Optimal. Leaf size=251 \[ -\frac {\sqrt {\pi } a^2 \text {erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } a^2 \text {erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b^3}-\frac {\sqrt {\pi } a \text {erfi}\left (1-\cosh ^{-1}(a+b x)\right )}{4 e b^3}-\frac {\sqrt {\pi } a \text {erfi}\left (\cosh ^{-1}(a+b x)+1\right )}{4 e b^3}-\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)-3\right )\right )}{16 e^{9/4} b^3}-\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)-1\right )\right )}{16 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)+1\right )\right )}{16 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)+3\right )\right )}{16 e^{9/4} b^3} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.49, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5899, 6741, 12, 6742, 5512, 2234, 2204, 5514} \[ -\frac {\sqrt {\pi } a^2 \text {Erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } a^2 \text {Erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b^3}-\frac {\sqrt {\pi } a \text {Erfi}\left (1-\cosh ^{-1}(a+b x)\right )}{4 e b^3}-\frac {\sqrt {\pi } a \text {Erfi}\left (\cosh ^{-1}(a+b x)+1\right )}{4 e b^3}-\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)-3\right )\right )}{16 e^{9/4} b^3}-\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)-1\right )\right )}{16 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)+1\right )\right )}{16 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)+3\right )\right )}{16 e^{9/4} b^3} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 12
Rule 2204
Rule 2234
Rule 5512
Rule 5514
Rule 5899
Rule 6741
Rule 6742
Rubi steps
\begin {align*} \int e^{\cosh ^{-1}(a+b x)^2} x^2 \, dx &=\frac {\operatorname {Subst}\left (\int e^{x^2} \left (-\frac {a}{b}+\frac {\cosh (x)}{b}\right )^2 \sinh (x) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {e^{x^2} (a-\cosh (x))^2 \sinh (x)}{b^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} (a-\cosh (x))^2 \sinh (x) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^2 e^{x^2} \sinh (x)-2 a e^{x^2} \cosh (x) \sinh (x)+e^{x^2} \cosh ^2(x) \sinh (x)\right ) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cosh ^2(x) \sinh (x) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int e^{x^2} \cosh (x) \sinh (x) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \operatorname {Subst}\left (\int e^{x^2} \sinh (x) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{8} e^{-3 x+x^2}-\frac {1}{8} e^{-x+x^2}+\frac {e^{x+x^2}}{8}+\frac {1}{8} e^{3 x+x^2}\right ) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int \left (-\frac {1}{4} e^{-2 x+x^2}+\frac {1}{4} e^{2 x+x^2}\right ) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \operatorname {Subst}\left (\int \left (-\frac {1}{2} e^{-x+x^2}+\frac {e^{x+x^2}}{2}\right ) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b^3}\\ &=-\frac {\operatorname {Subst}\left (\int e^{-3 x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{8 b^3}-\frac {\operatorname {Subst}\left (\int e^{-x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{8 b^3}+\frac {\operatorname {Subst}\left (\int e^{x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{8 b^3}+\frac {\operatorname {Subst}\left (\int e^{3 x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{8 b^3}+\frac {a \operatorname {Subst}\left (\int e^{-2 x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{2 b^3}-\frac {a \operatorname {Subst}\left (\int e^{2 x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{2 b^3}-\frac {a^2 \operatorname {Subst}\left (\int e^{-x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{2 b^3}+\frac {a^2 \operatorname {Subst}\left (\int e^{x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{2 b^3}\\ &=-\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (-3+2 x)^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{8 b^3 e^{9/4}}+\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (3+2 x)^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{8 b^3 e^{9/4}}+\frac {a \operatorname {Subst}\left (\int e^{\frac {1}{4} (-2+2 x)^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{2 b^3 e}-\frac {a \operatorname {Subst}\left (\int e^{\frac {1}{4} (2+2 x)^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{2 b^3 e}-\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (-1+2 x)^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{8 b^3 \sqrt [4]{e}}+\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (1+2 x)^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{8 b^3 \sqrt [4]{e}}-\frac {a^2 \operatorname {Subst}\left (\int e^{\frac {1}{4} (-1+2 x)^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{2 b^3 \sqrt [4]{e}}+\frac {a^2 \operatorname {Subst}\left (\int e^{\frac {1}{4} (1+2 x)^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{2 b^3 \sqrt [4]{e}}\\ &=-\frac {a \sqrt {\pi } \text {erfi}\left (1-\cosh ^{-1}(a+b x)\right )}{4 b^3 e}-\frac {a \sqrt {\pi } \text {erfi}\left (1+\cosh ^{-1}(a+b x)\right )}{4 b^3 e}-\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-3+2 \cosh ^{-1}(a+b x)\right )\right )}{16 b^3 e^{9/4}}-\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-1+2 \cosh ^{-1}(a+b x)\right )\right )}{16 b^3 \sqrt [4]{e}}-\frac {a^2 \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-1+2 \cosh ^{-1}(a+b x)\right )\right )}{4 b^3 \sqrt [4]{e}}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (1+2 \cosh ^{-1}(a+b x)\right )\right )}{16 b^3 \sqrt [4]{e}}+\frac {a^2 \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (1+2 \cosh ^{-1}(a+b x)\right )\right )}{4 b^3 \sqrt [4]{e}}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (3+2 \cosh ^{-1}(a+b x)\right )\right )}{16 b^3 e^{9/4}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.23, size = 136, normalized size = 0.54 \[ \frac {\sqrt {\pi } \left (4 e^2 a^2 \text {erfi}\left (\cosh ^{-1}(a+b x)+\frac {1}{2}\right )+e^2 \left (4 a^2+1\right ) \text {erfi}\left (\frac {1}{2}-\cosh ^{-1}(a+b x)\right )-4 e^{5/4} a \text {erfi}\left (1-\cosh ^{-1}(a+b x)\right )-4 e^{5/4} a \text {erfi}\left (\cosh ^{-1}(a+b x)+1\right )+\text {erfi}\left (\frac {3}{2}-\cosh ^{-1}(a+b x)\right )+e^2 \text {erfi}\left (\cosh ^{-1}(a+b x)+\frac {1}{2}\right )+\text {erfi}\left (\cosh ^{-1}(a+b x)+\frac {3}{2}\right )\right )}{16 e^{9/4} b^3} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} e^{\left (\operatorname {arcosh}\left (b x + a\right )^{2}\right )}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} e^{\left (\operatorname {arcosh}\left (b x + a\right )^{2}\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 0.01, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{\mathrm {arccosh}\left (b x +a \right )^{2}} x^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} e^{\left (\operatorname {arcosh}\left (b x + a\right )^{2}\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\mathrm {e}}^{{\mathrm {acosh}\left (a+b\,x\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} e^{\operatorname {acosh}^{2}{\left (a + b x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________