3.29 \(\int (d+e x) \sinh (a+b x+c x^2) \, dx\)

Optimal. Leaf size=128 \[ -\frac {\sqrt {\pi } e^{\frac {b^2}{4 c}-a} (2 c d-b e) \text {erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{8 c^{3/2}}+\frac {\sqrt {\pi } e^{a-\frac {b^2}{4 c}} (2 c d-b e) \text {erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{8 c^{3/2}}+\frac {e \cosh \left (a+b x+c x^2\right )}{2 c} \]

[Out]

1/2*e*cosh(c*x^2+b*x+a)/c-1/8*(-b*e+2*c*d)*exp(-a+1/4*b^2/c)*erf(1/2*(2*c*x+b)/c^(1/2))*Pi^(1/2)/c^(3/2)+1/8*(
-b*e+2*c*d)*exp(a-1/4*b^2/c)*erfi(1/2*(2*c*x+b)/c^(1/2))*Pi^(1/2)/c^(3/2)

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Rubi [A]  time = 0.06, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5382, 5374, 2234, 2204, 2205} \[ -\frac {\sqrt {\pi } e^{\frac {b^2}{4 c}-a} (2 c d-b e) \text {Erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{8 c^{3/2}}+\frac {\sqrt {\pi } e^{a-\frac {b^2}{4 c}} (2 c d-b e) \text {Erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{8 c^{3/2}}+\frac {e \cosh \left (a+b x+c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*Sinh[a + b*x + c*x^2],x]

[Out]

(e*Cosh[a + b*x + c*x^2])/(2*c) - ((2*c*d - b*e)*E^(-a + b^2/(4*c))*Sqrt[Pi]*Erf[(b + 2*c*x)/(2*Sqrt[c])])/(8*
c^(3/2)) + ((2*c*d - b*e)*E^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[(b + 2*c*x)/(2*Sqrt[c])])/(8*c^(3/2))

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 5374

Int[Sinh[(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 5382

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

Rubi steps

\begin {align*} \int (d+e x) \sinh \left (a+b x+c x^2\right ) \, dx &=\frac {e \cosh \left (a+b x+c x^2\right )}{2 c}-\frac {(-2 c d+b e) \int \sinh \left (a+b x+c x^2\right ) \, dx}{2 c}\\ &=\frac {e \cosh \left (a+b x+c x^2\right )}{2 c}-\frac {(2 c d-b e) \int e^{-a-b x-c x^2} \, dx}{4 c}+\frac {(2 c d-b e) \int e^{a+b x+c x^2} \, dx}{4 c}\\ &=\frac {e \cosh \left (a+b x+c x^2\right )}{2 c}+\frac {\left ((2 c d-b e) e^{a-\frac {b^2}{4 c}}\right ) \int e^{\frac {(b+2 c x)^2}{4 c}} \, dx}{4 c}-\frac {\left ((2 c d-b e) e^{-a+\frac {b^2}{4 c}}\right ) \int e^{-\frac {(-b-2 c x)^2}{4 c}} \, dx}{4 c}\\ &=\frac {e \cosh \left (a+b x+c x^2\right )}{2 c}-\frac {(2 c d-b e) e^{-a+\frac {b^2}{4 c}} \sqrt {\pi } \text {erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{8 c^{3/2}}+\frac {(2 c d-b e) e^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{8 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 146, normalized size = 1.14 \[ \frac {\sqrt {\pi } (2 c d-b e) \text {erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right ) \left (\sinh \left (a-\frac {b^2}{4 c}\right )-\cosh \left (a-\frac {b^2}{4 c}\right )\right )+\sqrt {\pi } (2 c d-b e) \text {erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right ) \left (\sinh \left (a-\frac {b^2}{4 c}\right )+\cosh \left (a-\frac {b^2}{4 c}\right )\right )+4 \sqrt {c} e \cosh (a+x (b+c x))}{8 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*Sinh[a + b*x + c*x^2],x]

[Out]

(4*Sqrt[c]*e*Cosh[a + x*(b + c*x)] + (2*c*d - b*e)*Sqrt[Pi]*Erf[(b + 2*c*x)/(2*Sqrt[c])]*(-Cosh[a - b^2/(4*c)]
+ Sinh[a - b^2/(4*c)]) + (2*c*d - b*e)*Sqrt[Pi]*Erfi[(b + 2*c*x)/(2*Sqrt[c])]*(Cosh[a - b^2/(4*c)] + Sinh[a -
b^2/(4*c)]))/(8*c^(3/2))

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fricas [B]  time = 0.57, size = 413, normalized size = 3.23 \[ \frac {2 \, c e \cosh \left (c x^{2} + b x + a\right )^{2} + 4 \, c e \cosh \left (c x^{2} + b x + a\right ) \sinh \left (c x^{2} + b x + a\right ) + 2 \, c e \sinh \left (c x^{2} + b x + a\right )^{2} - \sqrt {\pi } {\left ({\left (2 \, c d - b e\right )} \cosh \left (c x^{2} + b x + a\right ) \cosh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + {\left (2 \, c d - b e\right )} \cosh \left (c x^{2} + b x + a\right ) \sinh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + {\left ({\left (2 \, c d - b e\right )} \cosh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + {\left (2 \, c d - b e\right )} \sinh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sinh \left (c x^{2} + b x + a\right )\right )} \sqrt {-c} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, c}\right ) - \sqrt {\pi } {\left ({\left (2 \, c d - b e\right )} \cosh \left (c x^{2} + b x + a\right ) \cosh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - {\left (2 \, c d - b e\right )} \cosh \left (c x^{2} + b x + a\right ) \sinh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + {\left ({\left (2 \, c d - b e\right )} \cosh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - {\left (2 \, c d - b e\right )} \sinh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sinh \left (c x^{2} + b x + a\right )\right )} \sqrt {c} \operatorname {erf}\left (\frac {2 \, c x + b}{2 \, \sqrt {c}}\right ) + 2 \, c e}{8 \, {\left (c^{2} \cosh \left (c x^{2} + b x + a\right ) + c^{2} \sinh \left (c x^{2} + b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*sinh(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

1/8*(2*c*e*cosh(c*x^2 + b*x + a)^2 + 4*c*e*cosh(c*x^2 + b*x + a)*sinh(c*x^2 + b*x + a) + 2*c*e*sinh(c*x^2 + b*
x + a)^2 - sqrt(pi)*((2*c*d - b*e)*cosh(c*x^2 + b*x + a)*cosh(-1/4*(b^2 - 4*a*c)/c) + (2*c*d - b*e)*cosh(c*x^2
+ b*x + a)*sinh(-1/4*(b^2 - 4*a*c)/c) + ((2*c*d - b*e)*cosh(-1/4*(b^2 - 4*a*c)/c) + (2*c*d - b*e)*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)*((2*c*d - b*e)*cos
h(c*x^2 + b*x + a)*cosh(-1/4*(b^2 - 4*a*c)/c) - (2*c*d - b*e)*cosh(c*x^2 + b*x + a)*sinh(-1/4*(b^2 - 4*a*c)/c)
+ ((2*c*d - b*e)*cosh(-1/4*(b^2 - 4*a*c)/c) - (2*c*d - b*e)*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)) + 2*c*e)/(c^2*cosh(c*x^2 + b*x + a) + c^2*sinh(c*x^2 + b*x + a))

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giac [B]  time = 0.13, size = 209, normalized size = 1.63 \[ \frac {\sqrt {\pi } d \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (\frac {b^{2} - 4 \, a c}{4 \, c}\right )}}{4 \, \sqrt {c}} - \frac {\sqrt {\pi } d \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )}}{4 \, \sqrt {-c}} - \frac {\frac {\sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (\frac {b^{2} - 4 \, a c + 4 \, c}{4 \, c}\right )}}{\sqrt {c}} - 2 \, e^{\left (-c x^{2} - b x - a + 1\right )}}{8 \, c} + \frac {\frac {\sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} - 4 \, a c - 4 \, c}{4 \, c}\right )}}{\sqrt {-c}} + 2 \, e^{\left (c x^{2} + b x + a + 1\right )}}{8 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*sinh(c*x^2+b*x+a),x, algorithm="giac")

[Out]

1/4*sqrt(pi)*d*erf(-1/2*sqrt(c)*(2*x + b/c))*e^(1/4*(b^2 - 4*a*c)/c)/sqrt(c) - 1/4*sqrt(pi)*d*erf(-1/2*sqrt(-c
)*(2*x + b/c))*e^(-1/4*(b^2 - 4*a*c)/c)/sqrt(-c) - 1/8*(sqrt(pi)*b*erf(-1/2*sqrt(c)*(2*x + b/c))*e^(1/4*(b^2 -
4*a*c + 4*c)/c)/sqrt(c) - 2*e^(-c*x^2 - b*x - a + 1))/c + 1/8*(sqrt(pi)*b*erf(-1/2*sqrt(-c)*(2*x + b/c))*e^(-
1/4*(b^2 - 4*a*c - 4*c)/c)/sqrt(-c) + 2*e^(c*x^2 + b*x + a + 1))/c

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maple [B]  time = 0.05, size = 211, normalized size = 1.65 \[ -\frac {d \sqrt {\pi }\, {\mathrm e}^{-\frac {4 a c -b^{2}}{4 c}} \erf \left (\sqrt {c}\, x +\frac {b}{2 \sqrt {c}}\right )}{4 \sqrt {c}}+\frac {e \,{\mathrm e}^{-c \,x^{2}-b x -a}}{4 c}+\frac {e b \sqrt {\pi }\, {\mathrm e}^{-\frac {4 a c -b^{2}}{4 c}} \erf \left (\sqrt {c}\, x +\frac {b}{2 \sqrt {c}}\right )}{8 c^{\frac {3}{2}}}-\frac {d \sqrt {\pi }\, {\mathrm e}^{\frac {4 a c -b^{2}}{4 c}} \erf \left (-\sqrt {-c}\, x +\frac {b}{2 \sqrt {-c}}\right )}{4 \sqrt {-c}}+\frac {e \,{\mathrm e}^{c \,x^{2}+b x +a}}{4 c}+\frac {e b \sqrt {\pi }\, {\mathrm e}^{\frac {4 a c -b^{2}}{4 c}} \erf \left (-\sqrt {-c}\, x +\frac {b}{2 \sqrt {-c}}\right )}{8 c \sqrt {-c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*sinh(c*x^2+b*x+a),x)

[Out]

-1/4*d*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*e/c*exp(-c*x^2-b*x-a)+1/8*e*b
/c^(3/2)*Pi^(1/2)*exp(-1/4*(4*a*c-b^2)/c)*erf(c^(1/2)*x+1/2*b/c^(1/2))-1/4*d*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*e/c*exp(c*x^2+b*x+a)+1/8*e*b/c*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))

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maxima [B]  time = 0.86, size = 254, normalized size = 1.98 \[ \frac {\sqrt {\pi } d \operatorname {erf}\left (\sqrt {-c} x - \frac {b}{2 \, \sqrt {-c}}\right ) e^{\left (a - \frac {b^{2}}{4 \, c}\right )}}{4 \, \sqrt {-c}} - \frac {\sqrt {\pi } d \operatorname {erf}\left (\sqrt {c} x + \frac {b}{2 \, \sqrt {c}}\right ) e^{\left (-a + \frac {b^{2}}{4 \, c}\right )}}{4 \, \sqrt {c}} - \frac {{\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {{\left (2 \, c x + b\right )}^{2}}{c}}\right ) - 1\right )}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2}}{c}} c^{\frac {3}{2}}} - \frac {2 \, e^{\left (\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}\right )}}{\sqrt {c}}\right )} e e^{\left (a - \frac {b^{2}}{4 \, c}\right )}}{8 \, \sqrt {c}} + \frac {{\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {{\left (2 \, c x + b\right )}^{2}}{c}}\right ) - 1\right )}}{\sqrt {\frac {{\left (2 \, c x + b\right )}^{2}}{c}} \left (-c\right )^{\frac {3}{2}}} + \frac {2 \, c e^{\left (-\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}\right )}}{\left (-c\right )^{\frac {3}{2}}}\right )} e e^{\left (-a + \frac {b^{2}}{4 \, c}\right )}}{8 \, \sqrt {-c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*sinh(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

1/4*sqrt(pi)*d*erf(sqrt(-c)*x - 1/2*b/sqrt(-c))*e^(a - 1/4*b^2/c)/sqrt(-c) - 1/4*sqrt(pi)*d*erf(sqrt(c)*x + 1/
2*b/sqrt(c))*e^(-a + 1/4*b^2/c)/sqrt(c) - 1/8*(sqrt(pi)*(2*c*x + b)*b*(erf(1/2*sqrt(-(2*c*x + b)^2/c)) - 1)/(s
qrt(-(2*c*x + b)^2/c)*c^(3/2)) - 2*e^(1/4*(2*c*x + b)^2/c)/sqrt(c))*e*e^(a - 1/4*b^2/c)/sqrt(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))*e*e^(-a + 1/4*b^2/c)/sqrt(-c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x + c*x^2)*(d + e*x),x)

[Out]

int(sinh(a + b*x + c*x^2)*(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*sinh(c*x**2+b*x+a),x)

[Out]

Integral((d + e*x)*sinh(a + b*x + c*x**2), x)

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