3.19 \(\int (c+d x) \sinh ^3(a+b x) \, dx\)

Optimal. Leaf size=75 \[ -\frac {d \sinh ^3(a+b x)}{9 b^2}+\frac {2 d \sinh (a+b x)}{3 b^2}-\frac {2 (c+d x) \cosh (a+b x)}{3 b}+\frac {(c+d x) \sinh ^2(a+b x) \cosh (a+b x)}{3 b} \]

[Out]

-2/3*(d*x+c)*cosh(b*x+a)/b+2/3*d*sinh(b*x+a)/b^2+1/3*(d*x+c)*cosh(b*x+a)*sinh(b*x+a)^2/b-1/9*d*sinh(b*x+a)^3/b
^2

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Rubi [A]  time = 0.06, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3310, 3296, 2637} \[ -\frac {d \sinh ^3(a+b x)}{9 b^2}+\frac {2 d \sinh (a+b x)}{3 b^2}-\frac {2 (c+d x) \cosh (a+b x)}{3 b}+\frac {(c+d x) \sinh ^2(a+b x) \cosh (a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sinh[a + b*x]^3,x]

[Out]

(-2*(c + d*x)*Cosh[a + b*x])/(3*b) + (2*d*Sinh[a + b*x])/(3*b^2) + ((c + d*x)*Cosh[a + b*x]*Sinh[a + b*x]^2)/(
3*b) - (d*Sinh[a + b*x]^3)/(9*b^2)

Rule 2637

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

Rule 3296

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 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rubi steps

\begin {align*} \int (c+d x) \sinh ^3(a+b x) \, dx &=\frac {(c+d x) \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac {d \sinh ^3(a+b x)}{9 b^2}-\frac {2}{3} \int (c+d x) \sinh (a+b x) \, dx\\ &=-\frac {2 (c+d x) \cosh (a+b x)}{3 b}+\frac {(c+d x) \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac {d \sinh ^3(a+b x)}{9 b^2}+\frac {(2 d) \int \cosh (a+b x) \, dx}{3 b}\\ &=-\frac {2 (c+d x) \cosh (a+b x)}{3 b}+\frac {2 d \sinh (a+b x)}{3 b^2}+\frac {(c+d x) \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac {d \sinh ^3(a+b x)}{9 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 59, normalized size = 0.79 \[ \frac {-27 b (c+d x) \cosh (a+b x)+3 b (c+d x) \cosh (3 (a+b x))+d (27 \sinh (a+b x)-\sinh (3 (a+b x)))}{36 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Sinh[a + b*x]^3,x]

[Out]

(-27*b*(c + d*x)*Cosh[a + b*x] + 3*b*(c + d*x)*Cosh[3*(a + b*x)] + d*(27*Sinh[a + b*x] - Sinh[3*(a + b*x)]))/(
36*b^2)

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fricas [A]  time = 0.59, size = 97, normalized size = 1.29 \[ \frac {3 \, {\left (b d x + b c\right )} \cosh \left (b x + a\right )^{3} + 9 \, {\left (b d x + b c\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - d \sinh \left (b x + a\right )^{3} - 27 \, {\left (b d x + b c\right )} \cosh \left (b x + a\right ) - 3 \, {\left (d \cosh \left (b x + a\right )^{2} - 9 \, d\right )} \sinh \left (b x + a\right )}{36 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/36*(3*(b*d*x + b*c)*cosh(b*x + a)^3 + 9*(b*d*x + b*c)*cosh(b*x + a)*sinh(b*x + a)^2 - d*sinh(b*x + a)^3 - 27
*(b*d*x + b*c)*cosh(b*x + a) - 3*(d*cosh(b*x + a)^2 - 9*d)*sinh(b*x + a))/b^2

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giac [A]  time = 0.20, size = 98, normalized size = 1.31 \[ \frac {{\left (3 \, b d x + 3 \, b c - d\right )} e^{\left (3 \, b x + 3 \, a\right )}}{72 \, b^{2}} - \frac {3 \, {\left (b d x + b c - d\right )} e^{\left (b x + a\right )}}{8 \, b^{2}} - \frac {3 \, {\left (b d x + b c + d\right )} e^{\left (-b x - a\right )}}{8 \, b^{2}} + \frac {{\left (3 \, b d x + 3 \, b c + d\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{72 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

1/72*(3*b*d*x + 3*b*c - d)*e^(3*b*x + 3*a)/b^2 - 3/8*(b*d*x + b*c - d)*e^(b*x + a)/b^2 - 3/8*(b*d*x + b*c + d)
*e^(-b*x - a)/b^2 + 1/72*(3*b*d*x + 3*b*c + d)*e^(-3*b*x - 3*a)/b^2

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maple [A]  time = 0.02, size = 109, normalized size = 1.45 \[ \frac {\frac {d \left (-\frac {2 \left (b x +a \right ) \cosh \left (b x +a \right )}{3}+\frac {\left (b x +a \right ) \cosh \left (b x +a \right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{3}+\frac {2 \sinh \left (b x +a \right )}{3}-\frac {\left (\sinh ^{3}\left (b x +a \right )\right )}{9}\right )}{b}-\frac {d a \left (-\frac {2}{3}+\frac {\left (\sinh ^{2}\left (b x +a \right )\right )}{3}\right ) \cosh \left (b x +a \right )}{b}+c \left (-\frac {2}{3}+\frac {\left (\sinh ^{2}\left (b x +a \right )\right )}{3}\right ) \cosh \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*sinh(b*x+a)^3,x)

[Out]

1/b*(1/b*d*(-2/3*(b*x+a)*cosh(b*x+a)+1/3*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)^2+2/3*sinh(b*x+a)-1/9*sinh(b*x+a)^3)-
1/b*d*a*(-2/3+1/3*sinh(b*x+a)^2)*cosh(b*x+a)+c*(-2/3+1/3*sinh(b*x+a)^2)*cosh(b*x+a))

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maxima [B]  time = 0.45, size = 141, normalized size = 1.88 \[ \frac {1}{72} \, d {\left (\frac {{\left (3 \, b x e^{\left (3 \, a\right )} - e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{b^{2}} - \frac {27 \, {\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{b^{2}} - \frac {27 \, {\left (b x + 1\right )} e^{\left (-b x - a\right )}}{b^{2}} + \frac {{\left (3 \, b x + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{b^{2}}\right )} + \frac {1}{24} \, c {\left (\frac {e^{\left (3 \, b x + 3 \, a\right )}}{b} - \frac {9 \, e^{\left (b x + a\right )}}{b} - \frac {9 \, e^{\left (-b x - a\right )}}{b} + \frac {e^{\left (-3 \, b x - 3 \, a\right )}}{b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

1/72*d*((3*b*x*e^(3*a) - e^(3*a))*e^(3*b*x)/b^2 - 27*(b*x*e^a - e^a)*e^(b*x)/b^2 - 27*(b*x + 1)*e^(-b*x - a)/b
^2 + (3*b*x + 1)*e^(-3*b*x - 3*a)/b^2) + 1/24*c*(e^(3*b*x + 3*a)/b - 9*e^(b*x + a)/b - 9*e^(-b*x - a)/b + e^(-
3*b*x - 3*a)/b)

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mupad [B]  time = 0.16, size = 79, normalized size = 1.05 \[ \frac {7\,d\,\mathrm {sinh}\left (a+b\,x\right )}{9\,b^2}-\frac {c\,\mathrm {cosh}\left (a+b\,x\right )-\frac {c\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{3}+d\,x\,\mathrm {cosh}\left (a+b\,x\right )-\frac {d\,x\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{3}}{b}-\frac {d\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {sinh}\left (a+b\,x\right )}{9\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)^3*(c + d*x),x)

[Out]

(7*d*sinh(a + b*x))/(9*b^2) - (c*cosh(a + b*x) - (c*cosh(a + b*x)^3)/3 + d*x*cosh(a + b*x) - (d*x*cosh(a + b*x
)^3)/3)/b - (d*cosh(a + b*x)^2*sinh(a + b*x))/(9*b^2)

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sympy [A]  time = 0.93, size = 126, normalized size = 1.68 \[ \begin {cases} \frac {c \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{b} - \frac {2 c \cosh ^{3}{\left (a + b x \right )}}{3 b} + \frac {d x \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{b} - \frac {2 d x \cosh ^{3}{\left (a + b x \right )}}{3 b} - \frac {7 d \sinh ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac {2 d \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sinh ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sinh(b*x+a)**3,x)

[Out]

Piecewise((c*sinh(a + b*x)**2*cosh(a + b*x)/b - 2*c*cosh(a + b*x)**3/(3*b) + d*x*sinh(a + b*x)**2*cosh(a + b*x
)/b - 2*d*x*cosh(a + b*x)**3/(3*b) - 7*d*sinh(a + b*x)**3/(9*b**2) + 2*d*sinh(a + b*x)*cosh(a + b*x)**2/(3*b**
2), Ne(b, 0)), ((c*x + d*x**2/2)*sinh(a)**3, True))

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