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

Optimal. Leaf size=177 \[ -\frac{10 b e^{5/2} (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right ),\frac{1}{2}\right )}{147 d \sqrt{(c+d x)^2+1}}+\frac{2 (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{7 d e}+\frac{20 b e^2 \sqrt{(c+d x)^2+1} \sqrt{e (c+d x)}}{147 d}-\frac{4 b \sqrt{(c+d x)^2+1} (e (c+d x))^{5/2}}{49 d} \]

[Out]

(20*b*e^2*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(147*d) - (4*b*(e*(c + d*x))^(5/2)*Sqrt[1 + (c + d*x)^2])/(
49*d) + (2*(e*(c + d*x))^(7/2)*(a + b*ArcSinh[c + d*x]))/(7*d*e) - (10*b*e^(5/2)*(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])/(147*d*Sqrt[1 + (c + d*x)^2])

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Rubi [A]  time = 0.159405, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {5865, 5661, 321, 329, 220} \[ \frac{2 (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{7 d e}+\frac{20 b e^2 \sqrt{(c+d x)^2+1} \sqrt{e (c+d x)}}{147 d}-\frac{10 b e^{5/2} (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )|\frac{1}{2}\right )}{147 d \sqrt{(c+d x)^2+1}}-\frac{4 b \sqrt{(c+d x)^2+1} (e (c+d x))^{5/2}}{49 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(20*b*e^2*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(147*d) - (4*b*(e*(c + d*x))^(5/2)*Sqrt[1 + (c + d*x)^2])/(
49*d) + (2*(e*(c + d*x))^(7/2)*(a + b*ArcSinh[c + d*x]))/(7*d*e) - (10*b*e^(5/2)*(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])/(147*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 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]

Rubi steps

\begin{align*} \int (c e+d e x)^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int (e x)^{5/2} \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{2 (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{7 d e}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{(e x)^{7/2}}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{7 d e}\\ &=-\frac{4 b (e (c+d x))^{5/2} \sqrt{1+(c+d x)^2}}{49 d}+\frac{2 (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{7 d e}+\frac{(10 b e) \operatorname{Subst}\left (\int \frac{(e x)^{3/2}}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{49 d}\\ &=\frac{20 b e^2 \sqrt{e (c+d x)} \sqrt{1+(c+d x)^2}}{147 d}-\frac{4 b (e (c+d x))^{5/2} \sqrt{1+(c+d x)^2}}{49 d}+\frac{2 (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{7 d e}-\frac{\left (10 b e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{e x} \sqrt{1+x^2}} \, dx,x,c+d x\right )}{147 d}\\ &=\frac{20 b e^2 \sqrt{e (c+d x)} \sqrt{1+(c+d x)^2}}{147 d}-\frac{4 b (e (c+d x))^{5/2} \sqrt{1+(c+d x)^2}}{49 d}+\frac{2 (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{7 d e}-\frac{\left (20 b e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^4}{e^2}}} \, dx,x,\sqrt{e (c+d x)}\right )}{147 d}\\ &=\frac{20 b e^2 \sqrt{e (c+d x)} \sqrt{1+(c+d x)^2}}{147 d}-\frac{4 b (e (c+d x))^{5/2} \sqrt{1+(c+d x)^2}}{49 d}+\frac{2 (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{7 d e}-\frac{10 b e^{5/2} (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 )}{147 d \sqrt{1+(c+d x)^2}}\\ \end{align*}

Mathematica [C]  time = 0.179269, size = 113, normalized size = 0.64 \[ \frac{2 (e (c+d x))^{5/2} \left (-10 b \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-(c+d x)^2\right )+21 a (c+d x)^3-6 b \sqrt{(c+d x)^2+1} (c+d x)^2+10 b \sqrt{(c+d x)^2+1}+21 b (c+d x)^3 \sinh ^{-1}(c+d x)\right )}{147 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(5/2)*(21*a*(c + d*x)^3 + 10*b*Sqrt[1 + (c + d*x)^2] - 6*b*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]
+ 21*b*(c + d*x)^3*ArcSinh[c + d*x] - 10*b*Hypergeometric2F1[1/4, 1/2, 5/4, -(c + d*x)^2]))/(147*d*(c + d*x)^2
)

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Maple [C]  time = 0.01, size = 212, normalized size = 1.2 \begin{align*} 2\,{\frac{1}{de} \left ( 1/7\, \left ( dex+ce \right ) ^{7/2}a+b \left ( 1/7\, \left ( dex+ce \right ) ^{7/2}{\it Arcsinh} \left ({\frac{dex+ce}{e}} \right ) -2/7\,{\frac{1}{e} \left ( 1/7\,{e}^{2} \left ( dex+ce \right ) ^{5/2}\sqrt{{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}-{\frac{5\,{e}^{4}\sqrt{dex+ce}}{21}\sqrt{{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}}+{\frac{5\,{e}^{4}}{21}\sqrt{1-{\frac{i \left ( dex+ce \right ) }{e}}}\sqrt{1+{\frac{i \left ( dex+ce \right ) }{e}}}{\it EllipticF} \left ( \sqrt{dex+ce}\sqrt{{\frac{i}{e}}},i \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)^(5/2)*(a+b*arcsinh(d*x+c)),x)

[Out]

2/d/e*(1/7*(d*e*x+c*e)^(7/2)*a+b*(1/7*(d*e*x+c*e)^(7/2)*arcsinh(1/e*(d*e*x+c*e))-2/7/e*(1/7*e^2*(d*e*x+c*e)^(5
/2)*(1/e^2*(d*e*x+c*e)^2+1)^(1/2)-5/21*e^4*(d*e*x+c*e)^(1/2)*(1/e^2*(d*e*x+c*e)^2+1)^(1/2)+5/21*e^4/(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))))

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d e x + c e\right )}^{\frac{5}{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)^(5/2)*(a+b*arcsinh(d*x+c)),x, algorithm="giac")

[Out]

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