (d e x +c e \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (d x +c \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^(5/2)*(a+b*arcsinh(d*x+c))^2,x)

[Out]

int((d*e*x+c*e)^(5/2)*(a+b*arcsinh(d*x+c))^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(5/2)*(a+b*arcsinh(d*x+c))^2,x, algorithm="maxima")

[Out]

2/7*(d*x*e + c*e)^(7/2)*a^2*e^(-1)/d + 2/7*(b^2*d^3*x^3*e^(5/2) + 3*b^2*c*d^2*x^2*e^(5/2) + 3*b^2*c^2*d*x*e^(5
/2) + b^2*c^3*e^(5/2))*sqrt(d*x + c)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^2/d + integrate(2/7*(((7
*a*b*d^4 - 2*b^2*d^4)*x^4*e^(5/2) + 4*(7*a*b*c*d^3 - 2*b^2*c*d^3)*x^3*e^(5/2) - (12*b^2*c^2*d^2 - 7*(6*c^2*d^2
+ d^2)*a*b)*x^2*e^(5/2) - 2*(4*b^2*c^3*d - 7*(2*c^3*d + c*d)*a*b)*x*e^(5/2) - (2*b^2*c^4 - 7*(c^4 + c^2)*a*b)
*e^(5/2))*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*sqrt(d*x + c) + ((7*a*b*d^5 - 2*b^2*d^5)*x^5*e^(5/2) + 5*(7*a*b*c*
d^4 - 2*b^2*c*d^4)*x^4*e^(5/2) + (7*(10*c^2*d^3 + d^3)*a*b - 2*(10*c^2*d^3 + d^3)*b^2)*x^3*e^(5/2) + (7*(10*c^
3*d^2 + 3*c*d^2)*a*b - 2*(10*c^3*d^2 + 3*c*d^2)*b^2)*x^2*e^(5/2) + (7*(5*c^4*d + 3*c^2*d)*a*b - 2*(5*c^4*d + 3
*c^2*d)*b^2)*x*e^(5/2) + (7*(c^5 + c^3)*a*b - 2*(c^5 + c^3)*b^2)*e^(5/2))*sqrt(d*x + c))*log(d*x + c + sqrt(d^
2*x^2 + 2*c*d*x + c^2 + 1))/(d^3*x^3 + 3*c*d^2*x^2 + c^3 + (3*c^2*d + d)*x + (d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/
2) + c), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(5/2)*(a+b*arcsinh(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*arcsinh(d*x + c)^2*e^2 + 2*(a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^
2)*arcsinh(d*x + c)*e^2 + (a^2*d^2*x^2 + 2*a^2*c*d*x + a^2*c^2)*e^2)*sqrt(d*x + c)*e^(1/2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e \left (c + d x\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**(5/2)*(a+b*asinh(d*x+c))**2,x)

[Out]

Integral((e*(c + d*x))**(5/2)*(a + b*asinh(c + d*x))**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(5/2)*(a+b*arcsinh(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^(5/2)*(b*arcsinh(d*x + c) + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (c\,e+d\,e\,x\right )}^{5/2}\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^(5/2)*(a + b*asinh(c + d*x))^2,x)

[Out]

int((c*e + d*e*x)^(5/2)*(a + b*asinh(c + d*x))^2, x)

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