3.61 \(\int \cosh (a+b \sqrt {c+d x}) \, dx\)

Optimal. Leaf size=54 \[ \frac {2 \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b d}-\frac {2 \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5311, 5305, 3296, 2638} \[ \frac {2 \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b d}-\frac {2 \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*Sqrt[c + d*x]],x]

[Out]

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 50, normalized size = 0.93 \[ \frac {2 \left (b \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )-\cosh \left (a+b \sqrt {c+d x}\right )\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*Sqrt[c + d*x]],x]

[Out]

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

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fricas [A]  time = 0.47, size = 44, normalized size = 0.81 \[ \frac {2 \, {\left (\sqrt {d x + c} b \sinh \left (\sqrt {d x + c} b + a\right ) - \cosh \left (\sqrt {d x + c} b + a\right )\right )}}{b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*(d*x+c)^(1/2)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.12, size = 65, normalized size = 1.20 \[ \frac {{\left (\sqrt {d x + c} b - 1\right )} e^{\left (\sqrt {d x + c} b + a\right )}}{b^{2} d} - \frac {{\left (\sqrt {d x + c} b + 1\right )} e^{\left (-\sqrt {d x + c} b - a\right )}}{b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*(d*x+c)^(1/2)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 63, normalized size = 1.17 \[ \frac {2 \sinh \left (a +b \sqrt {d x +c}\right ) \left (a +b \sqrt {d x +c}\right )-2 \cosh \left (a +b \sqrt {d x +c}\right )-2 a \sinh \left (a +b \sqrt {d x +c}\right )}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a+b*(d*x+c)^(1/2)),x)

[Out]

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

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maxima [B]  time = 0.31, size = 110, normalized size = 2.04 \[ -\frac {b {\left (\frac {{\left ({\left (d x + c\right )} b^{2} e^{a} - 2 \, \sqrt {d x + c} b e^{a} + 2 \, e^{a}\right )} e^{\left (\sqrt {d x + c} b\right )}}{b^{3}} + \frac {{\left ({\left (d x + c\right )} b^{2} + 2 \, \sqrt {d x + c} b + 2\right )} e^{\left (-\sqrt {d x + c} b - a\right )}}{b^{3}}\right )} - 2 \, {\left (d x + c\right )} \cosh \left (\sqrt {d x + c} b + a\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*(d*x+c)^(1/2)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.96, size = 43, normalized size = 0.80 \[ -\frac {2\,\left (\mathrm {cosh}\left (a+b\,\sqrt {c+d\,x}\right )-b\,\mathrm {sinh}\left (a+b\,\sqrt {c+d\,x}\right )\,\sqrt {c+d\,x}\right )}{b^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*(c + d*x)^(1/2)),x)

[Out]

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

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sympy [A]  time = 0.47, size = 65, normalized size = 1.20 \[ \begin {cases} x \cosh {\relax (a )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee d = 0\right ) \\x \cosh {\left (a + b \sqrt {c} \right )} & \text {for}\: d = 0 \\\frac {2 \sqrt {c + d x} \sinh {\left (a + b \sqrt {c + d x} \right )}}{b d} - \frac {2 \cosh {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b*(d*x+c)**(1/2)),x)

[Out]

Piecewise((x*cosh(a), Eq(b, 0) & (Eq(b, 0) | Eq(d, 0))), (x*cosh(a + b*sqrt(c)), Eq(d, 0)), (2*sqrt(c + d*x)*s
inh(a + b*sqrt(c + d*x))/(b*d) - 2*cosh(a + b*sqrt(c + d*x))/(b**2*d), True))

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