\(\int x^m (d+c^2 d x^2)^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx\) [322]

Optimal result

Integrand size = 28, antiderivative size = 28 \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {2 b^2 c^2 d x^{3+m} \sqrt {d+c^2 d x^2}}{(4+m)^3}-\frac {6 b c d x^{2+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(2+m)^2 (4+m) \sqrt {1+c^2 x^2}}-\frac {2 b c d x^{2+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{\left (8+6 m+m^2\right ) \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^{4+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(4+m)^2 \sqrt {1+c^2 x^2}}+\frac {3 d x^{1+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2}{8+6 m+m^2}+\frac {x^{1+m} \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{4+m}+\frac {6 b^2 c^2 d x^{3+m} \sqrt {d+c^2 d x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},-c^2 x^2\right )}{(2+m)^2 (3+m) (4+m) \sqrt {1+c^2 x^2}}+\frac {2 b^2 c^2 d (10+3 m) x^{3+m} \sqrt {d+c^2 d x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},-c^2 x^2\right )}{(2+m) (3+m) (4+m)^3 \sqrt {1+c^2 x^2}}+\frac {3 d^2 \text {Int}\left (\frac {x^m (a+b \text {arcsinh}(c x))^2}{\sqrt {d+c^2 d x^2}},x\right )}{8+6 m+m^2} \]

[Out]

x^(1+m)*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2/(4+m)+2*b^2*c^2*d*x^(3+m)*(c^2*d*x^2+d)^(1/2)/(4+m)^3+3*d*x^(
1+m)*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/(m^2+6*m+8)-6*b*c*d*x^(2+m)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/
2)/(2+m)^2/(4+m)/(c^2*x^2+1)^(1/2)-2*b*c*d*x^(2+m)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(m^2+6*m+8)/(c^2*x^2
+1)^(1/2)-2*b*c^3*d*x^(4+m)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(4+m)^2/(c^2*x^2+1)^(1/2)+2*b^2*c^2*d*(10+3
*m)*x^(3+m)*hypergeom([1/2, 3/2+1/2*m],[5/2+1/2*m],-c^2*x^2)*(c^2*d*x^2+d)^(1/2)/(4+m)^3/(m^2+5*m+6)/(c^2*x^2+
1)^(1/2)+6*b^2*c^2*d*x^(3+m)*hypergeom([1/2, 3/2+1/2*m],[5/2+1/2*m],-c^2*x^2)*(c^2*d*x^2+d)^(1/2)/(2+m)^2/(m^2
+7*m+12)/(c^2*x^2+1)^(1/2)+3*d^2*Unintegrable(x^m*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(1/2),x)/(m^2+6*m+8)

Rubi [N/A]

Not integrable

Time = 0.42 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx \]

[In]

Int[x^m*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2,x]

[Out]

(2*b^2*c^2*d*x^(3 + m)*Sqrt[d + c^2*d*x^2])/(4 + m)^3 - (6*b*c*d*x^(2 + m)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[
c*x]))/((2 + m)^2*(4 + m)*Sqrt[1 + c^2*x^2]) - (2*b*c*d*x^(2 + m)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/((
8 + 6*m + m^2)*Sqrt[1 + c^2*x^2]) - (2*b*c^3*d*x^(4 + m)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/((4 + m)^2*
Sqrt[1 + c^2*x^2]) + (3*d*x^(1 + m)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(8 + 6*m + m^2) + (x^(1 + m)*(
d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2)/(4 + m) + (6*b^2*c^2*d*x^(3 + m)*Sqrt[d + c^2*d*x^2]*Hypergeometr
ic2F1[1/2, (3 + m)/2, (5 + m)/2, -(c^2*x^2)])/((2 + m)^2*(3 + m)*(4 + m)*Sqrt[1 + c^2*x^2]) + (2*b^2*c^2*d*(10
+ 3*m)*x^(3 + m)*Sqrt[d + c^2*d*x^2]*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, -(c^2*x^2)])/((2 + m)*(3 +
m)*(4 + m)^3*Sqrt[1 + c^2*x^2]) + (3*d^2*Defer[Int][(x^m*(a + b*ArcSinh[c*x])^2)/Sqrt[d + c^2*d*x^2], x])/(8 +
6*m + m^2)

Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{4+m}+\frac {(3 d) \int x^m \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2 \, dx}{4+m}-\frac {\left (2 b c d \sqrt {d+c^2 d x^2}\right ) \int x^{1+m} \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x)) \, dx}{(4+m) \sqrt {1+c^2 x^2}} \\ & = -\frac {2 b c d x^{2+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{\left (8+6 m+m^2\right ) \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^{4+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(4+m)^2 \sqrt {1+c^2 x^2}}+\frac {3 d x^{1+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2}{8+6 m+m^2}+\frac {x^{1+m} \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{4+m}+\frac {\left (3 d^2\right ) \int \frac {x^m (a+b \text {arcsinh}(c x))^2}{\sqrt {d+c^2 d x^2}} \, dx}{8+6 m+m^2}+\frac {\left (2 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x^{2+m} \left (\frac {1}{2+m}+\frac {c^2 x^2}{4+m}\right )}{\sqrt {1+c^2 x^2}} \, dx}{(4+m) \sqrt {1+c^2 x^2}}-\frac {\left (6 b c d \sqrt {d+c^2 d x^2}\right ) \int x^{1+m} (a+b \text {arcsinh}(c x)) \, dx}{(2+m) (4+m) \sqrt {1+c^2 x^2}} \\ & = \frac {2 b^2 c^2 d x^{3+m} \sqrt {d+c^2 d x^2}}{(4+m)^3}-\frac {6 b c d x^{2+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(2+m)^2 (4+m) \sqrt {1+c^2 x^2}}-\frac {2 b c d x^{2+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{\left (8+6 m+m^2\right ) \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^{4+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(4+m)^2 \sqrt {1+c^2 x^2}}+\frac {3 d x^{1+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2}{8+6 m+m^2}+\frac {x^{1+m} \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{4+m}+\frac {\left (3 d^2\right ) \int \frac {x^m (a+b \text {arcsinh}(c x))^2}{\sqrt {d+c^2 d x^2}} \, dx}{8+6 m+m^2}+\frac {\left (6 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x^{2+m}}{\sqrt {1+c^2 x^2}} \, dx}{(2+m)^2 (4+m) \sqrt {1+c^2 x^2}}+\frac {\left (2 b^2 c^2 d (10+3 m) \sqrt {d+c^2 d x^2}\right ) \int \frac {x^{2+m}}{\sqrt {1+c^2 x^2}} \, dx}{(2+m) (4+m)^3 \sqrt {1+c^2 x^2}} \\ & = \frac {2 b^2 c^2 d x^{3+m} \sqrt {d+c^2 d x^2}}{(4+m)^3}-\frac {6 b c d x^{2+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(2+m)^2 (4+m) \sqrt {1+c^2 x^2}}-\frac {2 b c d x^{2+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{\left (8+6 m+m^2\right ) \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^{4+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(4+m)^2 \sqrt {1+c^2 x^2}}+\frac {3 d x^{1+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2}{8+6 m+m^2}+\frac {x^{1+m} \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{4+m}+\frac {6 b^2 c^2 d x^{3+m} \sqrt {d+c^2 d x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},-c^2 x^2\right )}{(2+m)^2 (3+m) (4+m) \sqrt {1+c^2 x^2}}+\frac {2 b^2 c^2 d (10+3 m) x^{3+m} \sqrt {d+c^2 d x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},-c^2 x^2\right )}{(2+m) (3+m) (4+m)^3 \sqrt {1+c^2 x^2}}+\frac {\left (3 d^2\right ) \int \frac {x^m (a+b \text {arcsinh}(c x))^2}{\sqrt {d+c^2 d x^2}} \, dx}{8+6 m+m^2} \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 0.91 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx \]

[In]

Integrate[x^m*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2,x]

[Out]

Integrate[x^m*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2, x]

Maple [N/A] (verified)

Not integrable

Time = 0.85 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93

\[\int x^{m} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{2}d x\]

[In]

int(x^m*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x)

[Out]

int(x^m*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x)

Fricas [N/A]

Not integrable

Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.86 \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2} x^{m} \,d x } \]

[In]

integrate(x^m*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\text {Timed out} \]

[In]

integrate(x**m*(c**2*d*x**2+d)**(3/2)*(a+b*asinh(c*x))**2,x)

[Out]

Timed out

Maxima [N/A]

Not integrable

Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2} x^{m} \,d x } \]

[In]

integrate(x^m*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x, algorithm="maxima")

[Out]

integrate((c^2*d*x^2 + d)^(3/2)*(b*arcsinh(c*x) + a)^2*x^m, x)

Giac [F(-2)]

Exception generated. \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^m*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*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 [N/A]

Not integrable

Time = 3.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int x^m\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d\,c^2\,x^2+d\right )}^{3/2} \,d x \]

[In]

int(x^m*(a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(3/2),x)

[Out]

int(x^m*(a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(3/2), x)