3.66 \(\int \cosh (a+b \sqrt [3]{c+d x}) \, dx\)

Optimal. Leaf size=85 \[ \frac{6 \sinh \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac{6 \sqrt [3]{c+d x} \cosh \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac{3 (c+d x)^{2/3} \sinh \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

[Out]

(-6*(c + d*x)^(1/3)*Cosh[a + b*(c + d*x)^(1/3)])/(b^2*d) + (6*Sinh[a + b*(c + d*x)^(1/3)])/(b^3*d) + (3*(c + d
*x)^(2/3)*Sinh[a + b*(c + d*x)^(1/3)])/(b*d)

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Rubi [A]  time = 0.0797936, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5311, 5305, 3296, 2637} \[ \frac{6 \sinh \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac{6 \sqrt [3]{c+d x} \cosh \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac{3 (c+d x)^{2/3} \sinh \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*(c + d*x)^(1/3)],x]

[Out]

(-6*(c + d*x)^(1/3)*Cosh[a + b*(c + d*x)^(1/3)])/(b^2*d) + (6*Sinh[a + b*(c + d*x)^(1/3)])/(b^3*d) + (3*(c + d
*x)^(2/3)*Sinh[a + b*(c + d*x)^(1/3)])/(b*d)

Rule 5311

Int[((a_.) + Cosh[(c_.) + (d_.)*(u_)^(n_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/Coefficient[u, x, 1], Subst[Int[(
a + b*Cosh[c + d*x^n])^p, x], x, u], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[p] && LinearQ[u, x] && NeQ[u,
x]

Rule 5305

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.), x_Symbol] :> Module[{k = Denominator[n]}, Dist[k, Sub
st[Int[x^(k - 1)*(a + b*Cosh[c + d*x^(k*n)])^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && FractionQ[n]
&& IntegerQ[p]

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 2637

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

Rubi steps

\begin{align*} \int \cosh \left (a+b \sqrt [3]{c+d x}\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \cosh \left (a+b \sqrt [3]{x}\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{3 \operatorname{Subst}\left (\int x^2 \cosh (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{d}\\ &=\frac{3 (c+d x)^{2/3} \sinh \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac{6 \operatorname{Subst}\left (\int x \sinh (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b d}\\ &=-\frac{6 \sqrt [3]{c+d x} \cosh \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac{3 (c+d x)^{2/3} \sinh \left (a+b \sqrt [3]{c+d x}\right )}{b d}+\frac{6 \operatorname{Subst}\left (\int \cosh (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^2 d}\\ &=-\frac{6 \sqrt [3]{c+d x} \cosh \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac{6 \sinh \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac{3 (c+d x)^{2/3} \sinh \left (a+b \sqrt [3]{c+d x}\right )}{b d}\\ \end{align*}

Mathematica [A]  time = 0.0767914, size = 65, normalized size = 0.76 \[ \frac{3 \left (b^2 (c+d x)^{2/3}+2\right ) \sinh \left (a+b \sqrt [3]{c+d x}\right )-6 b \sqrt [3]{c+d x} \cosh \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*(c + d*x)^(1/3)],x]

[Out]

(-6*b*(c + d*x)^(1/3)*Cosh[a + b*(c + d*x)^(1/3)] + 3*(2 + b^2*(c + d*x)^(2/3))*Sinh[a + b*(c + d*x)^(1/3)])/(
b^3*d)

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Maple [A]  time = 0.011, size = 133, normalized size = 1.6 \begin{align*} 3\,{\frac{ \left ( a+b\sqrt [3]{dx+c} \right ) ^{2}\sinh \left ( a+b\sqrt [3]{dx+c} \right ) -2\, \left ( a+b\sqrt [3]{dx+c} \right ) \cosh \left ( a+b\sqrt [3]{dx+c} \right ) +2\,\sinh \left ( a+b\sqrt [3]{dx+c} \right ) -2\,a \left ( \left ( a+b\sqrt [3]{dx+c} \right ) \sinh \left ( a+b\sqrt [3]{dx+c} \right ) -\cosh \left ( a+b\sqrt [3]{dx+c} \right ) \right ) +{a}^{2}\sinh \left ( a+b\sqrt [3]{dx+c} \right ) }{d{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a+b*(d*x+c)^(1/3)),x)

[Out]

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

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Maxima [A]  time = 1.07512, size = 184, normalized size = 2.16 \begin{align*} -\frac{b{\left (\frac{{\left ({\left (d x + c\right )} b^{3} e^{a} - 3 \,{\left (d x + c\right )}^{\frac{2}{3}} b^{2} e^{a} + 6 \,{\left (d x + c\right )}^{\frac{1}{3}} b e^{a} - 6 \, e^{a}\right )} e^{\left ({\left (d x + c\right )}^{\frac{1}{3}} b\right )}}{b^{4}} + \frac{{\left ({\left (d x + c\right )} b^{3} + 3 \,{\left (d x + c\right )}^{\frac{2}{3}} b^{2} + 6 \,{\left (d x + c\right )}^{\frac{1}{3}} b + 6\right )} e^{\left (-{\left (d x + c\right )}^{\frac{1}{3}} b - a\right )}}{b^{4}}\right )} - 2 \,{\left (d x + c\right )} \cosh \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*(d*x+c)^(1/3)),x, algorithm="maxima")

[Out]

-1/2*(b*(((d*x + c)*b^3*e^a - 3*(d*x + c)^(2/3)*b^2*e^a + 6*(d*x + c)^(1/3)*b*e^a - 6*e^a)*e^((d*x + c)^(1/3)*
b)/b^4 + ((d*x + c)*b^3 + 3*(d*x + c)^(2/3)*b^2 + 6*(d*x + c)^(1/3)*b + 6)*e^(-(d*x + c)^(1/3)*b - a)/b^4) - 2
*(d*x + c)*cosh((d*x + c)^(1/3)*b + a))/d

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Fricas [A]  time = 1.7728, size = 159, normalized size = 1.87 \begin{align*} -\frac{3 \,{\left (2 \,{\left (d x + c\right )}^{\frac{1}{3}} b \cosh \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right ) -{\left ({\left (d x + c\right )}^{\frac{2}{3}} b^{2} + 2\right )} \sinh \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )\right )}}{b^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*(d*x+c)^(1/3)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.58895, size = 95, normalized size = 1.12 \begin{align*} \begin{cases} x \cosh{\left (a \right )} & \text{for}\: b = 0 \wedge d = 0 \\x \cosh{\left (a + b \sqrt [3]{c} \right )} & \text{for}\: d = 0 \\x \cosh{\left (a \right )} & \text{for}\: b = 0 \\\frac{3 \left (c + d x\right )^{\frac{2}{3}} \sinh{\left (a + b \sqrt [3]{c + d x} \right )}}{b d} - \frac{6 \sqrt [3]{c + d x} \cosh{\left (a + b \sqrt [3]{c + d x} \right )}}{b^{2} d} + \frac{6 \sinh{\left (a + b \sqrt [3]{c + d x} \right )}}{b^{3} d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*(d*x+c)**(1/3)),x)

[Out]

Piecewise((x*cosh(a), Eq(b, 0) & Eq(d, 0)), (x*cosh(a + b*c**(1/3)), Eq(d, 0)), (x*cosh(a), Eq(b, 0)), (3*(c +
d*x)**(2/3)*sinh(a + b*(c + d*x)**(1/3))/(b*d) - 6*(c + d*x)**(1/3)*cosh(a + b*(c + d*x)**(1/3))/(b**2*d) + 6
*sinh(a + b*(c + d*x)**(1/3))/(b**3*d), True))

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Giac [A]  time = 1.41726, size = 173, normalized size = 2.04 \begin{align*} \frac{3 \,{\left ({\left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )}^{2} - 2 \,{\left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )} a + a^{2} - 2 \,{\left (d x + c\right )}^{\frac{1}{3}} b + 2\right )} e^{\left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )}}{2 \, b^{3} d} - \frac{3 \,{\left ({\left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )}^{2} - 2 \,{\left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )} a + a^{2} + 2 \,{\left (d x + c\right )}^{\frac{1}{3}} b + 2\right )} e^{\left (-{\left (d x + c\right )}^{\frac{1}{3}} b - a\right )}}{2 \, b^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*(d*x+c)^(1/3)),x, algorithm="giac")

[Out]

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