3.1.18 \(\int x^3 \text {Shi}(a+b x) \, dx\) [18]

Optimal. Leaf size=184 \[ \frac {a \cosh (a+b x)}{2 b^4}+\frac {a^3 \cosh (a+b x)}{4 b^4}-\frac {3 x \cosh (a+b x)}{2 b^3}-\frac {a^2 x \cosh (a+b x)}{4 b^3}+\frac {a x^2 \cosh (a+b x)}{4 b^2}-\frac {x^3 \cosh (a+b x)}{4 b}+\frac {3 \sinh (a+b x)}{2 b^4}+\frac {a^2 \sinh (a+b x)}{4 b^4}-\frac {a x \sinh (a+b x)}{2 b^3}+\frac {3 x^2 \sinh (a+b x)}{4 b^2}-\frac {a^4 \text {Shi}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \text {Shi}(a+b x) \]

[Out]

1/2*a*cosh(b*x+a)/b^4+1/4*a^3*cosh(b*x+a)/b^4-3/2*x*cosh(b*x+a)/b^3-1/4*a^2*x*cosh(b*x+a)/b^3+1/4*a*x^2*cosh(b
*x+a)/b^2-1/4*x^3*cosh(b*x+a)/b-1/4*a^4*Shi(b*x+a)/b^4+1/4*x^4*Shi(b*x+a)+3/2*sinh(b*x+a)/b^4+1/4*a^2*sinh(b*x
+a)/b^4-1/2*a*x*sinh(b*x+a)/b^3+3/4*x^2*sinh(b*x+a)/b^2

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Rubi [A]
time = 0.28, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6667, 6874, 2718, 3377, 2717, 3379} \begin {gather*} -\frac {a^4 \text {Shi}(a+b x)}{4 b^4}+\frac {a^3 \cosh (a+b x)}{4 b^4}+\frac {a^2 \sinh (a+b x)}{4 b^4}-\frac {a^2 x \cosh (a+b x)}{4 b^3}+\frac {3 \sinh (a+b x)}{2 b^4}+\frac {a \cosh (a+b x)}{2 b^4}-\frac {a x \sinh (a+b x)}{2 b^3}-\frac {3 x \cosh (a+b x)}{2 b^3}+\frac {3 x^2 \sinh (a+b x)}{4 b^2}+\frac {a x^2 \cosh (a+b x)}{4 b^2}+\frac {1}{4} x^4 \text {Shi}(a+b x)-\frac {x^3 \cosh (a+b x)}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*SinhIntegral[a + b*x],x]

[Out]

(a*Cosh[a + b*x])/(2*b^4) + (a^3*Cosh[a + b*x])/(4*b^4) - (3*x*Cosh[a + b*x])/(2*b^3) - (a^2*x*Cosh[a + b*x])/
(4*b^3) + (a*x^2*Cosh[a + b*x])/(4*b^2) - (x^3*Cosh[a + b*x])/(4*b) + (3*Sinh[a + b*x])/(2*b^4) + (a^2*Sinh[a
+ b*x])/(4*b^4) - (a*x*Sinh[a + b*x])/(2*b^3) + (3*x^2*Sinh[a + b*x])/(4*b^2) - (a^4*SinhIntegral[a + b*x])/(4
*b^4) + (x^4*SinhIntegral[a + b*x])/4

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 6667

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

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int x^3 \text {Shi}(a+b x) \, dx &=\frac {1}{4} x^4 \text {Shi}(a+b x)-\frac {1}{4} b \int \frac {x^4 \sinh (a+b x)}{a+b x} \, dx\\ &=\frac {1}{4} x^4 \text {Shi}(a+b x)-\frac {1}{4} b \int \left (-\frac {a^3 \sinh (a+b x)}{b^4}+\frac {a^2 x \sinh (a+b x)}{b^3}-\frac {a x^2 \sinh (a+b x)}{b^2}+\frac {x^3 \sinh (a+b x)}{b}+\frac {a^4 \sinh (a+b x)}{b^4 (a+b x)}\right ) \, dx\\ &=\frac {1}{4} x^4 \text {Shi}(a+b x)-\frac {1}{4} \int x^3 \sinh (a+b x) \, dx+\frac {a^3 \int \sinh (a+b x) \, dx}{4 b^3}-\frac {a^4 \int \frac {\sinh (a+b x)}{a+b x} \, dx}{4 b^3}-\frac {a^2 \int x \sinh (a+b x) \, dx}{4 b^2}+\frac {a \int x^2 \sinh (a+b x) \, dx}{4 b}\\ &=\frac {a^3 \cosh (a+b x)}{4 b^4}-\frac {a^2 x \cosh (a+b x)}{4 b^3}+\frac {a x^2 \cosh (a+b x)}{4 b^2}-\frac {x^3 \cosh (a+b x)}{4 b}-\frac {a^4 \text {Shi}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \text {Shi}(a+b x)+\frac {a^2 \int \cosh (a+b x) \, dx}{4 b^3}-\frac {a \int x \cosh (a+b x) \, dx}{2 b^2}+\frac {3 \int x^2 \cosh (a+b x) \, dx}{4 b}\\ &=\frac {a^3 \cosh (a+b x)}{4 b^4}-\frac {a^2 x \cosh (a+b x)}{4 b^3}+\frac {a x^2 \cosh (a+b x)}{4 b^2}-\frac {x^3 \cosh (a+b x)}{4 b}+\frac {a^2 \sinh (a+b x)}{4 b^4}-\frac {a x \sinh (a+b x)}{2 b^3}+\frac {3 x^2 \sinh (a+b x)}{4 b^2}-\frac {a^4 \text {Shi}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \text {Shi}(a+b x)+\frac {a \int \sinh (a+b x) \, dx}{2 b^3}-\frac {3 \int x \sinh (a+b x) \, dx}{2 b^2}\\ &=\frac {a \cosh (a+b x)}{2 b^4}+\frac {a^3 \cosh (a+b x)}{4 b^4}-\frac {3 x \cosh (a+b x)}{2 b^3}-\frac {a^2 x \cosh (a+b x)}{4 b^3}+\frac {a x^2 \cosh (a+b x)}{4 b^2}-\frac {x^3 \cosh (a+b x)}{4 b}+\frac {a^2 \sinh (a+b x)}{4 b^4}-\frac {a x \sinh (a+b x)}{2 b^3}+\frac {3 x^2 \sinh (a+b x)}{4 b^2}-\frac {a^4 \text {Shi}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \text {Shi}(a+b x)+\frac {3 \int \cosh (a+b x) \, dx}{2 b^3}\\ &=\frac {a \cosh (a+b x)}{2 b^4}+\frac {a^3 \cosh (a+b x)}{4 b^4}-\frac {3 x \cosh (a+b x)}{2 b^3}-\frac {a^2 x \cosh (a+b x)}{4 b^3}+\frac {a x^2 \cosh (a+b x)}{4 b^2}-\frac {x^3 \cosh (a+b x)}{4 b}+\frac {3 \sinh (a+b x)}{2 b^4}+\frac {a^2 \sinh (a+b x)}{4 b^4}-\frac {a x \sinh (a+b x)}{2 b^3}+\frac {3 x^2 \sinh (a+b x)}{4 b^2}-\frac {a^4 \text {Shi}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \text {Shi}(a+b x)\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 94, normalized size = 0.51 \begin {gather*} \frac {\left (2 a+a^3-6 b x-a^2 b x+a b^2 x^2-b^3 x^3\right ) \cosh (a+b x)+\left (6+a^2-2 a b x+3 b^2 x^2\right ) \sinh (a+b x)+\left (-a^4+b^4 x^4\right ) \text {Shi}(a+b x)}{4 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*SinhIntegral[a + b*x],x]

[Out]

((2*a + a^3 - 6*b*x - a^2*b*x + a*b^2*x^2 - b^3*x^3)*Cosh[a + b*x] + (6 + a^2 - 2*a*b*x + 3*b^2*x^2)*Sinh[a +
b*x] + (-a^4 + b^4*x^4)*SinhIntegral[a + b*x])/(4*b^4)

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Maple [A]
time = 0.31, size = 156, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {\frac {\hyperbolicSineIntegral \left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \hyperbolicSineIntegral \left (b x +a \right )}{4}+a^{3} \cosh \left (b x +a \right )-\frac {3 a^{2} \left (\left (b x +a \right ) \cosh \left (b x +a \right )-\sinh \left (b x +a \right )\right )}{2}+a \left (\left (b x +a \right )^{2} \cosh \left (b x +a \right )-2 \left (b x +a \right ) \sinh \left (b x +a \right )+2 \cosh \left (b x +a \right )\right )-\frac {\left (b x +a \right )^{3} \cosh \left (b x +a \right )}{4}+\frac {3 \left (b x +a \right )^{2} \sinh \left (b x +a \right )}{4}-\frac {3 \left (b x +a \right ) \cosh \left (b x +a \right )}{2}+\frac {3 \sinh \left (b x +a \right )}{2}}{b^{4}}\)	\(156\)
default	\(\frac {\frac {\hyperbolicSineIntegral \left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \hyperbolicSineIntegral \left (b x +a \right )}{4}+a^{3} \cosh \left (b x +a \right )-\frac {3 a^{2} \left (\left (b x +a \right ) \cosh \left (b x +a \right )-\sinh \left (b x +a \right )\right )}{2}+a \left (\left (b x +a \right )^{2} \cosh \left (b x +a \right )-2 \left (b x +a \right ) \sinh \left (b x +a \right )+2 \cosh \left (b x +a \right )\right )-\frac {\left (b x +a \right )^{3} \cosh \left (b x +a \right )}{4}+\frac {3 \left (b x +a \right )^{2} \sinh \left (b x +a \right )}{4}-\frac {3 \left (b x +a \right ) \cosh \left (b x +a \right )}{2}+\frac {3 \sinh \left (b x +a \right )}{2}}{b^{4}}\)	\(156\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*Shi(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/b^4*(1/4*Shi(b*x+a)*b^4*x^4-1/4*a^4*Shi(b*x+a)+a^3*cosh(b*x+a)-3/2*a^2*((b*x+a)*cosh(b*x+a)-sinh(b*x+a))+a*(
(b*x+a)^2*cosh(b*x+a)-2*(b*x+a)*sinh(b*x+a)+2*cosh(b*x+a))-1/4*(b*x+a)^3*cosh(b*x+a)+3/4*(b*x+a)^2*sinh(b*x+a)
-3/2*(b*x+a)*cosh(b*x+a)+3/2*sinh(b*x+a))

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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(x^3*Shi(b*x+a),x, algorithm="maxima")

[Out]

integrate(x^3*Shi(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(x^3*Shi(b*x+a),x, algorithm="fricas")

[Out]

integral(x^3*sinh_integral(b*x + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \operatorname {Shi}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*Shi(b*x+a),x)

[Out]

Integral(x**3*Shi(a + b*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(x^3*Shi(b*x+a),x, algorithm="giac")

[Out]

integrate(x^3*Shi(b*x + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\mathrm {sinhint}\left (a+b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinhint(a + b*x),x)

[Out]

int(x^3*sinhint(a + b*x), x)

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