3.230 \(\int (c e+d e x)^{3/2} (a+b \sinh ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=261 \[ \frac{6 b e^{3/2} (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right ),\frac{1}{2}\right )}{25 d \sqrt{(c+d x)^2+1}}+\frac{2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}-\frac{12 b e^{3/2} (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )|\frac{1}{2}\right )}{25 d \sqrt{(c+d x)^2+1}}-\frac{4 b \sqrt{(c+d x)^2+1} (e (c+d x))^{3/2}}{25 d}+\frac{12 b e \sqrt{(c+d x)^2+1} \sqrt{e (c+d x)}}{25 d (c+d x+1)} \]

[Out]

(-4*b*(e*(c + d*x))^(3/2)*Sqrt[1 + (c + d*x)^2])/(25*d) + (12*b*e*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(25
*d*(1 + c + d*x)) + (2*(e*(c + d*x))^(5/2)*(a + b*ArcSinh[c + d*x]))/(5*d*e) - (12*b*e^(3/2)*(1 + c + d*x)*Sqr
t[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*EllipticE[2*ArcTan[Sqrt[e*(c + d*x)]/Sqrt[e]], 1/2])/(25*d*Sqrt[1 + (c +
d*x)^2]) + (6*b*e^(3/2)*(1 + c + d*x)*Sqrt[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*EllipticF[2*ArcTan[Sqrt[e*(c + d
*x)]/Sqrt[e]], 1/2])/(25*d*Sqrt[1 + (c + d*x)^2])

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Rubi [A]  time = 0.24261, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {5865, 5661, 321, 329, 305, 220, 1196} \[ \frac{2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}+\frac{6 b e^{3/2} (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )|\frac{1}{2}\right )}{25 d \sqrt{(c+d x)^2+1}}-\frac{12 b e^{3/2} (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )|\frac{1}{2}\right )}{25 d \sqrt{(c+d x)^2+1}}-\frac{4 b \sqrt{(c+d x)^2+1} (e (c+d x))^{3/2}}{25 d}+\frac{12 b e \sqrt{(c+d x)^2+1} \sqrt{e (c+d x)}}{25 d (c+d x+1)} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^(3/2)*(a + b*ArcSinh[c + d*x]),x]

[Out]

(-4*b*(e*(c + d*x))^(3/2)*Sqrt[1 + (c + d*x)^2])/(25*d) + (12*b*e*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(25
*d*(1 + c + d*x)) + (2*(e*(c + d*x))^(5/2)*(a + b*ArcSinh[c + d*x]))/(5*d*e) - (12*b*e^(3/2)*(1 + c + d*x)*Sqr
t[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*EllipticE[2*ArcTan[Sqrt[e*(c + d*x)]/Sqrt[e]], 1/2])/(25*d*Sqrt[1 + (c +
d*x)^2]) + (6*b*e^(3/2)*(1 + c + d*x)*Sqrt[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*EllipticF[2*ArcTan[Sqrt[e*(c + d
*x)]/Sqrt[e]], 1/2])/(25*d*Sqrt[1 + (c + d*x)^2])

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int (c e+d e x)^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int (e x)^{3/2} \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{(e x)^{5/2}}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{5 d e}\\ &=-\frac{4 b (e (c+d x))^{3/2} \sqrt{1+(c+d x)^2}}{25 d}+\frac{2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}+\frac{(6 b e) \operatorname{Subst}\left (\int \frac{\sqrt{e x}}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{25 d}\\ &=-\frac{4 b (e (c+d x))^{3/2} \sqrt{1+(c+d x)^2}}{25 d}+\frac{2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}+\frac{(12 b) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+\frac{x^4}{e^2}}} \, dx,x,\sqrt{e (c+d x)}\right )}{25 d}\\ &=-\frac{4 b (e (c+d x))^{3/2} \sqrt{1+(c+d x)^2}}{25 d}+\frac{2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}+\frac{(12 b e) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^4}{e^2}}} \, dx,x,\sqrt{e (c+d x)}\right )}{25 d}-\frac{(12 b e) \operatorname{Subst}\left (\int \frac{1-\frac{x^2}{e}}{\sqrt{1+\frac{x^4}{e^2}}} \, dx,x,\sqrt{e (c+d x)}\right )}{25 d}\\ &=-\frac{4 b (e (c+d x))^{3/2} \sqrt{1+(c+d x)^2}}{25 d}+\frac{12 b e \sqrt{e (c+d x)} \sqrt{1+(c+d x)^2}}{25 d (1+c+d x)}+\frac{2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}-\frac{12 b e^{3/2} (1+c+d x) \sqrt{\frac{1+(c+d x)^2}{(1+c+d x)^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )|\frac{1}{2}\right )}{25 d \sqrt{1+(c+d x)^2}}+\frac{6 b e^{3/2} (1+c+d x) \sqrt{\frac{1+(c+d x)^2}{(1+c+d x)^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )|\frac{1}{2}\right )}{25 d \sqrt{1+(c+d x)^2}}\\ \end{align*}

Mathematica [C]  time = 0.0465343, size = 87, normalized size = 0.33 \[ \frac{2 (e (c+d x))^{3/2} \left (2 b \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-(c+d x)^2\right )+5 a c+5 a d x-2 b \sqrt{(c+d x)^2+1}+5 b c \sinh ^{-1}(c+d x)+5 b d x \sinh ^{-1}(c+d x)\right )}{25 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^(3/2)*(a + b*ArcSinh[c + d*x]),x]

[Out]

(2*(e*(c + d*x))^(3/2)*(5*a*c + 5*a*d*x - 2*b*Sqrt[1 + (c + d*x)^2] + 5*b*c*ArcSinh[c + d*x] + 5*b*d*x*ArcSinh
[c + d*x] + 2*b*Hypergeometric2F1[1/2, 3/4, 7/4, -(c + d*x)^2]))/(25*d)

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Maple [C]  time = 0.01, size = 205, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{de} \left ( 1/5\, \left ( dex+ce \right ) ^{5/2}a+b \left ( 1/5\, \left ( dex+ce \right ) ^{5/2}{\it Arcsinh} \left ({\frac{dex+ce}{e}} \right ) -2/5\,{\frac{1}{e} \left ( 1/5\,{e}^{2} \left ( dex+ce \right ) ^{3/2}\sqrt{{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}-{3/5\,i{e}^{3}\sqrt{1-{\frac{i \left ( dex+ce \right ) }{e}}}\sqrt{1+{\frac{i \left ( dex+ce \right ) }{e}}} \left ({\it EllipticF} \left ( \sqrt{dex+ce}\sqrt{{\frac{i}{e}}},i \right ) -{\it EllipticE} \left ( \sqrt{dex+ce}\sqrt{{\frac{i}{e}}},i \right ) \right ){\frac{1}{\sqrt{{\frac{i}{e}}}}}{\frac{1}{\sqrt{{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}}}} \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^(3/2)*(a+b*arcsinh(d*x+c)),x)

[Out]

2/d/e*(1/5*(d*e*x+c*e)^(5/2)*a+b*(1/5*(d*e*x+c*e)^(5/2)*arcsinh(1/e*(d*e*x+c*e))-2/5/e*(1/5*e^2*(d*e*x+c*e)^(3
/2)*(1/e^2*(d*e*x+c*e)^2+1)^(1/2)-3/5*I*e^3/(I/e)^(1/2)*(1-I/e*(d*e*x+c*e))^(1/2)*(1+I/e*(d*e*x+c*e))^(1/2)/(1
/e^2*(d*e*x+c*e)^2+1)^(1/2)*(EllipticF((d*e*x+c*e)^(1/2)*(I/e)^(1/2),I)-EllipticE((d*e*x+c*e)^(1/2)*(I/e)^(1/2
),I)))))

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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)^(3/2)*(a+b*arcsinh(d*x+c)),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 (a d e x + a c e +{\left (b d e x + b c e\right )} \operatorname{arsinh}\left (d x + c\right )\right )} \sqrt{d e x + c e}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(3/2)*(a+b*arcsinh(d*x+c)),x, algorithm="fricas")

[Out]

integral((a*d*e*x + a*c*e + (b*d*e*x + b*c*e)*arcsinh(d*x + c))*sqrt(d*e*x + c*e), 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 )^{\frac{3}{2}} \left (a + b \operatorname{asinh}{\left (c + d x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**(3/2)*(a+b*asinh(d*x+c)),x)

[Out]

Integral((e*(c + d*x))**(3/2)*(a + b*asinh(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d e x + c e\right )}^{\frac{3}{2}}{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(3/2)*(a+b*arcsinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^(3/2)*(b*arcsinh(d*x + c) + a), x)