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

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

[Out]

2/7*(e*(d*x+c))^(7/2)*(a+b*arcsinh(d*x+c))/d/e-4/49*b*(e*(d*x+c))^(5/2)*(1+(d*x+c)^2)^(1/2)/d+20/147*b*e^2*(e*
(d*x+c))^(1/2)*(1+(d*x+c)^2)^(1/2)/d-10/147*b*e^(5/2)*(d*x+c+1)*(cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))^2)^(
1/2)/cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))*EllipticF(sin(2*arctan((e*(d*x+c))^(1/2)/e^(1/2))),1/2*2^(1/2))*
((1+(d*x+c)^2)/(d*x+c+1)^2)^(1/2)/d/(1+(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 177, normalized size of antiderivative = 1.00, 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 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 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 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 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)^{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*}

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Mathematica [C]  time = 0.19, size = 113, normalized size = 0.64 \[ \frac {2 (e (c+d x))^{5/2} \left (21 a (c+d x)^3-10 b \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-(c+d x)^2\right )-6 b \sqrt {(c+d x)^2+1} (c+d x)^2+10 b \sqrt {(c+d x)^2+1}+21 b (c+d x)^3 \sinh ^{-1}(c+d x)\right )}{147 d (c+d x)^2} \]

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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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a d^{2} e^{2} x^{2} + 2 \, a c d e^{2} x + a c^{2} e^{2} + {\left (b d^{2} e^{2} x^{2} + 2 \, b c d e^{2} x + b c^{2} e^{2}\right )} \operatorname {arsinh}\left (d x + c\right )\right )} \sqrt {d e x + c e}, x\right ) \]

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

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)

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maple [C]  time = 0.01, size = 212, normalized size = 1.20 \[ \frac {\frac {2 \left (d e x +c e \right )^{\frac {7}{2}} a}{7}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {7}{2}} \arcsinh \left (\frac {d e x +c e}{e}\right )}{7}-\frac {2 \left (\frac {e^{2} \left (d e x +c e \right )^{\frac {5}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{7}-\frac {5 e^{4} \sqrt {d e x +c e}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{21}+\frac {5 e^{4} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{21 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{7 e}\right )}{d e} \]

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((d*e*x+c*e)/e)-2/7/e*(1/7*e^2*(d*e*x+c*e)^(5/2
)*((d*e*x+c*e)^2/e^2+1)^(1/2)-5/21*e^4*(d*e*x+c*e)^(1/2)*((d*e*x+c*e)^2/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)/((d*e*x+c*e)^2/e^2+1)^(1/2)*EllipticF((d*e*x+c*e)^(1/2)*(I/e)^
(1/2),I))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, {\left (d e x + c e\right )}^{\frac {7}{2}} a}{7 \, d e} + \frac {1}{294} \, {\left (\frac {84 \, {\left (d^{3} e^{\frac {5}{2}} x^{3} + 3 \, c d^{2} e^{\frac {5}{2}} x^{2} + 3 \, c^{2} d e^{\frac {5}{2}} x + c^{3} e^{\frac {5}{2}}\right )} \sqrt {d x + c} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{d} - \frac {24 \, {\left (d x + c\right )}^{\frac {7}{2}} e^{\frac {5}{2}} - 56 \, {\left (d x + c\right )}^{\frac {3}{2}} e^{\frac {5}{2}} - 21 \, {\left (i \, \sqrt {2} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right )\right )} - i \, \sqrt {2} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right )\right )} + \sqrt {2} \log \left (d x + \sqrt {2} \sqrt {d x + c} + c + 1\right ) - \sqrt {2} \log \left (d x - \sqrt {2} \sqrt {d x + c} + c + 1\right )\right )} e^{\frac {5}{2}}}{d} - 294 \, \int \frac {2 \, {\left (d^{3} e^{\frac {5}{2}} x^{3} + 3 \, c d^{2} e^{\frac {5}{2}} x^{2} + 3 \, c^{2} d e^{\frac {5}{2}} x + c^{3} e^{\frac {5}{2}}\right )} \sqrt {d x + c}}{7 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + c^{3} + {\left (3 \, c^{2} d + d\right )} x + {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}^{\frac {3}{2}} + c\right )}}\,{d x}\right )} b \]

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]

2/7*(d*e*x + c*e)^(7/2)*a/(d*e) + 1/294*(84*(d^3*e^(5/2)*x^3 + 3*c*d^2*e^(5/2)*x^2 + 3*c^2*d*e^(5/2)*x + 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))/d - (24*(d*x + c)^(7/2)*e^(5/2) - 56*(d
*x + c)^(3/2)*e^(5/2) - 21*(I*sqrt(2)*(log(1/2*I*sqrt(2)*(sqrt(2) + 2*sqrt(d*x + c)) + 1) - log(-1/2*I*sqrt(2)
*(sqrt(2) + 2*sqrt(d*x + c)) + 1)) - I*sqrt(2)*(log(1/2*I*sqrt(2)*(sqrt(2) - 2*sqrt(d*x + c)) + 1) - log(-1/2*
I*sqrt(2)*(sqrt(2) - 2*sqrt(d*x + c)) + 1)) + sqrt(2)*log(d*x + sqrt(2)*sqrt(d*x + c) + c + 1) - sqrt(2)*log(d
*x - sqrt(2)*sqrt(d*x + c) + c + 1))*e^(5/2))/d - 294*integrate(2/7*(d^3*e^(5/2)*x^3 + 3*c*d^2*e^(5/2)*x^2 + 3
*c^2*d*e^(5/2)*x + c^3*e^(5/2))*sqrt(d*x + c)/(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))*b

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

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)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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