3.228 \(\int (c e+d e x)^{7/2} (a+b \sinh ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=298 \[ \frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}-\frac {14 b e^{7/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{135 d \sqrt {(c+d x)^2+1}}+\frac {28 b e^{7/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{135 d \sqrt {(c+d x)^2+1}}-\frac {28 b e^3 \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{135 d (c+d x+1)}+\frac {28 b e^2 \sqrt {(c+d x)^2+1} (e (c+d x))^{3/2}}{405 d}-\frac {4 b \sqrt {(c+d x)^2+1} (e (c+d x))^{7/2}}{81 d} \]

[Out]

2/9*(e*(d*x+c))^(9/2)*(a+b*arcsinh(d*x+c))/d/e+28/405*b*e^2*(e*(d*x+c))^(3/2)*(1+(d*x+c)^2)^(1/2)/d-4/81*b*(e*
(d*x+c))^(7/2)*(1+(d*x+c)^2)^(1/2)/d-28/135*b*e^3*(e*(d*x+c))^(1/2)*(1+(d*x+c)^2)^(1/2)/d/(d*x+c+1)+28/135*b*e
^(7/2)*(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)))*E
llipticE(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)-14/135*b*e^(7/2)*(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 = 0.33, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}+\frac {28 b e^2 \sqrt {(c+d x)^2+1} (e (c+d x))^{3/2}}{405 d}-\frac {28 b e^3 \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{135 d (c+d x+1)}-\frac {14 b e^{7/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{135 d \sqrt {(c+d x)^2+1}}+\frac {28 b e^{7/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{135 d \sqrt {(c+d x)^2+1}}-\frac {4 b \sqrt {(c+d x)^2+1} (e (c+d x))^{7/2}}{81 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^(7/2)*(a + b*ArcSinh[c + d*x]),x]

[Out]

(28*b*e^2*(e*(c + d*x))^(3/2)*Sqrt[1 + (c + d*x)^2])/(405*d) - (4*b*(e*(c + d*x))^(7/2)*Sqrt[1 + (c + d*x)^2])
/(81*d) - (28*b*e^3*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(135*d*(1 + c + d*x)) + (2*(e*(c + d*x))^(9/2)*(a
+ b*ArcSinh[c + d*x]))/(9*d*e) + (28*b*e^(7/2)*(1 + c + d*x)*Sqrt[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*Elliptic
E[2*ArcTan[Sqrt[e*(c + d*x)]/Sqrt[e]], 1/2])/(135*d*Sqrt[1 + (c + d*x)^2]) - (14*b*e^(7/2)*(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])/(135*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 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 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 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]

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)^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int (e x)^{7/2} \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {(e x)^{9/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{9 d e}\\ &=-\frac {4 b (e (c+d x))^{7/2} \sqrt {1+(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}+\frac {(14 b e) \operatorname {Subst}\left (\int \frac {(e x)^{5/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{81 d}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{405 d}-\frac {4 b (e (c+d x))^{7/2} \sqrt {1+(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}-\frac {\left (14 b e^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {e x}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{135 d}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{405 d}-\frac {4 b (e (c+d x))^{7/2} \sqrt {1+(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}-\frac {\left (28 b e^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{135 d}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{405 d}-\frac {4 b (e (c+d x))^{7/2} \sqrt {1+(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}-\frac {\left (28 b e^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{135 d}+\frac {\left (28 b e^3\right ) \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 )}{135 d}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{405 d}-\frac {4 b (e (c+d x))^{7/2} \sqrt {1+(c+d x)^2}}{81 d}-\frac {28 b e^3 \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{135 d (1+c+d x)}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}+\frac {28 b e^{7/2} (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 )}{135 d \sqrt {1+(c+d x)^2}}-\frac {14 b e^{7/2} (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 )}{135 d \sqrt {1+(c+d x)^2}}\\ \end {align*}

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Mathematica [C]  time = 0.21, size = 113, normalized size = 0.38 \[ \frac {2 (e (c+d x))^{7/2} \left (45 a (c+d x)^3-14 b \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-(c+d x)^2\right )-10 b \sqrt {(c+d x)^2+1} (c+d x)^2+14 b \sqrt {(c+d x)^2+1}+45 b (c+d x)^3 \sinh ^{-1}(c+d x)\right )}{405 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^(7/2)*(a + b*ArcSinh[c + d*x]),x]

[Out]

(2*(e*(c + d*x))^(7/2)*(45*a*(c + d*x)^3 + 14*b*Sqrt[1 + (c + d*x)^2] - 10*b*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]
+ 45*b*(c + d*x)^3*ArcSinh[c + d*x] - 14*b*Hypergeometric2F1[1/2, 3/4, 7/4, -(c + d*x)^2]))/(405*d*(c + d*x)^
2)

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a d^{3} e^{3} x^{3} + 3 \, a c d^{2} e^{3} x^{2} + 3 \, a c^{2} d e^{3} x + a c^{3} e^{3} + {\left (b d^{3} e^{3} x^{3} + 3 \, b c d^{2} e^{3} x^{2} + 3 \, b c^{2} d e^{3} x + b c^{3} e^{3}\right )} \operatorname {arsinh}\left (d x + c\right )\right )} \sqrt {d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(7/2)*(a+b*arcsinh(d*x+c)),x, algorithm="fricas")

[Out]

integral((a*d^3*e^3*x^3 + 3*a*c*d^2*e^3*x^2 + 3*a*c^2*d*e^3*x + a*c^3*e^3 + (b*d^3*e^3*x^3 + 3*b*c*d^2*e^3*x^2
+ 3*b*c^2*d*e^3*x + b*c^3*e^3)*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 \[ \int {\left (d e x + c e\right )}^{\frac {7}{2}} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(7/2)*(a+b*arcsinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^(7/2)*(b*arcsinh(d*x + c) + a), x)

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maple [C]  time = 0.03, size = 238, normalized size = 0.80 \[ \frac {\frac {2 \left (d e x +c e \right )^{\frac {9}{2}} a}{9}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {9}{2}} \arcsinh \left (\frac {d e x +c e}{e}\right )}{9}-\frac {2 \left (\frac {e^{2} \left (d e x +c e \right )^{\frac {7}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{9}-\frac {7 e^{4} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{45}+\frac {7 i e^{5} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\EllipticE \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{15 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{9 e}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^(7/2)*(a+b*arcsinh(d*x+c)),x)

[Out]

2/d/e*(1/9*(d*e*x+c*e)^(9/2)*a+b*(1/9*(d*e*x+c*e)^(9/2)*arcsinh((d*e*x+c*e)/e)-2/9/e*(1/9*e^2*(d*e*x+c*e)^(7/2
)*((d*e*x+c*e)^2/e^2+1)^(1/2)-7/45*e^4*(d*e*x+c*e)^(3/2)*((d*e*x+c*e)^2/e^2+1)^(1/2)+7/15*I*e^5/(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)-EllipticE((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 \[ \frac {2 \, {\left (d e x + c e\right )}^{\frac {9}{2}} a}{9 \, d e} + \frac {1}{810} \, {\left (\frac {180 \, {\left (d^{4} e^{\frac {7}{2}} x^{4} + 4 \, c d^{3} e^{\frac {7}{2}} x^{3} + 6 \, c^{2} d^{2} e^{\frac {7}{2}} x^{2} + 4 \, c^{3} d e^{\frac {7}{2}} x + c^{4} e^{\frac {7}{2}}\right )} \sqrt {d x + c} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{d} - \frac {40 \, {\left (d x + c\right )}^{\frac {9}{2}} e^{\frac {7}{2}} - 72 \, {\left (d x + c\right )}^{\frac {5}{2}} e^{\frac {7}{2}} + 360 \, \sqrt {d x + c} e^{\frac {7}{2}} + 45 \, {\left (i \, \sqrt {2} e^{3} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right )\right )} - i \, \sqrt {2} e^{3} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right )\right )} - \sqrt {2} e^{3} \log \left (d x + \sqrt {2} \sqrt {d x + c} + c + 1\right ) + \sqrt {2} e^{3} \log \left (d x - \sqrt {2} \sqrt {d x + c} + c + 1\right )\right )} \sqrt {e}}{d} - 810 \, \int \frac {2 \, {\left (d^{4} e^{\frac {7}{2}} x^{4} + 4 \, c d^{3} e^{\frac {7}{2}} x^{3} + 6 \, c^{2} d^{2} e^{\frac {7}{2}} x^{2} + 4 \, c^{3} d e^{\frac {7}{2}} x + c^{4} e^{\frac {7}{2}}\right )} \sqrt {d x + c}}{9 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + c^{3} + {\left (3 \, c^{2} d + d\right )} x + {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}^{\frac {3}{2}} + c\right )}}\,{d x}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(7/2)*(a+b*arcsinh(d*x+c)),x, algorithm="maxima")

[Out]

2/9*(d*e*x + c*e)^(9/2)*a/(d*e) + 1/810*(180*(d^4*e^(7/2)*x^4 + 4*c*d^3*e^(7/2)*x^3 + 6*c^2*d^2*e^(7/2)*x^2 +
4*c^3*d*e^(7/2)*x + c^4*e^(7/2))*sqrt(d*x + c)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/d - (40*(d*x +
c)^(9/2)*e^(7/2) - 72*(d*x + c)^(5/2)*e^(7/2) + 360*sqrt(d*x + c)*e^(7/2) + 45*(I*sqrt(2)*e^3*(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)*e^3*(l
og(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)) - sqr
t(2)*e^3*log(d*x + sqrt(2)*sqrt(d*x + c) + c + 1) + sqrt(2)*e^3*log(d*x - sqrt(2)*sqrt(d*x + c) + c + 1))*sqrt
(e))/d - 810*integrate(2/9*(d^4*e^(7/2)*x^4 + 4*c*d^3*e^(7/2)*x^3 + 6*c^2*d^2*e^(7/2)*x^2 + 4*c^3*d*e^(7/2)*x
+ c^4*e^(7/2))*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 = -0.00 \[ \int {\left (c\,e+d\,e\,x\right )}^{7/2}\,\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^(7/2)*(a + b*asinh(c + d*x)),x)

[Out]

int((c*e + d*e*x)^(7/2)*(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)**(7/2)*(a+b*asinh(d*x+c)),x)

[Out]

Timed out

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