3.56 \(\int \cosh ((a+b x)^2) \, dx\)

Optimal. Leaf size=37 \[ \frac {\sqrt {\pi } \text {erf}(a+b x)}{4 b}+\frac {\sqrt {\pi } \text {erfi}(a+b x)}{4 b} \]

[Out]

1/4*erf(b*x+a)*Pi^(1/2)/b+1/4*erfi(b*x+a)*Pi^(1/2)/b

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Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5311, 5299, 2204, 2205} \[ \frac {\sqrt {\pi } \text {Erf}(a+b x)}{4 b}+\frac {\sqrt {\pi } \text {Erfi}(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[(a + b*x)^2],x]

[Out]

(Sqrt[Pi]*Erf[a + b*x])/(4*b) + (Sqrt[Pi]*Erfi[a + b*x])/(4*b)

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 5299

Int[Cosh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[1/2, Int[E^(c + d*x^n), x], x] + Dist[1/2, Int[E^(-c - d*
x^n), x], x] /; FreeQ[{c, d}, x] && IGtQ[n, 1]

Rule 5311

Int[((a_.) + Cosh[(c_.) + (d_.)*(u_)^(n_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/Coefficient[u, x, 1], Subst[Int[(
a + b*Cosh[c + d*x^n])^p, x], x, u], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[p] && LinearQ[u, x] && NeQ[u,
x]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 25, normalized size = 0.68 \[ \frac {\sqrt {\pi } (\text {erf}(a+b x)+\text {erfi}(a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[(a + b*x)^2],x]

[Out]

(Sqrt[Pi]*(Erf[a + b*x] + Erfi[a + b*x]))/(4*b)

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fricas [A]  time = 1.14, size = 54, normalized size = 1.46 \[ \frac {\sqrt {\pi } \sqrt {b^{2}} \operatorname {erf}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + \sqrt {\pi } \sqrt {b^{2}} \operatorname {erfi}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right )}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(pi)*sqrt(b^2)*erf(sqrt(b^2)*(b*x + a)/b) + sqrt(pi)*sqrt(b^2)*erfi(sqrt(b^2)*(b*x + a)/b))/b^2

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giac [C]  time = 0.13, size = 39, normalized size = 1.05 \[ -\frac {i \, \sqrt {\pi } \operatorname {erf}\left (i \, b {\left (x + \frac {a}{b}\right )}\right )}{4 \, b} - \frac {\sqrt {\pi } \operatorname {erf}\left (-b {\left (x + \frac {a}{b}\right )}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*sqrt(pi)*erf(I*b*(x + a/b))/b - 1/4*sqrt(pi)*erf(-b*(x + a/b))/b

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maple [C]  time = 0.12, size = 36, normalized size = 0.97 \[ \frac {\erf \left (b x +a \right ) \sqrt {\pi }}{4 b}-\frac {i \sqrt {\pi }\, \erf \left (i b x +i a \right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh((b*x+a)^2),x)

[Out]

1/4*erf(b*x+a)*Pi^(1/2)/b-1/4*I*Pi^(1/2)/b*erf(I*b*x+I*a)

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maxima [B]  time = 0.61, size = 478, normalized size = 12.92 \[ -\frac {1}{2} \, {\left (\frac {{\left (\frac {\sqrt {\pi } {\left (b^{2} x + a b\right )} a b^{2} {\left (\operatorname {erf}\left (\frac {\sqrt {{\left (b^{2} x + a b\right )}^{2}}}{b}\right ) - 1\right )}}{\sqrt {{\left (b^{2} x + a b\right )}^{2}} \left (-b^{2}\right )^{\frac {3}{2}}} + \frac {b^{2} e^{\left (-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}}{\left (-b^{2}\right )^{\frac {3}{2}}}\right )} a}{\sqrt {-b^{2}}} + \frac {{\left (\frac {\sqrt {\pi } {\left (b^{2} x + a b\right )} a^{2} b^{3} {\left (\operatorname {erf}\left (\frac {\sqrt {{\left (b^{2} x + a b\right )}^{2}}}{b}\right ) - 1\right )}}{\sqrt {{\left (b^{2} x + a b\right )}^{2}} \left (-b^{2}\right )^{\frac {5}{2}}} - \frac {{\left (b^{2} x + a b\right )}^{3} b^{3} \Gamma \left (\frac {3}{2}, \frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}{{\left ({\left (b^{2} x + a b\right )}^{2}\right )}^{\frac {3}{2}} \left (-b^{2}\right )^{\frac {5}{2}}} + \frac {2 \, a b^{3} e^{\left (-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}}{\left (-b^{2}\right )^{\frac {5}{2}}}\right )} b}{\sqrt {-b^{2}}} - \frac {a {\left (\frac {\sqrt {\pi } {\left (b^{2} x + a b\right )} a {\left (\operatorname {erf}\left (\sqrt {-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}}\right ) - 1\right )}}{b^{2} \sqrt {-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}}} - \frac {e^{\left (\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}}{b}\right )}}{b} + \frac {\sqrt {\pi } {\left (b^{2} x + a b\right )} a^{2} {\left (\operatorname {erf}\left (\sqrt {-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}}\right ) - 1\right )}}{b^{3} \sqrt {-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}}} - \frac {2 \, a e^{\left (\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}}{b^{2}} - \frac {{\left (b^{2} x + a b\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}{b^{5} \left (-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )^{\frac {3}{2}}}\right )} b + x \cosh \left ({\left (b x + a\right )}^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((sqrt(pi)*(b^2*x + a*b)*a*b^2*(erf(sqrt((b^2*x + a*b)^2)/b) - 1)/(sqrt((b^2*x + a*b)^2)*(-b^2)^(3/2)) +
b^2*e^(-(b^2*x + a*b)^2/b^2)/(-b^2)^(3/2))*a/sqrt(-b^2) + (sqrt(pi)*(b^2*x + a*b)*a^2*b^3*(erf(sqrt((b^2*x + a
*b)^2)/b) - 1)/(sqrt((b^2*x + a*b)^2)*(-b^2)^(5/2)) - (b^2*x + a*b)^3*b^3*gamma(3/2, (b^2*x + a*b)^2/b^2)/(((b
^2*x + a*b)^2)^(3/2)*(-b^2)^(5/2)) + 2*a*b^3*e^(-(b^2*x + a*b)^2/b^2)/(-b^2)^(5/2))*b/sqrt(-b^2) - a*(sqrt(pi)
*(b^2*x + a*b)*a*(erf(sqrt(-(b^2*x + a*b)^2/b^2)) - 1)/(b^2*sqrt(-(b^2*x + a*b)^2/b^2)) - e^((b^2*x + a*b)^2/b
^2)/b)/b + sqrt(pi)*(b^2*x + a*b)*a^2*(erf(sqrt(-(b^2*x + a*b)^2/b^2)) - 1)/(b^3*sqrt(-(b^2*x + a*b)^2/b^2)) -
2*a*e^((b^2*x + a*b)^2/b^2)/b^2 - (b^2*x + a*b)^3*gamma(3/2, -(b^2*x + a*b)^2/b^2)/(b^5*(-(b^2*x + a*b)^2/b^2
)^(3/2)))*b + x*cosh((b*x + a)^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \mathrm {cosh}\left ({\left (a+b\,x\right )}^2\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh((a + b*x)^2),x)

[Out]

int(cosh((a + b*x)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh((b*x+a)**2),x)

[Out]

Integral(cosh((a + b*x)**2), x)

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