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

Optimal. Leaf size=43 \[ \frac {2 \sqrt {\sinh (c+d x)}}{b d}-\frac {2 a \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{b^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3223, 190, 43} \[ \frac {2 \sqrt {\sinh (c+d x)}}{b d}-\frac {2 a \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[c + d*x]/(a + b*Sqrt[Sinh[c + d*x]]),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 190

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 3223

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x]
, x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0
] || IGtQ[p, 0] || IntegersQ[m, p])

Rubi steps

\begin {align*} \int \frac {\cosh (c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{a+b \sqrt {x}} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {x}{a+b x} \, dx,x,\sqrt {\sinh (c+d x)}\right )}{d}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (\frac {1}{b}-\frac {a}{b (a+b x)}\right ) \, dx,x,\sqrt {\sinh (c+d x)}\right )}{d}\\ &=-\frac {2 a \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{b^2 d}+\frac {2 \sqrt {\sinh (c+d x)}}{b d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 41, normalized size = 0.95 \[ \frac {2 \left (\frac {\sqrt {\sinh (c+d x)}}{b}-\frac {a \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{b^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[c + d*x]/(a + b*Sqrt[Sinh[c + d*x]]),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 2.31, size = 225, normalized size = 5.23 \[ \frac {a d x + a \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a^{2} \cosh \left (d x + c\right ) - b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a^{2}\right )} \sinh \left (d x + c\right ) - 4 \, {\left (a b \cosh \left (d x + c\right ) + a b \sinh \left (d x + c\right )\right )} \sqrt {\sinh \left (d x + c\right )}}{b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} - 2 \, a^{2} \cosh \left (d x + c\right ) - b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) - a^{2}\right )} \sinh \left (d x + c\right )}\right ) - a \log \left (\frac {2 \, {\left (b^{2} \sinh \left (d x + c\right ) - a^{2}\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 2 \, b \sqrt {\sinh \left (d x + c\right )}}{b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*d*x + a*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a^2*cosh(d*x + c) - b^2 + 2*(b^2*cosh(d*x + c) +
a^2)*sinh(d*x + c) - 4*(a*b*cosh(d*x + c) + a*b*sinh(d*x + c))*sqrt(sinh(d*x + c)))/(b^2*cosh(d*x + c)^2 + b^
2*sinh(d*x + c)^2 - 2*a^2*cosh(d*x + c) - b^2 + 2*(b^2*cosh(d*x + c) - a^2)*sinh(d*x + c))) - a*log(2*(b^2*sin
h(d*x + c) - a^2)/(cosh(d*x + c) - sinh(d*x + c))) + 2*b*sqrt(sinh(d*x + c)))/(b^2*d)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh \left (d x + c\right )}{b \sqrt {\sinh \left (d x + c\right )} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.03, size = 89, normalized size = 2.07 \[ \frac {2 \left (\sqrt {\sinh }\left (d x +c \right )\right )}{b d}+\frac {a \ln \left (b \left (\sqrt {\sinh }\left (d x +c \right )\right )-a \right )}{d \,b^{2}}-\frac {a \ln \left (a +b \left (\sqrt {\sinh }\left (d x +c \right )\right )\right )}{b^{2} d}-\frac {a \ln \left (b^{2} \sinh \left (d x +c \right )-a^{2}\right )}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sinh(d*x+c)^(1/2)/b/d+1/d/b^2*a*ln(b*sinh(d*x+c)^(1/2)-a)-a*ln(a+b*sinh(d*x+c)^(1/2))/b^2/d-1/d*a*ln(b^2*sin
h(d*x+c)-a^2)/b^2

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh \left (d x + c\right )}{b \sqrt {\sinh \left (d x + c\right )} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.01, size = 39, normalized size = 0.91 \[ \frac {2\,\sqrt {\mathrm {sinh}\left (c+d\,x\right )}}{b\,d}-\frac {2\,a\,\ln \left (a+b\,\sqrt {\mathrm {sinh}\left (c+d\,x\right )}\right )}{b^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.82, size = 68, normalized size = 1.58 \[ \begin {cases} \frac {x \cosh {\relax (c )}}{a} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sinh {\left (c + d x \right )}}{a d} & \text {for}\: b = 0 \\\frac {x \cosh {\relax (c )}}{a + b \sqrt {\sinh {\relax (c )}}} & \text {for}\: d = 0 \\- \frac {2 a \log {\left (\frac {a}{b} + \sqrt {\sinh {\left (c + d x \right )}} \right )}}{b^{2} d} + \frac {2 \sqrt {\sinh {\left (c + d x \right )}}}{b d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________