\(\int \sinh (a+b \sqrt [3]{c+d x}) \, dx\) [100]

Optimal result

Integrand size = 14, antiderivative size = 85 \[ \int \sinh \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {6 \cosh \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac {3 (c+d x)^{2/3} \cosh \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac {6 \sqrt [3]{c+d x} \sinh \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5418, 5412, 3377, 2718} \[ \int \sinh \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {6 \cosh \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac {6 \sqrt [3]{c+d x} \sinh \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac {3 (c+d x)^{2/3} \cosh \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

[In]

Int[Sinh[a + b*(c + d*x)^(1/3)],x]

[Out]

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

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 5412

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

Rule 5418

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \sinh \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {3 \left (2+b^2 (c+d x)^{2/3}\right ) \cosh \left (a+b \sqrt [3]{c+d x}\right )-6 b \sqrt [3]{c+d x} \sinh \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.56

method	result	size
derivativedivides	\(\frac {3 a^{2} \cosh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-6 a \left (\left (a +b \left (d x +c \right )^{\frac {1}{3}}\right ) \cosh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-\sinh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )\right )+3 \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )^{2} \cosh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-6 \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right ) \sinh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )+6 \cosh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )}{b^{3} d}\)	\(133\)
default	\(\frac {3 a^{2} \cosh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-6 a \left (\left (a +b \left (d x +c \right )^{\frac {1}{3}}\right ) \cosh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-\sinh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )\right )+3 \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )^{2} \cosh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-6 \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right ) \sinh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )+6 \cosh \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )}{b^{3} d}\)	\(133\)

[In]

int(sinh(a+b*(d*x+c)^(1/3)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68 \[ \int \sinh \left (a+b \sqrt [3]{c+d x}\right ) \, dx=-\frac {3 \, {\left (2 \, {\left (d x + c\right )}^{\frac {1}{3}} b \sinh \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) - {\left ({\left (d x + c\right )}^{\frac {2}{3}} b^{2} + 2\right )} \cosh \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )}}{b^{3} d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.11 \[ \int \sinh \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\begin {cases} x \sinh {\left (a \right )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee d = 0\right ) \\x \sinh {\left (a + b \sqrt [3]{c} \right )} & \text {for}\: d = 0 \\\frac {3 \left (c + d x\right )^{\frac {2}{3}} \cosh {\left (a + b \sqrt [3]{c + d x} \right )}}{b d} - \frac {6 \sqrt [3]{c + d x} \sinh {\left (a + b \sqrt [3]{c + d x} \right )}}{b^{2} d} + \frac {6 \cosh {\left (a + b \sqrt [3]{c + d x} \right )}}{b^{3} d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.61 \[ \int \sinh \left (a+b \sqrt [3]{c+d x}\right ) \, dx=-\frac {b {\left (\frac {{\left ({\left (d x + c\right )} b^{3} e^{a} - 3 \, {\left (d x + c\right )}^{\frac {2}{3}} b^{2} e^{a} + 6 \, {\left (d x + c\right )}^{\frac {1}{3}} b e^{a} - 6 \, e^{a}\right )} e^{\left ({\left (d x + c\right )}^{\frac {1}{3}} b\right )}}{b^{4}} - \frac {{\left ({\left (d x + c\right )} b^{3} + 3 \, {\left (d x + c\right )}^{\frac {2}{3}} b^{2} + 6 \, {\left (d x + c\right )}^{\frac {1}{3}} b + 6\right )} e^{\left (-{\left (d x + c\right )}^{\frac {1}{3}} b - a\right )}}{b^{4}}\right )} - 2 \, {\left (d x + c\right )} \sinh \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}{2 \, d} \]

[In]

integrate(sinh(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)*sinh((d*x + c)^(1/3)*b + a))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.51 \[ \int \sinh \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {3 \, {\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} - 2 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} a + a^{2} - 2 \, {\left (d x + c\right )}^{\frac {1}{3}} b + 2\right )} e^{\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}}{2 \, b^{3} d} + \frac {3 \, {\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} - 2 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} a + a^{2} + 2 \, {\left (d x + c\right )}^{\frac {1}{3}} b + 2\right )} e^{\left (-{\left (d x + c\right )}^{\frac {1}{3}} b - a\right )}}{2 \, b^{3} d} \]

[In]

integrate(sinh(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)

Mupad [B] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \sinh \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {6\,\mathrm {cosh}\left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )}{b^3\,d}+\frac {3\,\mathrm {cosh}\left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )\,{\left (c+d\,x\right )}^{2/3}}{b\,d}-\frac {6\,\mathrm {sinh}\left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )\,{\left (c+d\,x\right )}^{1/3}}{b^2\,d} \]

[In]

int(sinh(a + b*(c + d*x)^(1/3)),x)

[Out]

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