\(\int \frac {(d+c^2 d x^2)^{5/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx\) [141]

Optimal result

Integrand size = 26, antiderivative size = 257 \[ \int \frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx=-\frac {9 b c^3 d^2 x^2 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}-\frac {b c^5 d^2 x^4 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}+\frac {15}{8} c^2 d^2 x \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))+\frac {5}{4} c^2 d x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))-\frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x}+\frac {15 c d^2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2}{16 b \sqrt {1+c^2 x^2}}+\frac {b c d^2 \sqrt {d+c^2 d x^2} \log (x)}{\sqrt {1+c^2 x^2}} \]

[Out]

5/4*c^2*d*x*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))-(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))/x+15/8*c^2*d^2*x*(a+
b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)-9/16*b*c^3*d^2*x^2*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-1/16*b*c^5*d^2*x^
4*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)+15/16*c*d^2*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/b/(c^2*x^2+1)^(1/
2)+b*c*d^2*ln(x)*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5807, 5786, 5785, 5783, 30, 14, 272, 45} \[ \int \frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {15}{8} c^2 d^2 x \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))+\frac {15 c d^2 \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2}{16 b \sqrt {c^2 x^2+1}}+\frac {5}{4} c^2 d x \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))-\frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))}{x}+\frac {b c d^2 \log (x) \sqrt {c^2 d x^2+d}}{\sqrt {c^2 x^2+1}}-\frac {b c^5 d^2 x^4 \sqrt {c^2 d x^2+d}}{16 \sqrt {c^2 x^2+1}}-\frac {9 b c^3 d^2 x^2 \sqrt {c^2 d x^2+d}}{16 \sqrt {c^2 x^2+1}} \]

[In]

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

[Out]

(-9*b*c^3*d^2*x^2*Sqrt[d + c^2*d*x^2])/(16*Sqrt[1 + c^2*x^2]) - (b*c^5*d^2*x^4*Sqrt[d + c^2*d*x^2])/(16*Sqrt[1
+ c^2*x^2]) + (15*c^2*d^2*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/8 + (5*c^2*d*x*(d + c^2*d*x^2)^(3/2)*(a
+ b*ArcSinh[c*x]))/4 - ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x + (15*c*d^2*Sqrt[d + c^2*d*x^2]*(a + b*
ArcSinh[c*x])^2)/(16*b*Sqrt[1 + c^2*x^2]) + (b*c*d^2*Sqrt[d + c^2*d*x^2]*Log[x])/Sqrt[1 + c^2*x^2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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 5783

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

Rule 5785

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

Rule 5786

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

Rule 5807

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*((a + b*ArcSinh[c*x])^n/(f*(m + 1))), x] + (-Dist[2*e*(p/(f^2*(m + 1))), Int[(f*x
)^(m + 2)*(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*c*(n/(f*(m + 1)))*Simp[(d + e*x^2)^p/(1
+ c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b,
c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x}+\left (5 c^2 d\right ) \int \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx+\frac {\left (b c d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {\left (1+c^2 x^2\right )^2}{x} \, dx}{\sqrt {1+c^2 x^2}} \\ & = \frac {5}{4} c^2 d x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))-\frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x}+\frac {1}{4} \left (15 c^2 d^2\right ) \int \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x)) \, dx+\frac {\left (b c d^2 \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int \frac {\left (1+c^2 x\right )^2}{x} \, dx,x,x^2\right )}{2 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c^3 d^2 \sqrt {d+c^2 d x^2}\right ) \int x \left (1+c^2 x^2\right ) \, dx}{4 \sqrt {1+c^2 x^2}} \\ & = \frac {15}{8} c^2 d^2 x \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))+\frac {5}{4} c^2 d x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))-\frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x}+\frac {\left (b c d^2 \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int \left (2 c^2+\frac {1}{x}+c^4 x\right ) \, dx,x,x^2\right )}{2 \sqrt {1+c^2 x^2}}+\frac {\left (15 c^2 d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{8 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c^3 d^2 \sqrt {d+c^2 d x^2}\right ) \int \left (x+c^2 x^3\right ) \, dx}{4 \sqrt {1+c^2 x^2}}-\frac {\left (15 b c^3 d^2 \sqrt {d+c^2 d x^2}\right ) \int x \, dx}{8 \sqrt {1+c^2 x^2}} \\ & = -\frac {9 b c^3 d^2 x^2 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}-\frac {b c^5 d^2 x^4 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}+\frac {15}{8} c^2 d^2 x \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))+\frac {5}{4} c^2 d x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))-\frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x}+\frac {15 c d^2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2}{16 b \sqrt {1+c^2 x^2}}+\frac {b c d^2 \sqrt {d+c^2 d x^2} \log (x)}{\sqrt {1+c^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.05 \[ \int \frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {1}{128} d^2 \left (\frac {16 a \sqrt {d+c^2 d x^2} \left (-8+9 c^2 x^2+2 c^4 x^4\right )}{x}+\frac {64 b \sqrt {d+c^2 d x^2} \left (-2 \sqrt {1+c^2 x^2} \text {arcsinh}(c x)+c x \text {arcsinh}(c x)^2+2 c x \log (c x)\right )}{x \sqrt {1+c^2 x^2}}+240 a c \sqrt {d} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )+\frac {32 b c \sqrt {d+c^2 d x^2} (-\cosh (2 \text {arcsinh}(c x))+2 \text {arcsinh}(c x) (\text {arcsinh}(c x)+\sinh (2 \text {arcsinh}(c x))))}{\sqrt {1+c^2 x^2}}-\frac {b c \sqrt {d+c^2 d x^2} \left (8 \text {arcsinh}(c x)^2+\cosh (4 \text {arcsinh}(c x))-4 \text {arcsinh}(c x) \sinh (4 \text {arcsinh}(c x))\right )}{\sqrt {1+c^2 x^2}}\right ) \]

[In]

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

[Out]

(d^2*((16*a*Sqrt[d + c^2*d*x^2]*(-8 + 9*c^2*x^2 + 2*c^4*x^4))/x + (64*b*Sqrt[d + c^2*d*x^2]*(-2*Sqrt[1 + c^2*x
^2]*ArcSinh[c*x] + c*x*ArcSinh[c*x]^2 + 2*c*x*Log[c*x]))/(x*Sqrt[1 + c^2*x^2]) + 240*a*c*Sqrt[d]*Log[c*d*x + S
qrt[d]*Sqrt[d + c^2*d*x^2]] + (32*b*c*Sqrt[d + c^2*d*x^2]*(-Cosh[2*ArcSinh[c*x]] + 2*ArcSinh[c*x]*(ArcSinh[c*x
] + Sinh[2*ArcSinh[c*x]])))/Sqrt[1 + c^2*x^2] - (b*c*Sqrt[d + c^2*d*x^2]*(8*ArcSinh[c*x]^2 + Cosh[4*ArcSinh[c*
x]] - 4*ArcSinh[c*x]*Sinh[4*ArcSinh[c*x]]))/Sqrt[1 + c^2*x^2]))/128

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.11

method	result	size
default	\(-\frac {a \left (c^{2} d \,x^{2}+d \right )^{\frac {7}{2}}}{d x}+a \,c^{2} x \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}+\frac {5 \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} a \,c^{2} d x}{4}+\frac {15 a \,d^{2} \sqrt {c^{2} d \,x^{2}+d}\, c^{2} x}{8}+\frac {15 a \,c^{2} d^{3} \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{8 \sqrt {c^{2} d}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (32 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{4} c^{4}-8 c^{5} x^{5}+144 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{2} c^{2}-72 c^{3} x^{3}+120 \operatorname {arcsinh}\left (c x \right )^{2} x c -128 \,\operatorname {arcsinh}\left (c x \right ) c x +128 \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right ) x c -128 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}-33 c x \right ) d^{2}}{128 \sqrt {c^{2} x^{2}+1}\, x}\)	\(285\)
parts	\(-\frac {a \left (c^{2} d \,x^{2}+d \right )^{\frac {7}{2}}}{d x}+a \,c^{2} x \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}+\frac {5 \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} a \,c^{2} d x}{4}+\frac {15 a \,d^{2} \sqrt {c^{2} d \,x^{2}+d}\, c^{2} x}{8}+\frac {15 a \,c^{2} d^{3} \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{8 \sqrt {c^{2} d}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (32 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{4} c^{4}-8 c^{5} x^{5}+144 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{2} c^{2}-72 c^{3} x^{3}+120 \operatorname {arcsinh}\left (c x \right )^{2} x c -128 \,\operatorname {arcsinh}\left (c x \right ) c x +128 \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right ) x c -128 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}-33 c x \right ) d^{2}}{128 \sqrt {c^{2} x^{2}+1}\, x}\)	\(285\)

[In]

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

[Out]

-a/d/x*(c^2*d*x^2+d)^(7/2)+a*c^2*x*(c^2*d*x^2+d)^(5/2)+5/4*(c^2*d*x^2+d)^(3/2)*a*c^2*d*x+15/8*a*d^2*(c^2*d*x^2
+d)^(1/2)*c^2*x+15/8*a*c^2*d^3*ln(c^2*d*x/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)+1/128*b*(d*(c^2*x^2
+1))^(1/2)/(c^2*x^2+1)^(1/2)/x*(32*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^4*c^4-8*c^5*x^5+144*arcsinh(c*x)*(c^2*x^2+
1)^(1/2)*x^2*c^2-72*c^3*x^3+120*arcsinh(c*x)^2*x*c-128*arcsinh(c*x)*c*x+128*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)*x*
c-128*arcsinh(c*x)*(c^2*x^2+1)^(1/2)-33*c*x)*d^2

Fricas [F]

\[ \int \frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx=\int { \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{x^{2}} \,d x } \]

[In]

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

[Out]

integral((a*c^4*d^2*x^4 + 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 + 2*b*c^2*d^2*x^2 + b*d^2)*arcsinh(c*x))*sq
rt(c^2*d*x^2 + d)/x^2, x)

Sympy [F]

\[ \int \frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx=\int \frac {\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{x^{2}}\, dx \]

[In]

integrate((c**2*d*x**2+d)**(5/2)*(a+b*asinh(c*x))/x**2,x)

[Out]

Integral((d*(c**2*x**2 + 1))**(5/2)*(a + b*asinh(c*x))/x**2, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx=\int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^{5/2}}{x^2} \,d x \]

[In]

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

[Out]

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