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

Optimal. Leaf size=261 \[ -\frac {\sqrt {\pi } e^{\frac {b^2}{4 c}-a} (2 c d-b e)^2 \text {erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{16 c^{5/2}}+\frac {\sqrt {\pi } e^{a-\frac {b^2}{4 c}} (2 c d-b e)^2 \text {erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{16 c^{5/2}}-\frac {\sqrt {\pi } e^2 e^{\frac {b^2}{4 c}-a} \text {erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{8 c^{3/2}}-\frac {\sqrt {\pi } e^2 e^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{8 c^{3/2}}+\frac {e (2 c d-b e) \cosh \left (a+b x+c x^2\right )}{4 c^2}+\frac {e (d+e x) \cosh \left (a+b x+c x^2\right )}{2 c} \]

[Out]

1/4*e*(-b*e+2*c*d)*cosh(c*x^2+b*x+a)/c^2+1/2*e*(e*x+d)*cosh(c*x^2+b*x+a)/c-1/8*e^2*exp(-a+1/4*b^2/c)*erf(1/2*(
2*c*x+b)/c^(1/2))*Pi^(1/2)/c^(3/2)-1/16*(-b*e+2*c*d)^2*exp(-a+1/4*b^2/c)*erf(1/2*(2*c*x+b)/c^(1/2))*Pi^(1/2)/c
^(5/2)-1/8*e^2*exp(a-1/4*b^2/c)*erfi(1/2*(2*c*x+b)/c^(1/2))*Pi^(1/2)/c^(3/2)+1/16*(-b*e+2*c*d)^2*exp(a-1/4*b^2
/c)*erfi(1/2*(2*c*x+b)/c^(1/2))*Pi^(1/2)/c^(5/2)

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {5386, 5375, 2234, 2204, 2205, 5382, 5374} \[ -\frac {\sqrt {\pi } e^{\frac {b^2}{4 c}-a} (2 c d-b e)^2 \text {Erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{16 c^{5/2}}+\frac {\sqrt {\pi } e^{a-\frac {b^2}{4 c}} (2 c d-b e)^2 \text {Erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{16 c^{5/2}}-\frac {\sqrt {\pi } e^2 e^{\frac {b^2}{4 c}-a} \text {Erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{8 c^{3/2}}-\frac {\sqrt {\pi } e^2 e^{a-\frac {b^2}{4 c}} \text {Erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{8 c^{3/2}}+\frac {e (2 c d-b e) \cosh \left (a+b x+c x^2\right )}{4 c^2}+\frac {e (d+e x) \cosh \left (a+b x+c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(2*c*d - b*e)*Cosh[a + b*x + c*x^2])/(4*c^2) + (e*(d + e*x)*Cosh[a + b*x + c*x^2])/(2*c) - (e^2*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)^2*E^(-a + b^2/(4*c))*Sqrt[Pi]*Erf[
(b + 2*c*x)/(2*Sqrt[c])])/(16*c^(5/2)) - (e^2*E^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[(b + 2*c*x)/(2*Sqrt[c])])/(8*c^(
3/2)) + ((2*c*d - b*e)^2*E^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[(b + 2*c*x)/(2*Sqrt[c])])/(16*c^(5/2))

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 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 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]

Rule 5386

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

Rubi steps

\begin {align*} \int (d+e x)^2 \sinh \left (a+b x+c x^2\right ) \, dx &=\frac {e (d+e x) \cosh \left (a+b x+c x^2\right )}{2 c}-\frac {e^2 \int \cosh \left (a+b x+c x^2\right ) \, dx}{2 c}-\frac {(-2 c d+b e) \int (d+e x) \sinh \left (a+b x+c x^2\right ) \, dx}{2 c}\\ &=\frac {e (2 c d-b e) \cosh \left (a+b x+c x^2\right )}{4 c^2}+\frac {e (d+e x) \cosh \left (a+b x+c x^2\right )}{2 c}-\frac {e^2 \int e^{-a-b x-c x^2} \, dx}{4 c}-\frac {e^2 \int e^{a+b x+c x^2} \, dx}{4 c}+\frac {(2 c d-b e)^2 \int \sinh \left (a+b x+c x^2\right ) \, dx}{4 c^2}\\ &=\frac {e (2 c d-b e) \cosh \left (a+b x+c x^2\right )}{4 c^2}+\frac {e (d+e x) \cosh \left (a+b x+c x^2\right )}{2 c}-\frac {(2 c d-b e)^2 \int e^{-a-b x-c x^2} \, dx}{8 c^2}+\frac {(2 c d-b e)^2 \int e^{a+b x+c x^2} \, dx}{8 c^2}-\frac {\left (e^2 e^{a-\frac {b^2}{4 c}}\right ) \int e^{\frac {(b+2 c x)^2}{4 c}} \, dx}{4 c}-\frac {\left (e^2 e^{-a+\frac {b^2}{4 c}}\right ) \int e^{-\frac {(-b-2 c x)^2}{4 c}} \, dx}{4 c}\\ &=\frac {e (2 c d-b e) \cosh \left (a+b x+c x^2\right )}{4 c^2}+\frac {e (d+e x) \cosh \left (a+b x+c x^2\right )}{2 c}-\frac {e^2 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 {e^2 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}}+\frac {\left ((2 c d-b e)^2 e^{a-\frac {b^2}{4 c}}\right ) \int e^{\frac {(b+2 c x)^2}{4 c}} \, dx}{8 c^2}-\frac {\left ((2 c d-b e)^2 e^{-a+\frac {b^2}{4 c}}\right ) \int e^{-\frac {(-b-2 c x)^2}{4 c}} \, dx}{8 c^2}\\ &=\frac {e (2 c d-b e) \cosh \left (a+b x+c x^2\right )}{4 c^2}+\frac {e (d+e x) \cosh \left (a+b x+c x^2\right )}{2 c}-\frac {e^2 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)^2 e^{-a+\frac {b^2}{4 c}} \sqrt {\pi } \text {erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{16 c^{5/2}}-\frac {e^2 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}}+\frac {(2 c d-b e)^2 e^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{16 c^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.58, size = 194, normalized size = 0.74 \[ \frac {\sqrt {\pi } \left (b^2 e^2+2 c e (e-2 b d)+4 c^2 d^2\right ) \text {erf}\left (\frac {b+2 c x}{2 \sqrt {c}}\right ) \left (\sinh \left (a-\frac {b^2}{4 c}\right )-\cosh \left (a-\frac {b^2}{4 c}\right )\right )+\sqrt {\pi } \left (b^2 e^2-2 c e (2 b d+e)+4 c^2 d^2\right ) \text {erfi}\left (\frac {b+2 c x}{2 \sqrt {c}}\right ) \left (\sinh \left (a-\frac {b^2}{4 c}\right )+\cosh \left (a-\frac {b^2}{4 c}\right )\right )+4 \sqrt {c} e \cosh (a+x (b+c x)) (-b e+4 c d+2 c e x)}{16 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[c]*e*(4*c*d - b*e + 2*c*e*x)*Cosh[a + x*(b + c*x)] + (4*c^2*d^2 + b^2*e^2 + 2*c*e*(-2*b*d + e))*Sqrt[P
i]*Erf[(b + 2*c*x)/(2*Sqrt[c])]*(-Cosh[a - b^2/(4*c)] + Sinh[a - b^2/(4*c)]) + (4*c^2*d^2 + b^2*e^2 - 2*c*e*(2
*b*d + e))*Sqrt[Pi]*Erfi[(b + 2*c*x)/(2*Sqrt[c])]*(Cosh[a - b^2/(4*c)] + Sinh[a - b^2/(4*c)]))/(16*c^(5/2))

________________________________________________________________________________________

fricas [B]  time = 0.57, size = 634, normalized size = 2.43 \[ \frac {4 \, c^{2} e^{2} x + 8 \, c^{2} d e - 2 \, b c e^{2} + 2 \, {\left (2 \, c^{2} e^{2} x + 4 \, c^{2} d e - b c e^{2}\right )} \cosh \left (c x^{2} + b x + a\right )^{2} - \sqrt {\pi } {\left ({\left (4 \, c^{2} d^{2} - 4 \, b c d e + {\left (b^{2} - 2 \, c\right )} e^{2}\right )} \cosh \left (c x^{2} + b x + a\right ) \cosh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + {\left (4 \, c^{2} d^{2} - 4 \, b c d e + {\left (b^{2} - 2 \, c\right )} e^{2}\right )} \cosh \left (c x^{2} + b x + a\right ) \sinh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + {\left ({\left (4 \, c^{2} d^{2} - 4 \, b c d e + {\left (b^{2} - 2 \, c\right )} e^{2}\right )} \cosh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + {\left (4 \, c^{2} d^{2} - 4 \, b c d e + {\left (b^{2} - 2 \, c\right )} e^{2}\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 (4 \, c^{2} d^{2} - 4 \, b c d e + {\left (b^{2} + 2 \, c\right )} e^{2}\right )} \cosh \left (c x^{2} + b x + a\right ) \cosh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - {\left (4 \, c^{2} d^{2} - 4 \, b c d e + {\left (b^{2} + 2 \, c\right )} e^{2}\right )} \cosh \left (c x^{2} + b x + a\right ) \sinh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + {\left ({\left (4 \, c^{2} d^{2} - 4 \, b c d e + {\left (b^{2} + 2 \, c\right )} e^{2}\right )} \cosh \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - {\left (4 \, c^{2} d^{2} - 4 \, b c d e + {\left (b^{2} + 2 \, c\right )} e^{2}\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 ) + 4 \, {\left (2 \, c^{2} e^{2} x + 4 \, c^{2} d e - b c e^{2}\right )} \cosh \left (c x^{2} + b x + a\right ) \sinh \left (c x^{2} + b x + a\right ) + 2 \, {\left (2 \, c^{2} e^{2} x + 4 \, c^{2} d e - b c e^{2}\right )} \sinh \left (c x^{2} + b x + a\right )^{2}}{16 \, {\left (c^{3} \cosh \left (c x^{2} + b x + a\right ) + c^{3} \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)^2*sinh(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

1/16*(4*c^2*e^2*x + 8*c^2*d*e - 2*b*c*e^2 + 2*(2*c^2*e^2*x + 4*c^2*d*e - b*c*e^2)*cosh(c*x^2 + b*x + a)^2 - sq
rt(pi)*((4*c^2*d^2 - 4*b*c*d*e + (b^2 - 2*c)*e^2)*cosh(c*x^2 + b*x + a)*cosh(-1/4*(b^2 - 4*a*c)/c) + (4*c^2*d^
2 - 4*b*c*d*e + (b^2 - 2*c)*e^2)*cosh(c*x^2 + b*x + a)*sinh(-1/4*(b^2 - 4*a*c)/c) + ((4*c^2*d^2 - 4*b*c*d*e +
(b^2 - 2*c)*e^2)*cosh(-1/4*(b^2 - 4*a*c)/c) + (4*c^2*d^2 - 4*b*c*d*e + (b^2 - 2*c)*e^2)*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)*((4*c^2*d^2 - 4*b*c*d*e + (b^
2 + 2*c)*e^2)*cosh(c*x^2 + b*x + a)*cosh(-1/4*(b^2 - 4*a*c)/c) - (4*c^2*d^2 - 4*b*c*d*e + (b^2 + 2*c)*e^2)*cos
h(c*x^2 + b*x + a)*sinh(-1/4*(b^2 - 4*a*c)/c) + ((4*c^2*d^2 - 4*b*c*d*e + (b^2 + 2*c)*e^2)*cosh(-1/4*(b^2 - 4*
a*c)/c) - (4*c^2*d^2 - 4*b*c*d*e + (b^2 + 2*c)*e^2)*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)) + 4*(2*c^2*e^2*x + 4*c^2*d*e - b*c*e^2)*cosh(c*x^2 + b*x + a)*sinh(c*x^2 + b*x +
a) + 2*(2*c^2*e^2*x + 4*c^2*d*e - b*c*e^2)*sinh(c*x^2 + b*x + a)^2)/(c^3*cosh(c*x^2 + b*x + a) + c^3*sinh(c*x
^2 + b*x + a))

________________________________________________________________________________________

giac [A]  time = 0.17, size = 387, normalized size = 1.48 \[ \frac {\sqrt {\pi } d^{2} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (\frac {b^{2} - 4 \, a c}{4 \, c}\right )}}{4 \, \sqrt {c}} - \frac {\sqrt {\pi } d^{2} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )}}{4 \, \sqrt {-c}} - \frac {\frac {\sqrt {\pi } b d \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (\frac {b^{2} - 4 \, a c + 4 \, c}{4 \, c}\right )}}{\sqrt {c}} - 2 \, d e^{\left (-c x^{2} - b x - a + 1\right )}}{4 \, c} + \frac {\frac {\sqrt {\pi } b d \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} - 4 \, a c - 4 \, c}{4 \, c}\right )}}{\sqrt {-c}} + 2 \, d e^{\left (c x^{2} + b x + a + 1\right )}}{4 \, c} + \frac {\frac {\sqrt {\pi } {\left (b^{2} + 2 \, c\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (\frac {b^{2} - 4 \, a c + 8 \, c}{4 \, c}\right )}}{\sqrt {c}} + 2 \, {\left (c {\left (2 \, x + \frac {b}{c}\right )} - 2 \, b\right )} e^{\left (-c x^{2} - b x - a + 2\right )}}{16 \, c^{2}} - \frac {\frac {\sqrt {\pi } {\left (b^{2} - 2 \, c\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} - 4 \, a c - 8 \, c}{4 \, c}\right )}}{\sqrt {-c}} - 2 \, {\left (c {\left (2 \, x + \frac {b}{c}\right )} - 2 \, b\right )} e^{\left (c x^{2} + b x + a + 2\right )}}{16 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(pi)*d^2*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^2*erf(-1/2*sqr
t(-c)*(2*x + b/c))*e^(-1/4*(b^2 - 4*a*c)/c)/sqrt(-c) - 1/4*(sqrt(pi)*b*d*erf(-1/2*sqrt(c)*(2*x + b/c))*e^(1/4*
(b^2 - 4*a*c + 4*c)/c)/sqrt(c) - 2*d*e^(-c*x^2 - b*x - a + 1))/c + 1/4*(sqrt(pi)*b*d*erf(-1/2*sqrt(-c)*(2*x +
b/c))*e^(-1/4*(b^2 - 4*a*c - 4*c)/c)/sqrt(-c) + 2*d*e^(c*x^2 + b*x + a + 1))/c + 1/16*(sqrt(pi)*(b^2 + 2*c)*er
f(-1/2*sqrt(c)*(2*x + b/c))*e^(1/4*(b^2 - 4*a*c + 8*c)/c)/sqrt(c) + 2*(c*(2*x + b/c) - 2*b)*e^(-c*x^2 - b*x -
a + 2))/c^2 - 1/16*(sqrt(pi)*(b^2 - 2*c)*erf(-1/2*sqrt(-c)*(2*x + b/c))*e^(-1/4*(b^2 - 4*a*c - 8*c)/c)/sqrt(-c
) - 2*(c*(2*x + b/c) - 2*b)*e^(c*x^2 + b*x + a + 2))/c^2

________________________________________________________________________________________

maple [B]  time = 0.07, size = 493, normalized size = 1.89 \[ -\frac {d^{2} \sqrt {\pi }\, {\mathrm e}^{-\frac {4 a c -b^{2}}{4 c}} \erf \left (\sqrt {c}\, x +\frac {b}{2 \sqrt {c}}\right )}{4 \sqrt {c}}+\frac {e^{2} x \,{\mathrm e}^{-c \,x^{2}-b x -a}}{4 c}-\frac {e^{2} b \,{\mathrm e}^{-c \,x^{2}-b x -a}}{8 c^{2}}-\frac {e^{2} b^{2} \sqrt {\pi }\, {\mathrm e}^{-\frac {4 a c -b^{2}}{4 c}} \erf \left (\sqrt {c}\, x +\frac {b}{2 \sqrt {c}}\right )}{16 c^{\frac {5}{2}}}-\frac {e^{2} \sqrt {\pi }\, {\mathrm e}^{-\frac {4 a c -b^{2}}{4 c}} \erf \left (\sqrt {c}\, x +\frac {b}{2 \sqrt {c}}\right )}{8 c^{\frac {3}{2}}}+\frac {d e \,{\mathrm e}^{-c \,x^{2}-b x -a}}{2 c}+\frac {d e b \sqrt {\pi }\, {\mathrm e}^{-\frac {4 a c -b^{2}}{4 c}} \erf \left (\sqrt {c}\, x +\frac {b}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}-\frac {d^{2} \sqrt {\pi }\, {\mathrm e}^{\frac {4 a c -b^{2}}{4 c}} \erf \left (-\sqrt {-c}\, x +\frac {b}{2 \sqrt {-c}}\right )}{4 \sqrt {-c}}+\frac {e^{2} x \,{\mathrm e}^{c \,x^{2}+b x +a}}{4 c}-\frac {e^{2} b \,{\mathrm e}^{c \,x^{2}+b x +a}}{8 c^{2}}-\frac {e^{2} b^{2} \sqrt {\pi }\, {\mathrm e}^{\frac {4 a c -b^{2}}{4 c}} \erf \left (-\sqrt {-c}\, x +\frac {b}{2 \sqrt {-c}}\right )}{16 c^{2} \sqrt {-c}}+\frac {e^{2} \sqrt {\pi }\, {\mathrm e}^{\frac {4 a c -b^{2}}{4 c}} \erf \left (-\sqrt {-c}\, x +\frac {b}{2 \sqrt {-c}}\right )}{8 c \sqrt {-c}}+\frac {d e \,{\mathrm e}^{c \,x^{2}+b x +a}}{2 c}+\frac {d e b \sqrt {\pi }\, {\mathrm e}^{\frac {4 a c -b^{2}}{4 c}} \erf \left (-\sqrt {-c}\, x +\frac {b}{2 \sqrt {-c}}\right )}{4 c \sqrt {-c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*d^2*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^2/c*x*exp(-c*x^2-b*x-a)-1
/8*e^2/c^2*b*exp(-c*x^2-b*x-a)-1/16*e^2*b^2/c^(5/2)*Pi^(1/2)*exp(-1/4*(4*a*c-b^2)/c)*erf(c^(1/2)*x+1/2*b/c^(1/
2))-1/8*e^2/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))+1/2*d*e/c*exp(-c*x^2-b*x-a)+
1/4*d*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^2*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^2/c*x*exp(c*x^2+b*x+a)-1/8*e^2/c^2*b*exp(c*x^2+
b*x+a)-1/16*e^2/c^2*b^2*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))+1/8*e^2
/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))+1/2*d*e/c*exp(c*x^2+b*x+a)+1
/4*d*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))

________________________________________________________________________________________

maxima [B]  time = 1.72, size = 536, normalized size = 2.05 \[ \frac {\sqrt {\pi } d^{2} \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^{2} \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 )} d e e^{\left (a - \frac {b^{2}}{4 \, c}\right )}}{4 \, \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}} 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 )} e^{2} e^{\left (a - \frac {b^{2}}{4 \, c}\right )}}{16 \, \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 )} d e e^{\left (-a + \frac {b^{2}}{4 \, c}\right )}}{4 \, \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 )} e^{2} e^{\left (-a + \frac {b^{2}}{4 \, c}\right )}}{16 \, \sqrt {-c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(pi)*d^2*erf(sqrt(-c)*x - 1/2*b/sqrt(-c))*e^(a - 1/4*b^2/c)/sqrt(-c) - 1/4*sqrt(pi)*d^2*erf(sqrt(c)*x
+ 1/2*b/sqrt(c))*e^(-a + 1/4*b^2/c)/sqrt(c) - 1/4*(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))*d*e*e^(a - 1/4*b^2/c)/sqrt(c) + 1/16*(
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)))
*e^2*e^(a - 1/4*b^2/c)/sqrt(c) + 1/4*(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))*d*e*e^(-a + 1/4*b^2/c)/sqrt(-c) + 1/16*(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*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)))*e^2*e^(-a + 1/4*b^2/c)/sqrt(-c)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________