Integrand size = 11, antiderivative size = 91 \[ \int \cosh \left (a+b x+c x^2\right ) \, dx=\frac {e^{-a+\frac {b^2}{4 c}} \sqrt {\pi } \text {erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{4 \sqrt {c}}+\frac {e^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{4 \sqrt {c}} \]
1/4*exp(-a+1/4*b^2/c)*erf(1/2*(2*c*x+b)/c^(1/2))*Pi^(1/2)/c^(1/2)+1/4*exp( a-1/4*b^2/c)*erfi(1/2*(2*c*x+b)/c^(1/2))*Pi^(1/2)/c^(1/2)
Time = 0.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.15 \[ \int \cosh \left (a+b x+c x^2\right ) \, dx=\frac {\sqrt {\pi } \left (\text {erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right ) \left (\cosh \left (a-\frac {b^2}{4 c}\right )-\sinh \left (a-\frac {b^2}{4 c}\right )\right )+\text {erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right ) \left (\cosh \left (a-\frac {b^2}{4 c}\right )+\sinh \left (a-\frac {b^2}{4 c}\right )\right )\right )}{4 \sqrt {c}} \]
(Sqrt[Pi]*(Erf[(b + 2*c*x)/(2*Sqrt[c])]*(Cosh[a - b^2/(4*c)] - Sinh[a - b^ 2/(4*c)]) + Erfi[(b + 2*c*x)/(2*Sqrt[c])]*(Cosh[a - b^2/(4*c)] + Sinh[a - b^2/(4*c)])))/(4*Sqrt[c])
Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5898, 2664, 2633, 2634}
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.
\(\displaystyle \int \cosh \left (a+b x+c x^2\right ) \, dx\) |
\(\Big \downarrow \) 5898 |
\(\displaystyle \frac {1}{2} \int e^{-c x^2-b x-a}dx+\frac {1}{2} \int e^{c x^2+b x+a}dx\) |
\(\Big \downarrow \) 2664 |
\(\displaystyle \frac {1}{2} e^{\frac {b^2}{4 c}-a} \int e^{-\frac {(b+2 c x)^2}{4 c}}dx+\frac {1}{2} e^{a-\frac {b^2}{4 c}} \int e^{\frac {(b+2 c x)^2}{4 c}}dx\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {1}{2} e^{\frac {b^2}{4 c}-a} \int e^{-\frac {(b+2 c x)^2}{4 c}}dx+\frac {\sqrt {\pi } e^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\) |
\(\Big \downarrow \) 2634 |
\(\displaystyle \frac {\sqrt {\pi } e^{\frac {b^2}{4 c}-a} \text {erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{4 \sqrt {c}}+\frac {\sqrt {\pi } e^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\) |
(E^(-a + b^2/(4*c))*Sqrt[Pi]*Erf[(b + 2*c*x)/(2*Sqrt[c])])/(4*Sqrt[c]) + ( E^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[(b + 2*c*x)/(2*Sqrt[c])])/(4*Sqrt[c])
3.1.3.3.1 Defintions of rubi rules used
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ F, a, b, c, d}, x] && PosQ[b]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F], 2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; Fr eeQ[{F, a, b, c, d}, x] && NegQ[b]
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[F^(a - b^2/ (4*c)) Int[F^((b + 2*c*x)^2/(4*c)), x], x] /; FreeQ[{F, a, b, c}, x]
Int[Cosh[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[1/2 Int[E^ (a + b*x + c*x^2), x], x] + Simp[1/2 Int[E^(-a - b*x - c*x^2), x], x] /; FreeQ[{a, b, c}, x]
Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {\sqrt {\pi }\, {\mathrm e}^{-\frac {4 a c -b^{2}}{4 c}} \operatorname {erf}\left (\sqrt {c}\, x +\frac {b}{2 \sqrt {c}}\right )}{4 \sqrt {c}}-\frac {\sqrt {\pi }\, {\mathrm e}^{\frac {4 a c -b^{2}}{4 c}} \operatorname {erf}\left (-\sqrt {-c}\, x +\frac {b}{2 \sqrt {-c}}\right )}{4 \sqrt {-c}}\) | \(83\) |
1/4*Pi^(1/2)*exp(-1/4*(4*a*c-b^2)/c)/c^(1/2)*erf(c^(1/2)*x+1/2*b/c^(1/2))- 1/4*Pi^(1/2)*exp(1/4*(4*a*c-b^2)/c)/(-c)^(1/2)*erf(-(-c)^(1/2)*x+1/2*b/(-c )^(1/2))
Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.23 \[ \int \cosh \left (a+b x+c x^2\right ) \, dx=-\frac {\sqrt {\pi } \sqrt {-c} {\left (\cosh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + \sinh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, c}\right ) - \sqrt {\pi } \sqrt {c} {\left (\cosh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - \sinh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \operatorname {erf}\left (\frac {2 \, c x + b}{2 \, \sqrt {c}}\right )}{4 \, c} \]
-1/4*(sqrt(pi)*sqrt(-c)*(cosh(-1/4*(b^2 - 4*a*c)/c) + sinh(-1/4*(b^2 - 4*a *c)/c))*erf(1/2*(2*c*x + b)*sqrt(-c)/c) - sqrt(pi)*sqrt(c)*(cosh(-1/4*(b^2 - 4*a*c)/c) - sinh(-1/4*(b^2 - 4*a*c)/c))*erf(1/2*(2*c*x + b)/sqrt(c)))/c
\[ \int \cosh \left (a+b x+c x^2\right ) \, dx=\int \cosh {\left (a + b x + c x^{2} \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 464 vs. \(2 (65) = 130\).
Time = 0.37 (sec) , antiderivative size = 464, normalized size of antiderivative = 5.10 \[ \int \cosh \left (a+b x+c x^2\right ) \, dx=\frac {{\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b {\left (\operatorname {erf}\left (\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 {2 \, e^{\left (\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}\right )}}{\sqrt {c}}\right )} b e^{\left (a - \frac {b^{2}}{4 \, c}\right )}}{8 \, \sqrt {c}} - \frac {1}{8} \, {\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b^{2} {\left (\operatorname {erf}\left (\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 {5}{2}}} - \frac {4 \, b e^{\left (\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}\right )}}{c^{\frac {3}{2}}} - \frac {4 \, {\left (2 \, c x + b\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}\right )}{\left (-\frac {{\left (2 \, c x + b\right )}^{2}}{c}\right )^{\frac {3}{2}} c^{\frac {5}{2}}}\right )} \sqrt {c} e^{\left (a - \frac {b^{2}}{4 \, c}\right )} - \frac {{\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b {\left (\operatorname {erf}\left (\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 {2 \, c e^{\left (-\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}\right )}}{\left (-c\right )^{\frac {3}{2}}}\right )} b e^{\left (-a + \frac {b^{2}}{4 \, c}\right )}}{8 \, \sqrt {-c}} - \frac {{\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b^{2} {\left (\operatorname {erf}\left (\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 {5}{2}}} + \frac {4 \, b c e^{\left (-\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}\right )}}{\left (-c\right )^{\frac {5}{2}}} - \frac {4 \, {\left (2 \, c x + b\right )}^{3} \Gamma \left (\frac {3}{2}, \frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}\right )}{\left (\frac {{\left (2 \, c x + b\right )}^{2}}{c}\right )^{\frac {3}{2}} \left (-c\right )^{\frac {5}{2}}}\right )} c e^{\left (-a + \frac {b^{2}}{4 \, c}\right )}}{8 \, \sqrt {-c}} + x \cosh \left (c x^{2} + b x + a\right ) \]
1/8*(sqrt(pi)*(2*c*x + b)*b*(erf(1/2*sqrt(-(2*c*x + b)^2/c)) - 1)/(sqrt(-( 2*c*x + b)^2/c)*c^(3/2)) - 2*e^(1/4*(2*c*x + b)^2/c)/sqrt(c))*b*e^(a - 1/4 *b^2/c)/sqrt(c) - 1/8*(sqrt(pi)*(2*c*x + b)*b^2*(erf(1/2*sqrt(-(2*c*x + b) ^2/c)) - 1)/(sqrt(-(2*c*x + b)^2/c)*c^(5/2)) - 4*b*e^(1/4*(2*c*x + b)^2/c) /c^(3/2) - 4*(2*c*x + b)^3*gamma(3/2, -1/4*(2*c*x + b)^2/c)/((-(2*c*x + b) ^2/c)^(3/2)*c^(5/2)))*sqrt(c)*e^(a - 1/4*b^2/c) - 1/8*(sqrt(pi)*(2*c*x + b )*b*(erf(1/2*sqrt((2*c*x + b)^2/c)) - 1)/(sqrt((2*c*x + b)^2/c)*(-c)^(3/2) ) + 2*c*e^(-1/4*(2*c*x + b)^2/c)/(-c)^(3/2))*b*e^(-a + 1/4*b^2/c)/sqrt(-c) - 1/8*(sqrt(pi)*(2*c*x + b)*b^2*(erf(1/2*sqrt((2*c*x + b)^2/c)) - 1)/(sqr t((2*c*x + b)^2/c)*(-c)^(5/2)) + 4*b*c*e^(-1/4*(2*c*x + b)^2/c)/(-c)^(5/2) - 4*(2*c*x + b)^3*gamma(3/2, 1/4*(2*c*x + b)^2/c)/(((2*c*x + b)^2/c)^(3/2 )*(-c)^(5/2)))*c*e^(-a + 1/4*b^2/c)/sqrt(-c) + x*cosh(c*x^2 + b*x + a)
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.87 \[ \int \cosh \left (a+b x+c x^2\right ) \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (\frac {b^{2} - 4 \, a c}{4 \, c}\right )}}{4 \, \sqrt {c}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )}}{4 \, \sqrt {-c}} \]
-1/4*sqrt(pi)*erf(-1/2*sqrt(c)*(2*x + b/c))*e^(1/4*(b^2 - 4*a*c)/c)/sqrt(c ) - 1/4*sqrt(pi)*erf(-1/2*sqrt(-c)*(2*x + b/c))*e^(-1/4*(b^2 - 4*a*c)/c)/s qrt(-c)
Timed out. \[ \int \cosh \left (a+b x+c x^2\right ) \, dx=\int \mathrm {cosh}\left (c\,x^2+b\,x+a\right ) \,d x \]