3.175 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^4 (d+c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=297 \[ \frac {2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3 \left (c^2 d x^2+d\right )^{3/2}}+\frac {16 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {c^2 d x^2+d}}+\frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {b c \sqrt {c^2 d x^2+d}}{6 d^3 x^2 \sqrt {c^2 x^2+1}}+\frac {b c^3 \sqrt {c^2 d x^2+d}}{6 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {8 b c^3 \log (x) \sqrt {c^2 d x^2+d}}{3 d^3 \sqrt {c^2 x^2+1}}-\frac {4 b c^3 \sqrt {c^2 d x^2+d} \log \left (c^2 x^2+1\right )}{3 d^3 \sqrt {c^2 x^2+1}} \]

[Out]

1/3*(-a-b*arcsinh(c*x))/d/x^3/(c^2*d*x^2+d)^(3/2)+2*c^2*(a+b*arcsinh(c*x))/d/x/(c^2*d*x^2+d)^(3/2)+8/3*c^4*x*(
a+b*arcsinh(c*x))/d/(c^2*d*x^2+d)^(3/2)+16/3*c^4*x*(a+b*arcsinh(c*x))/d^2/(c^2*d*x^2+d)^(1/2)+1/6*b*c^3*(c^2*d
*x^2+d)^(1/2)/d^3/(c^2*x^2+1)^(3/2)-1/6*b*c*(c^2*d*x^2+d)^(1/2)/d^3/x^2/(c^2*x^2+1)^(1/2)-8/3*b*c^3*ln(x)*(c^2
*d*x^2+d)^(1/2)/d^3/(c^2*x^2+1)^(1/2)-4/3*b*c^3*ln(c^2*x^2+1)*(c^2*d*x^2+d)^(1/2)/d^3/(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {5747, 5690, 5687, 260, 261, 266, 44} \[ \frac {16 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {c^2 d x^2+d}}+\frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (c^2 d x^2+d\right )^{3/2}}+\frac {2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3 \left (c^2 d x^2+d\right )^{3/2}}+\frac {b c^3}{6 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}-\frac {b c \sqrt {c^2 x^2+1}}{6 d^2 x^2 \sqrt {c^2 d x^2+d}}-\frac {8 b c^3 \sqrt {c^2 x^2+1} \log (x)}{3 d^2 \sqrt {c^2 d x^2+d}}-\frac {4 b c^3 \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{3 d^2 \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^(5/2)),x]

[Out]

(b*c^3)/(6*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]) - (b*c*Sqrt[1 + c^2*x^2])/(6*d^2*x^2*Sqrt[d + c^2*d*x^2]
) - (a + b*ArcSinh[c*x])/(3*d*x^3*(d + c^2*d*x^2)^(3/2)) + (2*c^2*(a + b*ArcSinh[c*x]))/(d*x*(d + c^2*d*x^2)^(
3/2)) + (8*c^4*x*(a + b*ArcSinh[c*x]))/(3*d*(d + c^2*d*x^2)^(3/2)) + (16*c^4*x*(a + b*ArcSinh[c*x]))/(3*d^2*Sq
rt[d + c^2*d*x^2]) - (8*b*c^3*Sqrt[1 + c^2*x^2]*Log[x])/(3*d^2*Sqrt[d + c^2*d*x^2]) - (4*b*c^3*Sqrt[1 + c^2*x^
2]*Log[1 + c^2*x^2])/(3*d^2*Sqrt[d + c^2*d*x^2])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 261

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5687

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSinh
[c*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
c*x])^(n - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +
b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[p]), Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 5747

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] + (-Dist[(c^2*(m + 2*p + 3))/(f^2
*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x^4 \left (d+c^2 d x^2\right )^{5/2}} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3 \left (d+c^2 d x^2\right )^{3/2}}-\left (2 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x^2 \left (d+c^2 d x^2\right )^{5/2}} \, dx+\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{x^3 \left (1+c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3 \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x \left (d+c^2 d x^2\right )^{3/2}}+\left (8 c^4\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^{5/2}} \, dx+\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (2 b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{x \left (1+c^2 x^2\right )^2} \, dx}{d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3 \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x \left (d+c^2 d x^2\right )^{3/2}}+\frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {\left (16 c^4\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^{3/2}} \, dx}{3 d}+\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x^2}-\frac {2 c^2}{x}+\frac {c^4}{\left (1+c^2 x\right )^2}+\frac {2 c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )}{6 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (b c^3 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )^2} \, dx,x,x^2\right )}{d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (8 b c^5 \sqrt {1+c^2 x^2}\right ) \int \frac {x}{\left (1+c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {7 b c^3}{6 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {b c \sqrt {1+c^2 x^2}}{6 d^2 x^2 \sqrt {d+c^2 d x^2}}-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3 \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x \left (d+c^2 d x^2\right )^{3/2}}+\frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {2 b c^3 \sqrt {1+c^2 x^2} \log (x)}{3 d^2 \sqrt {d+c^2 d x^2}}+\frac {b c^3 \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (b c^3 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x}-\frac {c^2}{\left (1+c^2 x\right )^2}-\frac {c^2}{1+c^2 x}\right ) \, dx,x,x^2\right )}{d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (16 b c^5 \sqrt {1+c^2 x^2}\right ) \int \frac {x}{1+c^2 x^2} \, dx}{3 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {b c^3}{6 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {b c \sqrt {1+c^2 x^2}}{6 d^2 x^2 \sqrt {d+c^2 d x^2}}-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3 \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x \left (d+c^2 d x^2\right )^{3/2}}+\frac {8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {8 b c^3 \sqrt {1+c^2 x^2} \log (x)}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {4 b c^3 \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{3 d^2 \sqrt {d+c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 267, normalized size = 0.90 \[ \frac {\sqrt {c^2 d x^2+d} \left (12 a c^2 x^2 \sqrt {c^2 x^2+1}-2 a \sqrt {c^2 x^2+1}+32 a c^6 x^6 \sqrt {c^2 x^2+1}+48 a c^4 x^4 \sqrt {c^2 x^2+1}-b c^3 x^3-16 b c^7 x^7 \log \left (c^2 x^2+1\right )-32 b c^5 x^5 \log \left (c^2 x^2+1\right )+8 b c^3 x^3 \left (c^2 x^2+1\right )^2 \log \left (\frac {1}{c^2 x^2}+1\right )-16 b c^3 x^3 \log \left (c^2 x^2+1\right )+2 b \sqrt {c^2 x^2+1} \left (16 c^6 x^6+24 c^4 x^4+6 c^2 x^2-1\right ) \sinh ^{-1}(c x)-b c x\right )}{6 d^3 x^3 \left (c^2 x^2+1\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^(5/2)),x]

[Out]

(Sqrt[d + c^2*d*x^2]*(-(b*c*x) - b*c^3*x^3 - 2*a*Sqrt[1 + c^2*x^2] + 12*a*c^2*x^2*Sqrt[1 + c^2*x^2] + 48*a*c^4
*x^4*Sqrt[1 + c^2*x^2] + 32*a*c^6*x^6*Sqrt[1 + c^2*x^2] + 2*b*Sqrt[1 + c^2*x^2]*(-1 + 6*c^2*x^2 + 24*c^4*x^4 +
16*c^6*x^6)*ArcSinh[c*x] + 8*b*c^3*x^3*(1 + c^2*x^2)^2*Log[1 + 1/(c^2*x^2)] - 16*b*c^3*x^3*Log[1 + c^2*x^2] -
32*b*c^5*x^5*Log[1 + c^2*x^2] - 16*b*c^7*x^7*Log[1 + c^2*x^2]))/(6*d^3*x^3*(1 + c^2*x^2)^(5/2))

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fricas [F]  time = 1.10, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c^{2} d x^{2} + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{c^{6} d^{3} x^{10} + 3 \, c^{4} d^{3} x^{8} + 3 \, c^{2} d^{3} x^{6} + d^{3} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)/(c^6*d^3*x^10 + 3*c^4*d^3*x^8 + 3*c^2*d^3*x^6 + d^3*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/((c^2*d*x^2 + d)^(5/2)*x^4), x)

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maple [B]  time = 0.29, size = 1790, normalized size = 6.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(5/2),x)

[Out]

32/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^3*arcsinh(c*x)*c^3+2*a*c^2/d/x/(c^2*d*x^2+d)^(3/2)-1/3*a/d/x^
3/(c^2*d*x^2+d)^(3/2)-64*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x^6*arcsi
nh(c*x)*(c^2*x^2+1)^(1/2)*c^9-128*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*
x^4*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^7-176/3*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*
x^2-1)/d^3*x^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^5-128/3*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*
x^4+10*c^2*x^2-1)/d^3*x^9*(c^2*x^2+1)*c^12-320/3*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*
c^2*x^2-1)/d^3*x^7*(c^2*x^2+1)*c^10+64*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)
/d^3*x^7*arcsinh(c*x)*c^10-80*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x^5*
(c^2*x^2+1)*c^8+160*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x^5*arcsinh(c*
x)*c^8-40/3*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x^3*(c^2*x^2+1)*c^6+34
4/3*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x^3*arcsinh(c*x)*c^6-2*b*(d*(c
^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x^2*c^5*(c^2*x^2+1)^(1/2)+16/3*b*(d*(c^2*
x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^3+8/3*b*(d*
(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x*(c^2*x^2+1)*c^4+12*b*(d*(c^2*x^2+1))^
(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x*arcsinh(c*x)*c^4-6*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8
*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3/x*arcsinh(c*x)*c^2+1/6*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6
*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3/x^2*c*(c^2*x^2+1)^(1/2)-2*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*
c^4*x^4+10*c^2*x^2-1)/d^3*c^3*(c^2*x^2+1)^(1/2)+128/3*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^
4+10*c^2*x^2-1)/d^3*x^11*c^14+448/3*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^
3*x^9*c^12+560/3*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x^7*c^10+280/3*b*
(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x^5*c^8+32/3*b*(d*(c^2*x^2+1))^(1/2)
/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x^3*c^6-8/3*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^
6+35*c^4*x^4+10*c^2*x^2-1)/d^3*x*c^4+1/3*b*(d*(c^2*x^2+1))^(1/2)/(12*c^8*x^8+36*c^6*x^6+35*c^4*x^4+10*c^2*x^2-
1)/d^3/x^3*arcsinh(c*x)-8/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^3*ln((c*x+(c^2*x^2+1)^(1/2))^4-1)*c^3+
16/3*a*c^4/d^2*x/(c^2*d*x^2+d)^(1/2)+8/3*a*c^4*x/d/(c^2*d*x^2+d)^(3/2)

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maxima [A]  time = 0.44, size = 236, normalized size = 0.79 \[ -\frac {1}{6} \, b c {\left (\frac {8 \, c^{2} \log \left (c^{2} x^{2} + 1\right )}{d^{\frac {5}{2}}} + \frac {16 \, c^{2} \log \relax (x)}{d^{\frac {5}{2}}} + \frac {1}{c^{2} d^{\frac {5}{2}} x^{4} + d^{\frac {5}{2}} x^{2}}\right )} + \frac {1}{3} \, {\left (\frac {16 \, c^{4} x}{\sqrt {c^{2} d x^{2} + d} d^{2}} + \frac {8 \, c^{4} x}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d} + \frac {6 \, c^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x} - \frac {1}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x^{3}}\right )} b \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, {\left (\frac {16 \, c^{4} x}{\sqrt {c^{2} d x^{2} + d} d^{2}} + \frac {8 \, c^{4} x}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d} + \frac {6 \, c^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x} - \frac {1}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x^{3}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/6*b*c*(8*c^2*log(c^2*x^2 + 1)/d^(5/2) + 16*c^2*log(x)/d^(5/2) + 1/(c^2*d^(5/2)*x^4 + d^(5/2)*x^2)) + 1/3*(1
6*c^4*x/(sqrt(c^2*d*x^2 + d)*d^2) + 8*c^4*x/((c^2*d*x^2 + d)^(3/2)*d) + 6*c^2/((c^2*d*x^2 + d)^(3/2)*d*x) - 1/
((c^2*d*x^2 + d)^(3/2)*d*x^3))*b*arcsinh(c*x) + 1/3*(16*c^4*x/(sqrt(c^2*d*x^2 + d)*d^2) + 8*c^4*x/((c^2*d*x^2
+ d)^(3/2)*d) + 6*c^2/((c^2*d*x^2 + d)^(3/2)*d*x) - 1/((c^2*d*x^2 + d)^(3/2)*d*x^3))*a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x^4\,{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/(x^4*(d + c^2*d*x^2)^(5/2)),x)

[Out]

int((a + b*asinh(c*x))/(x^4*(d + c^2*d*x^2)^(5/2)), 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((a+b*asinh(c*x))/x**4/(c**2*d*x**2+d)**(5/2),x)

[Out]

Timed out

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