Integrand size = 16, antiderivative size = 136 \[ \int x \cosh ^2\left (a+b x-c x^2\right ) \, dx=\frac {x^2}{4}-\frac {b e^{2 a+\frac {b^2}{2 c}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {b-2 c x}{\sqrt {2} \sqrt {c}}\right )}{16 c^{3/2}}-\frac {b e^{-2 a-\frac {b^2}{2 c}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {b-2 c x}{\sqrt {2} \sqrt {c}}\right )}{16 c^{3/2}}-\frac {\sinh \left (2 a+2 b x-2 c x^2\right )}{8 c} \] Output:
1/4*x^2-1/32*b*exp(2*a+1/2*b^2/c)*2^(1/2)*Pi^(1/2)*erf(1/2*(-2*c*x+b)*2^(1 /2)/c^(1/2))/c^(3/2)-1/32*b*exp(-2*a-1/2*b^2/c)*2^(1/2)*Pi^(1/2)*erfi(1/2* (-2*c*x+b)*2^(1/2)/c^(1/2))/c^(3/2)-1/8*sinh(-2*c*x^2+2*b*x+2*a)/c
Time = 0.27 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.18 \[ \int x \cosh ^2\left (a+b x-c x^2\right ) \, dx=\frac {b \sqrt {2 \pi } \text {erfi}\left (\frac {-b+2 c x}{\sqrt {2} \sqrt {c}}\right ) \left (\cosh \left (2 a+\frac {b^2}{2 c}\right )-\sinh \left (2 a+\frac {b^2}{2 c}\right )\right )+b \sqrt {2 \pi } \text {erf}\left (\frac {-b+2 c x}{\sqrt {2} \sqrt {c}}\right ) \left (\cosh \left (2 a+\frac {b^2}{2 c}\right )+\sinh \left (2 a+\frac {b^2}{2 c}\right )\right )+4 \sqrt {c} \left (2 c x^2-\sinh (2 (a+x (b-c x)))\right )}{32 c^{3/2}} \] Input:
Integrate[x*Cosh[a + b*x - c*x^2]^2,x]
Output:
(b*Sqrt[2*Pi]*Erfi[(-b + 2*c*x)/(Sqrt[2]*Sqrt[c])]*(Cosh[2*a + b^2/(2*c)] - Sinh[2*a + b^2/(2*c)]) + b*Sqrt[2*Pi]*Erf[(-b + 2*c*x)/(Sqrt[2]*Sqrt[c]) ]*(Cosh[2*a + b^2/(2*c)] + Sinh[2*a + b^2/(2*c)]) + 4*Sqrt[c]*(2*c*x^2 - S inh[2*(a + x*(b - c*x))]))/(32*c^(3/2))
Time = 0.31 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5918, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x \cosh ^2\left (a+b x-c x^2\right ) \, dx\) |
| \(\Big \downarrow \) 5918 |
| \(\displaystyle \int \left (\frac {1}{2} x \cosh \left (2 a+2 b x-2 c x^2\right )+\frac {x}{2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt {\frac {\pi }{2}} b e^{2 a+\frac {b^2}{2 c}} \text {erf}\left (\frac {b-2 c x}{\sqrt {2} \sqrt {c}}\right )}{16 c^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} b e^{-2 a-\frac {b^2}{2 c}} \text {erfi}\left (\frac {b-2 c x}{\sqrt {2} \sqrt {c}}\right )}{16 c^{3/2}}-\frac {\sinh \left (2 a+2 b x-2 c x^2\right )}{8 c}+\frac {x^2}{4}\) |
Input:
Int[x*Cosh[a + b*x - c*x^2]^2,x]
Output:
x^2/4 - (b*E^(2*a + b^2/(2*c))*Sqrt[Pi/2]*Erf[(b - 2*c*x)/(Sqrt[2]*Sqrt[c] )])/(16*c^(3/2)) - (b*E^(-2*a - b^2/(2*c))*Sqrt[Pi/2]*Erfi[(b - 2*c*x)/(Sq rt[2]*Sqrt[c])])/(16*c^(3/2)) - Sinh[2*a + 2*b*x - 2*c*x^2]/(8*c)
Int[Cosh[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]^(n_)*((d_.) + (e_.)*(x_))^(m_.) , x_Symbol] :> Int[ExpandTrigReduce[(d + e*x)^m, Cosh[a + b*x + c*x^2]^n, x ], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 1]
Time = 1.08 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01
| method | result | size |
| risch | \(\frac {x^{2}}{4}+\frac {{\mathrm e}^{2 c \,x^{2}-2 b x -2 a}}{16 c}+\frac {b \sqrt {\pi }\, {\mathrm e}^{-\frac {4 a c +b^{2}}{2 c}} \operatorname {erf}\left (\sqrt {-2 c}\, x +\frac {b}{\sqrt {-2 c}}\right )}{16 c \sqrt {-2 c}}-\frac {{\mathrm e}^{-2 c \,x^{2}+2 b x +2 a}}{16 c}-\frac {b \sqrt {\pi }\, {\mathrm e}^{\frac {4 a c +b^{2}}{2 c}} \sqrt {2}\, \operatorname {erf}\left (-\sqrt {2}\, \sqrt {c}\, x +\frac {b \sqrt {2}}{2 \sqrt {c}}\right )}{32 c^{\frac {3}{2}}}\) | \(137\) |
Input:
int(x*cosh(-c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/4*x^2+1/16/c*exp(2*c*x^2-2*b*x-2*a)+1/16*b/c*Pi^(1/2)*exp(-1/2*(4*a*c+b^ 2)/c)/(-2*c)^(1/2)*erf((-2*c)^(1/2)*x+b/(-2*c)^(1/2))-1/16/c*exp(-2*c*x^2+ 2*b*x+2*a)-1/32*b/c^(3/2)*Pi^(1/2)*exp(1/2*(4*a*c+b^2)/c)*2^(1/2)*erf(-2^( 1/2)*c^(1/2)*x+1/2*b*2^(1/2)/c^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 729 vs. \(2 (110) = 220\).
Time = 0.10 (sec) , antiderivative size = 729, normalized size of antiderivative = 5.36 \[ \int x \cosh ^2\left (a+b x-c x^2\right ) \, dx =\text {Too large to display} \] Input:
integrate(x*cosh(-c*x^2+b*x+a)^2,x, algorithm="fricas")
Output:
1/32*(8*c^2*x^2*cosh(c*x^2 - b*x - a)^2 + 2*c*cosh(c*x^2 - b*x - a)^4 + 8* c*cosh(c*x^2 - b*x - a)*sinh(c*x^2 - b*x - a)^3 + 2*c*sinh(c*x^2 - b*x - a )^4 - sqrt(2)*sqrt(pi)*(b*cosh(c*x^2 - b*x - a)^2*cosh(1/2*(b^2 + 4*a*c)/c ) - b*cosh(c*x^2 - b*x - a)^2*sinh(1/2*(b^2 + 4*a*c)/c) + (b*cosh(1/2*(b^2 + 4*a*c)/c) - b*sinh(1/2*(b^2 + 4*a*c)/c))*sinh(c*x^2 - b*x - a)^2 + 2*(b *cosh(c*x^2 - b*x - a)*cosh(1/2*(b^2 + 4*a*c)/c) - b*cosh(c*x^2 - b*x - a) *sinh(1/2*(b^2 + 4*a*c)/c))*sinh(c*x^2 - b*x - a))*sqrt(-c)*erf(1/2*sqrt(2 )*(2*c*x - b)*sqrt(-c)/c) + sqrt(2)*sqrt(pi)*(b*cosh(c*x^2 - b*x - a)^2*co sh(1/2*(b^2 + 4*a*c)/c) + b*cosh(c*x^2 - b*x - a)^2*sinh(1/2*(b^2 + 4*a*c) /c) + (b*cosh(1/2*(b^2 + 4*a*c)/c) + b*sinh(1/2*(b^2 + 4*a*c)/c))*sinh(c*x ^2 - b*x - a)^2 + 2*(b*cosh(c*x^2 - b*x - a)*cosh(1/2*(b^2 + 4*a*c)/c) + b *cosh(c*x^2 - b*x - a)*sinh(1/2*(b^2 + 4*a*c)/c))*sinh(c*x^2 - b*x - a))*s qrt(c)*erf(1/2*sqrt(2)*(2*c*x - b)/sqrt(c)) + 4*(2*c^2*x^2 + 3*c*cosh(c*x^ 2 - b*x - a)^2)*sinh(c*x^2 - b*x - a)^2 + 8*(2*c^2*x^2*cosh(c*x^2 - b*x - a) + c*cosh(c*x^2 - b*x - a)^3)*sinh(c*x^2 - b*x - a) - 2*c)/(c^2*cosh(c*x ^2 - b*x - a)^2 + 2*c^2*cosh(c*x^2 - b*x - a)*sinh(c*x^2 - b*x - a) + c^2* sinh(c*x^2 - b*x - a)^2)
\[ \int x \cosh ^2\left (a+b x-c x^2\right ) \, dx=\int x \cosh ^{2}{\left (a + b x - c x^{2} \right )}\, dx \] Input:
integrate(x*cosh(-c*x**2+b*x+a)**2,x)
Output:
Integral(x*cosh(a + b*x - c*x**2)**2, x)
Time = 0.15 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.59 \[ \int x \cosh ^2\left (a+b x-c x^2\right ) \, dx=\frac {1}{4} \, x^{2} + \frac {\sqrt {2} {\left (\frac {\sqrt {\pi } {\left (2 \, c x - b\right )} b {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2}} \sqrt {\frac {{\left (2 \, c x - b\right )}^{2}}{c}}\right ) - 1\right )}}{\sqrt {\frac {{\left (2 \, c x - b\right )}^{2}}{c}} \left (-c\right )^{\frac {3}{2}}} - \frac {\sqrt {2} c e^{\left (-\frac {{\left (2 \, c x - b\right )}^{2}}{2 \, c}\right )}}{\left (-c\right )^{\frac {3}{2}}}\right )} e^{\left (2 \, a + \frac {b^{2}}{2 \, c}\right )}}{32 \, \sqrt {-c}} + \frac {\sqrt {2} {\left (\frac {\sqrt {\pi } {\left (2 \, c x - b\right )} b {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2}} \sqrt {-\frac {{\left (2 \, c x - b\right )}^{2}}{c}}\right ) - 1\right )}}{\sqrt {-\frac {{\left (2 \, c x - b\right )}^{2}}{c}} c^{\frac {3}{2}}} + \frac {\sqrt {2} e^{\left (\frac {{\left (2 \, c x - b\right )}^{2}}{2 \, c}\right )}}{\sqrt {c}}\right )} e^{\left (-2 \, a - \frac {b^{2}}{2 \, c}\right )}}{32 \, \sqrt {c}} \] Input:
integrate(x*cosh(-c*x^2+b*x+a)^2,x, algorithm="maxima")
Output:
1/4*x^2 + 1/32*sqrt(2)*(sqrt(pi)*(2*c*x - b)*b*(erf(sqrt(1/2)*sqrt((2*c*x - b)^2/c)) - 1)/(sqrt((2*c*x - b)^2/c)*(-c)^(3/2)) - sqrt(2)*c*e^(-1/2*(2* c*x - b)^2/c)/(-c)^(3/2))*e^(2*a + 1/2*b^2/c)/sqrt(-c) + 1/32*sqrt(2)*(sqr t(pi)*(2*c*x - b)*b*(erf(sqrt(1/2)*sqrt(-(2*c*x - b)^2/c)) - 1)/(sqrt(-(2* c*x - b)^2/c)*c^(3/2)) + sqrt(2)*e^(1/2*(2*c*x - b)^2/c)/sqrt(c))*e^(-2*a - 1/2*b^2/c)/sqrt(c)
Time = 0.12 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.06 \[ \int x \cosh ^2\left (a+b x-c x^2\right ) \, dx=\frac {1}{4} \, x^{2} - \frac {\frac {\sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {c} {\left (2 \, x - \frac {b}{c}\right )}\right ) e^{\left (\frac {b^{2} + 4 \, a c}{2 \, c}\right )}}{\sqrt {c}} + 2 \, e^{\left (-2 \, c x^{2} + 2 \, b x + 2 \, a\right )}}{32 \, c} - \frac {\frac {\sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {-c} {\left (2 \, x - \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} + 4 \, a c}{2 \, c}\right )}}{\sqrt {-c}} - 2 \, e^{\left (2 \, c x^{2} - 2 \, b x - 2 \, a\right )}}{32 \, c} \] Input:
integrate(x*cosh(-c*x^2+b*x+a)^2,x, algorithm="giac")
Output:
1/4*x^2 - 1/32*(sqrt(2)*sqrt(pi)*b*erf(-1/2*sqrt(2)*sqrt(c)*(2*x - b/c))*e ^(1/2*(b^2 + 4*a*c)/c)/sqrt(c) + 2*e^(-2*c*x^2 + 2*b*x + 2*a))/c - 1/32*(s qrt(2)*sqrt(pi)*b*erf(-1/2*sqrt(2)*sqrt(-c)*(2*x - b/c))*e^(-1/2*(b^2 + 4* a*c)/c)/sqrt(-c) - 2*e^(2*c*x^2 - 2*b*x - 2*a))/c
Timed out. \[ \int x \cosh ^2\left (a+b x-c x^2\right ) \, dx=\int x\,{\mathrm {cosh}\left (-c\,x^2+b\,x+a\right )}^2 \,d x \] Input:
int(x*cosh(a + b*x - c*x^2)^2,x)
Output:
int(x*cosh(a + b*x - c*x^2)^2, x)
\[ \int x \cosh ^2\left (a+b x-c x^2\right ) \, dx=\frac {\sqrt {\pi }\, e^{\frac {4 c^{2} x^{2}+4 b c x +8 a c +b^{2}}{2 c}} \mathrm {erf}\left (\frac {2 c x -b}{\sqrt {c}\, \sqrt {2}}\right ) b -e^{4 b x +4 a} \sqrt {c}\, \sqrt {2}+4 e^{2 c \,x^{2}+2 b x +2 a} \sqrt {c}\, \sqrt {2}\, c \,x^{2}+2 e^{2 c \,x^{2}+2 b x} \sqrt {c}\, \sqrt {2}\, \left (\int \frac {e^{2 c \,x^{2}}}{e^{2 b x}}d x \right ) b +e^{4 c \,x^{2}} \sqrt {c}\, \sqrt {2}}{16 e^{2 c \,x^{2}+2 b x +2 a} \sqrt {c}\, \sqrt {2}\, c} \] Input:
int(x*cosh(-c*x^2+b*x+a)^2,x)
Output:
(sqrt(pi)*e**((8*a*c + b**2 + 4*b*c*x + 4*c**2*x**2)/(2*c))*erf(( - b + 2* c*x)/(sqrt(c)*sqrt(2)))*b - e**(4*a + 4*b*x)*sqrt(c)*sqrt(2) + 4*e**(2*a + 2*b*x + 2*c*x**2)*sqrt(c)*sqrt(2)*c*x**2 + 2*e**(2*b*x + 2*c*x**2)*sqrt(c )*sqrt(2)*int(e**(2*c*x**2)/e**(2*b*x),x)*b + e**(4*c*x**2)*sqrt(c)*sqrt(2 ))/(16*e**(2*a + 2*b*x + 2*c*x**2)*sqrt(c)*sqrt(2)*c)