3.1.64 \(\int \sinh (a+b x) \text {Shi}(c+d x) \, dx\) [64]

Optimal. Leaf size=153 \[ \frac {\text {Chi}\left (\frac {c (b-d)}{d}+(b-d) x\right ) \sinh \left (a-\frac {b c}{d}\right )}{2 b}-\frac {\text {Chi}\left (\frac {c (b+d)}{d}+(b+d) x\right ) \sinh \left (a-\frac {b c}{d}\right )}{2 b}+\frac {\cosh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (\frac {c (b-d)}{d}+(b-d) x\right )}{2 b}+\frac {\cosh (a+b x) \text {Shi}(c+d x)}{b}-\frac {\cosh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (\frac {c (b+d)}{d}+(b+d) x\right )}{2 b} \]

[Out]

1/2*cosh(a-b*c/d)*Shi(c*(b-d)/d+(b-d)*x)/b+cosh(b*x+a)*Shi(d*x+c)/b-1/2*cosh(a-b*c/d)*Shi(c*(b+d)/d+(b+d)*x)/b
+1/2*Chi(c*(b-d)/d+(b-d)*x)*sinh(a-b*c/d)/b-1/2*Chi(c*(b+d)/d+(b+d)*x)*sinh(a-b*c/d)/b

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Rubi [A]
time = 0.18, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6675, 5580, 3384, 3379, 3382} \begin {gather*} \frac {\sinh \left (a-\frac {b c}{d}\right ) \text {Chi}\left (x (b-d)+\frac {c (b-d)}{d}\right )}{2 b}-\frac {\sinh \left (a-\frac {b c}{d}\right ) \text {Chi}\left (x (b+d)+\frac {c (b+d)}{d}\right )}{2 b}+\frac {\cosh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (x (b-d)+\frac {c (b-d)}{d}\right )}{2 b}+\frac {\cosh (a+b x) \text {Shi}(c+d x)}{b}-\frac {\cosh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (x (b+d)+\frac {c (b+d)}{d}\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]*SinhIntegral[c + d*x],x]

[Out]

(CoshIntegral[(c*(b - d))/d + (b - d)*x]*Sinh[a - (b*c)/d])/(2*b) - (CoshIntegral[(c*(b + d))/d + (b + d)*x]*S
inh[a - (b*c)/d])/(2*b) + (Cosh[a - (b*c)/d]*SinhIntegral[(c*(b - d))/d + (b - d)*x])/(2*b) + (Cosh[a + b*x]*S
inhIntegral[c + d*x])/b - (Cosh[a - (b*c)/d]*SinhIntegral[(c*(b + d))/d + (b + d)*x])/(2*b)

Rule 3379

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[I*(SinhIntegral[c*f*(fz/
d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3382

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[c*f*(fz/d)
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 5580

Int[Cosh[(c_.) + (d_.)*(x_)]^(q_.)*((e_.) + (f_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Int
[ExpandTrigReduce[(e + f*x)^m, Sinh[a + b*x]^p*Cosh[c + d*x]^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && I
GtQ[p, 0] && IGtQ[q, 0]

Rule 6675

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

Rubi steps

\begin {align*} \int \sinh (a+b x) \text {Shi}(c+d x) \, dx &=\frac {\cosh (a+b x) \text {Shi}(c+d x)}{b}-\frac {d \int \frac {\cosh (a+b x) \sinh (c+d x)}{c+d x} \, dx}{b}\\ &=\frac {\cosh (a+b x) \text {Shi}(c+d x)}{b}-\frac {d \int \left (-\frac {\sinh (a-c+(b-d) x)}{2 (c+d x)}+\frac {\sinh (a+c+(b+d) x)}{2 (c+d x)}\right ) \, dx}{b}\\ &=\frac {\cosh (a+b x) \text {Shi}(c+d x)}{b}+\frac {d \int \frac {\sinh (a-c+(b-d) x)}{c+d x} \, dx}{2 b}-\frac {d \int \frac {\sinh (a+c+(b+d) x)}{c+d x} \, dx}{2 b}\\ &=\frac {\cosh (a+b x) \text {Shi}(c+d x)}{b}+\frac {\left (d \cosh \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sinh \left (\frac {c (b-d)}{d}+(b-d) x\right )}{c+d x} \, dx}{2 b}-\frac {\left (d \cosh \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sinh \left (\frac {c (b+d)}{d}+(b+d) x\right )}{c+d x} \, dx}{2 b}+\frac {\left (d \sinh \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cosh \left (\frac {c (b-d)}{d}+(b-d) x\right )}{c+d x} \, dx}{2 b}-\frac {\left (d \sinh \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cosh \left (\frac {c (b+d)}{d}+(b+d) x\right )}{c+d x} \, dx}{2 b}\\ &=\frac {\text {Chi}\left (\frac {c (b-d)}{d}+(b-d) x\right ) \sinh \left (a-\frac {b c}{d}\right )}{2 b}-\frac {\text {Chi}\left (\frac {c (b+d)}{d}+(b+d) x\right ) \sinh \left (a-\frac {b c}{d}\right )}{2 b}+\frac {\cosh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (\frac {c (b-d)}{d}+(b-d) x\right )}{2 b}+\frac {\cosh (a+b x) \text {Shi}(c+d x)}{b}-\frac {\cosh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (\frac {c (b+d)}{d}+(b+d) x\right )}{2 b}\\ \end {align*}

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Mathematica [A]
time = 1.61, size = 209, normalized size = 1.37 \begin {gather*} \frac {2 \text {Chi}\left (-\frac {(b-d) (c+d x)}{d}\right ) \sinh \left (a-\frac {b c}{d}\right )-2 \text {Chi}\left (\frac {(b+d) (c+d x)}{d}\right ) \sinh \left (a-\frac {b c}{d}\right )+4 \cosh (a+b x) \text {Shi}(c+d x)+\cosh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (\frac {(b-d) (c+d x)}{d}\right )+\sinh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (\frac {(b-d) (c+d x)}{d}\right )-2 \cosh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (\frac {(b+d) (c+d x)}{d}\right )-\cosh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (c-\frac {b c}{d}-b x+d x\right )+\sinh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (c-\frac {b c}{d}-b x+d x\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]*SinhIntegral[c + d*x],x]

[Out]

(2*CoshIntegral[-(((b - d)*(c + d*x))/d)]*Sinh[a - (b*c)/d] - 2*CoshIntegral[((b + d)*(c + d*x))/d]*Sinh[a - (
b*c)/d] + 4*Cosh[a + b*x]*SinhIntegral[c + d*x] + Cosh[a - (b*c)/d]*SinhIntegral[((b - d)*(c + d*x))/d] + Sinh
[a - (b*c)/d]*SinhIntegral[((b - d)*(c + d*x))/d] - 2*Cosh[a - (b*c)/d]*SinhIntegral[((b + d)*(c + d*x))/d] -
Cosh[a - (b*c)/d]*SinhIntegral[c - (b*c)/d - b*x + d*x] + Sinh[a - (b*c)/d]*SinhIntegral[c - (b*c)/d - b*x + d
*x])/(4*b)

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Maple [F]
time = 1.66, size = 0, normalized size = 0.00 \[\int \hyperbolicSineIntegral \left (d x +c \right ) \sinh \left (b x +a \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(Shi(d*x+c)*sinh(b*x+a),x)

[Out]

int(Shi(d*x+c)*sinh(b*x+a),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sinh(b*x + a)*sinh_integral(d*x + c), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sinh {\left (a + b x \right )} \operatorname {Shi}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \mathrm {sinhint}\left (c+d\,x\right )\,\mathrm {sinh}\left (a+b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinhint(c + d*x)*sinh(a + b*x),x)

[Out]

int(sinhint(c + d*x)*sinh(a + b*x), x)

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