3.1.4 \(\int (c+d x) \sinh (a+b x) \, dx\) [4]

Optimal. Leaf size=28 \[ \frac {(c+d x) \cosh (a+b x)}{b}-\frac {d \sinh (a+b x)}{b^2} \]

[Out]

(d*x+c)*cosh(b*x+a)/b-d*sinh(b*x+a)/b^2

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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3377, 2717} \begin {gather*} \frac {(c+d x) \cosh (a+b x)}{b}-\frac {d \sinh (a+b x)}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sinh[a + b*x],x]

[Out]

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 27, normalized size = 0.96 \begin {gather*} \frac {b (c+d x) \cosh (a+b x)-d \sinh (a+b x)}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Sinh[a + b*x],x]

[Out]

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

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Maple [A]
time = 0.42, size = 53, normalized size = 1.89

method	result	size
risch	\(\frac {\left (b d x +b c -d \right ) {\mathrm e}^{b x +a}}{2 b^{2}}+\frac {\left (b d x +b c +d \right ) {\mathrm e}^{-b x -a}}{2 b^{2}}\)	\(47\)
derivativedivides	\(\frac {\frac {d \left (\left (b x +a \right ) \cosh \left (b x +a \right )-\sinh \left (b x +a \right )\right )}{b}-\frac {d a \cosh \left (b x +a \right )}{b}+c \cosh \left (b x +a \right )}{b}\)	\(53\)
default	\(\frac {\frac {d \left (\left (b x +a \right ) \cosh \left (b x +a \right )-\sinh \left (b x +a \right )\right )}{b}-\frac {d a \cosh \left (b x +a \right )}{b}+c \cosh \left (b x +a \right )}{b}\)	\(53\)
meijerg	\(-\frac {2 d \sinh \left (a \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cosh \left (b x \right )}{2 \sqrt {\pi }}-\frac {b x \sinh \left (b x \right )}{2 \sqrt {\pi }}\right )}{b^{2}}+\frac {d \cosh \left (a \right ) \left (\cosh \left (b x \right ) b x -\sinh \left (b x \right )\right )}{b^{2}}+\frac {c \sinh \left (a \right ) \sinh \left (b x \right )}{b}-\frac {c \cosh \left (a \right ) \sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cosh \left (b x \right )}{\sqrt {\pi }}\right )}{b}\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*sinh(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (28) = 56\).
time = 0.26, size = 68, normalized size = 2.43 \begin {gather*} \frac {c e^{\left (b x + a\right )}}{2 \, b} + \frac {{\left (b x e^{a} - e^{a}\right )} d e^{\left (b x\right )}}{2 \, b^{2}} + \frac {c e^{\left (-b x - a\right )}}{2 \, b} + \frac {{\left (b x + 1\right )} d e^{\left (-b x - a\right )}}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c*e^(b*x + a)/b + 1/2*(b*x*e^a - e^a)*d*e^(b*x)/b^2 + 1/2*c*e^(-b*x - a)/b + 1/2*(b*x + 1)*d*e^(-b*x - a)/
b^2

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Fricas [A]
time = 0.37, size = 29, normalized size = 1.04 \begin {gather*} \frac {{\left (b d x + b c\right )} \cosh \left (b x + a\right ) - d \sinh \left (b x + a\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*d*x + b*c)*cosh(b*x + a) - d*sinh(b*x + a))/b^2

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Sympy [A]
time = 0.09, size = 46, normalized size = 1.64 \begin {gather*} \begin {cases} \frac {c \cosh {\left (a + b x \right )}}{b} + \frac {d x \cosh {\left (a + b x \right )}}{b} - \frac {d \sinh {\left (a + b x \right )}}{b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sinh {\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c*cosh(a + b*x)/b + d*x*cosh(a + b*x)/b - d*sinh(a + b*x)/b**2, Ne(b, 0)), ((c*x + d*x**2/2)*sinh(a
), True))

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Giac [A]
time = 0.41, size = 46, normalized size = 1.64 \begin {gather*} \frac {{\left (b d x + b c - d\right )} e^{\left (b x + a\right )}}{2 \, b^{2}} + \frac {{\left (b d x + b c + d\right )} e^{\left (-b x - a\right )}}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*d*x + b*c - d)*e^(b*x + a)/b^2 + 1/2*(b*d*x + b*c + d)*e^(-b*x - a)/b^2

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Mupad [B]
time = 0.10, size = 35, normalized size = 1.25 \begin {gather*} \frac {c\,\mathrm {cosh}\left (a+b\,x\right )+d\,x\,\mathrm {cosh}\left (a+b\,x\right )}{b}-\frac {d\,\mathrm {sinh}\left (a+b\,x\right )}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)*(c + d*x),x)

[Out]

(c*cosh(a + b*x) + d*x*cosh(a + b*x))/b - (d*sinh(a + b*x))/b^2

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