3.10 \(\int (-\frac {b \cosh (a+b x-c x^2)}{x}+\frac {\sinh (a+b x-c x^2)}{x^2}) \, dx\)

Optimal. Leaf size=108 \[ \frac {1}{2} \sqrt {\pi } \sqrt {c} e^{a+\frac {b^2}{4 c}} \text {erf}\left (\frac {b-2 c x}{2 \sqrt {c}}\right )+\frac {1}{2} \sqrt {\pi } \sqrt {c} e^{-a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {b-2 c x}{2 \sqrt {c}}\right )-\frac {\sinh \left (a+b x-c x^2\right )}{x} \]

[Out]

-sinh(-c*x^2+b*x+a)/x+1/2*exp(a+1/4*b^2/c)*erf(1/2*(-2*c*x+b)/c^(1/2))*c^(1/2)*Pi^(1/2)+1/2*exp(-a-1/4*b^2/c)*
erfi(1/2*(-2*c*x+b)/c^(1/2))*c^(1/2)*Pi^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5390, 5375, 2234, 2205, 2204} \[ \frac {1}{2} \sqrt {\pi } \sqrt {c} e^{a+\frac {b^2}{4 c}} \text {Erf}\left (\frac {b-2 c x}{2 \sqrt {c}}\right )+\frac {1}{2} \sqrt {\pi } \sqrt {c} e^{-a-\frac {b^2}{4 c}} \text {Erfi}\left (\frac {b-2 c x}{2 \sqrt {c}}\right )-\frac {\sinh \left (a+b x-c x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[-((b*Cosh[a + b*x - c*x^2])/x) + Sinh[a + b*x - c*x^2]/x^2,x]

[Out]

(Sqrt[c]*E^(a + b^2/(4*c))*Sqrt[Pi]*Erf[(b - 2*c*x)/(2*Sqrt[c])])/2 + (Sqrt[c]*E^(-a - b^2/(4*c))*Sqrt[Pi]*Erf
i[(b - 2*c*x)/(2*Sqrt[c])])/2 - Sinh[a + b*x - c*x^2]/x

Rule 2204

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]

Rule 2205

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] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 5375

Int[Cosh[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[1/2, Int[E^(a + b*x + c*x^2), x], x] + Dist[1/2
, Int[E^(-a - b*x - c*x^2), x], x] /; FreeQ[{a, b, c}, x]

Rule 5390

Int[((d_.) + (e_.)*(x_))^(m_)*Sinh[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[((d + e*x)^(m + 1)*Si
nh[a + b*x + c*x^2])/(e*(m + 1)), x] + (-Dist[(2*c)/(e^2*(m + 1)), Int[(d + e*x)^(m + 2)*Cosh[a + b*x + c*x^2]
, x], x] - Dist[(b*e - 2*c*d)/(e^2*(m + 1)), Int[(d + e*x)^(m + 1)*Cosh[a + b*x + c*x^2], x], x]) /; FreeQ[{a,
b, c, d, e}, x] && LtQ[m, -1] && NeQ[b*e - 2*c*d, 0]

Rubi steps

\begin {align*} \int \left (-\frac {b \cosh \left (a+b x-c x^2\right )}{x}+\frac {\sinh \left (a+b x-c x^2\right )}{x^2}\right ) \, dx &=-\left (b \int \frac {\cosh \left (a+b x-c x^2\right )}{x} \, dx\right )+\int \frac {\sinh \left (a+b x-c x^2\right )}{x^2} \, dx\\ &=-\frac {\sinh \left (a+b x-c x^2\right )}{x}-(2 c) \int \cosh \left (a+b x-c x^2\right ) \, dx\\ &=-\frac {\sinh \left (a+b x-c x^2\right )}{x}-c \int e^{a+b x-c x^2} \, dx-c \int e^{-a-b x+c x^2} \, dx\\ &=-\frac {\sinh \left (a+b x-c x^2\right )}{x}-\left (c e^{-a-\frac {b^2}{4 c}}\right ) \int e^{\frac {(-b+2 c x)^2}{4 c}} \, dx-\left (c e^{a+\frac {b^2}{4 c}}\right ) \int e^{-\frac {(b-2 c x)^2}{4 c}} \, dx\\ &=\frac {1}{2} \sqrt {c} e^{a+\frac {b^2}{4 c}} \sqrt {\pi } \text {erf}\left (\frac {b-2 c x}{2 \sqrt {c}}\right )+\frac {1}{2} \sqrt {c} e^{-a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {b-2 c x}{2 \sqrt {c}}\right )-\frac {\sinh \left (a+b x-c x^2\right )}{x}\\ \end {align*}

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Mathematica [A]  time = 0.68, size = 136, normalized size = 1.26 \[ \frac {1}{2} \left (-\sqrt {\pi } \sqrt {c} \text {erf}\left (\frac {2 c x-b}{2 \sqrt {c}}\right ) \left (\sinh \left (a+\frac {b^2}{4 c}\right )+\cosh \left (a+\frac {b^2}{4 c}\right )\right )+\sqrt {\pi } \sqrt {c} \text {erfi}\left (\frac {2 c x-b}{2 \sqrt {c}}\right ) \left (\sinh \left (a+\frac {b^2}{4 c}\right )-\cosh \left (a+\frac {b^2}{4 c}\right )\right )-\frac {2 \sinh (a+x (b-c x))}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[-((b*Cosh[a + b*x - c*x^2])/x) + Sinh[a + b*x - c*x^2]/x^2,x]

[Out]

(Sqrt[c]*Sqrt[Pi]*Erfi[(-b + 2*c*x)/(2*Sqrt[c])]*(-Cosh[a + b^2/(4*c)] + Sinh[a + b^2/(4*c)]) - Sqrt[c]*Sqrt[P
i]*Erf[(-b + 2*c*x)/(2*Sqrt[c])]*(Cosh[a + b^2/(4*c)] + Sinh[a + b^2/(4*c)]) - (2*Sinh[a + x*(b - c*x)])/x)/2

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fricas [B]  time = 0.63, size = 371, normalized size = 3.44 \[ \frac {\sqrt {\pi } {\left (x \cosh \left (c x^{2} - b x - a\right ) \cosh \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) - x \cosh \left (c x^{2} - b x - a\right ) \sinh \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + {\left (x \cosh \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) - x \sinh \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right )\right )} \sinh \left (c x^{2} - b x - a\right )\right )} \sqrt {-c} \operatorname {erf}\left (\frac {{\left (2 \, c x - b\right )} \sqrt {-c}}{2 \, c}\right ) - \sqrt {\pi } {\left (x \cosh \left (c x^{2} - b x - a\right ) \cosh \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + x \cosh \left (c x^{2} - b x - a\right ) \sinh \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + {\left (x \cosh \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + x \sinh \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right )\right )} \sinh \left (c x^{2} - b x - a\right )\right )} \sqrt {c} \operatorname {erf}\left (\frac {2 \, c x - b}{2 \, \sqrt {c}}\right ) + \cosh \left (c x^{2} - b x - a\right )^{2} + 2 \, \cosh \left (c x^{2} - b x - a\right ) \sinh \left (c x^{2} - b x - a\right ) + \sinh \left (c x^{2} - b x - a\right )^{2} - 1}{2 \, {\left (x \cosh \left (c x^{2} - b x - a\right ) + x \sinh \left (c x^{2} - b x - a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-b*cosh(-c*x^2+b*x+a)/x+sinh(-c*x^2+b*x+a)/x^2,x, algorithm="fricas")

[Out]

1/2*(sqrt(pi)*(x*cosh(c*x^2 - b*x - a)*cosh(1/4*(b^2 + 4*a*c)/c) - x*cosh(c*x^2 - b*x - a)*sinh(1/4*(b^2 + 4*a
*c)/c) + (x*cosh(1/4*(b^2 + 4*a*c)/c) - x*sinh(1/4*(b^2 + 4*a*c)/c))*sinh(c*x^2 - b*x - a))*sqrt(-c)*erf(1/2*(
2*c*x - b)*sqrt(-c)/c) - sqrt(pi)*(x*cosh(c*x^2 - b*x - a)*cosh(1/4*(b^2 + 4*a*c)/c) + x*cosh(c*x^2 - b*x - a)
*sinh(1/4*(b^2 + 4*a*c)/c) + (x*cosh(1/4*(b^2 + 4*a*c)/c) + x*sinh(1/4*(b^2 + 4*a*c)/c))*sinh(c*x^2 - b*x - a)
)*sqrt(c)*erf(1/2*(2*c*x - b)/sqrt(c)) + cosh(c*x^2 - b*x - a)^2 + 2*cosh(c*x^2 - b*x - a)*sinh(c*x^2 - b*x -
a) + sinh(c*x^2 - b*x - a)^2 - 1)/(x*cosh(c*x^2 - b*x - a) + x*sinh(c*x^2 - b*x - a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {b \cosh \left (-c x^{2} + b x + a\right )}{x} + \frac {\sinh \left (-c x^{2} + b x + a\right )}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-b*cosh(-c*x^2+b*x+a)/x+sinh(-c*x^2+b*x+a)/x^2,x, algorithm="giac")

[Out]

integrate(-b*cosh(-c*x^2 + b*x + a)/x + sinh(-c*x^2 + b*x + a)/x^2, x)

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int -\frac {b \cosh \left (-c \,x^{2}+b x +a \right )}{x}+\frac {\sinh \left (-c \,x^{2}+b x +a \right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-b*cosh(-c*x^2+b*x+a)/x+sinh(-c*x^2+b*x+a)/x^2,x)

[Out]

int(-b*cosh(-c*x^2+b*x+a)/x+sinh(-c*x^2+b*x+a)/x^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {b \cosh \left (c x^{2} - b x - a\right )}{x} - \frac {\sinh \left (c x^{2} - b x - a\right )}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-b*cosh(-c*x^2+b*x+a)/x+sinh(-c*x^2+b*x+a)/x^2,x, algorithm="maxima")

[Out]

integrate(-b*cosh(c*x^2 - b*x - a)/x - sinh(c*x^2 - b*x - a)/x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {sinh}\left (-c\,x^2+b\,x+a\right )}{x^2}-\frac {b\,\mathrm {cosh}\left (-c\,x^2+b\,x+a\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x - c*x^2)/x^2 - (b*cosh(a + b*x - c*x^2))/x,x)

[Out]

int(sinh(a + b*x - c*x^2)/x^2 - (b*cosh(a + b*x - c*x^2))/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {\sinh {\left (a + b x - c x^{2} \right )}}{x^{2}}\right )\, dx - \int \frac {b \cosh {\left (a + b x - c x^{2} \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-b*cosh(-c*x**2+b*x+a)/x+sinh(-c*x**2+b*x+a)/x**2,x)

[Out]

-Integral(-sinh(a + b*x - c*x**2)/x**2, x) - Integral(b*cosh(a + b*x - c*x**2)/x, x)

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