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

Optimal. Leaf size=167 \[ -\frac {12 \cosh \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}+\frac {12 \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b^3 d^2}-\frac {6 (c+d x) \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {2 c \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {2 (c+d x)^{3/2} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {2 c \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2} \]

[Out]

-12*cosh(a+b*(d*x+c)^(1/2))/b^4/d^2+2*c*cosh(a+b*(d*x+c)^(1/2))/b^2/d^2-6*(d*x+c)*cosh(a+b*(d*x+c)^(1/2))/b^2/
d^2+2*(d*x+c)^(3/2)*sinh(a+b*(d*x+c)^(1/2))/b/d^2+12*sinh(a+b*(d*x+c)^(1/2))*(d*x+c)^(1/2)/b^3/d^2-2*c*sinh(a+
b*(d*x+c)^(1/2))*(d*x+c)^(1/2)/b/d^2

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Rubi [A]  time = 0.19, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5365, 5287, 3296, 2638} \[ \frac {12 \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b^3 d^2}-\frac {6 (c+d x) \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {2 c \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}-\frac {12 \cosh \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}+\frac {2 (c+d x)^{3/2} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {2 c \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Cosh[a + b*Sqrt[c + d*x]],x]

[Out]

(-12*Cosh[a + b*Sqrt[c + d*x]])/(b^4*d^2) + (2*c*Cosh[a + b*Sqrt[c + d*x]])/(b^2*d^2) - (6*(c + d*x)*Cosh[a +
b*Sqrt[c + d*x]])/(b^2*d^2) + (12*Sqrt[c + d*x]*Sinh[a + b*Sqrt[c + d*x]])/(b^3*d^2) - (2*c*Sqrt[c + d*x]*Sinh
[a + b*Sqrt[c + d*x]])/(b*d^2) + (2*(c + d*x)^(3/2)*Sinh[a + b*Sqrt[c + d*x]])/(b*d^2)

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 5287

Int[Cosh[(c_.) + (d_.)*(x_)]*((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[Cosh[c + d*x], (e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rule 5365

Int[((a_.) + Cosh[(c_.) + (d_.)*(u_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/Coefficient[u, x, 1]^(
m + 1), Subst[Int[(x - Coefficient[u, x, 0])^m*(a + b*Cosh[c + d*x^n])^p, x], x, u], x] /; FreeQ[{a, b, c, d,
n, p}, x] && LinearQ[u, x] && NeQ[u, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int x \cosh \left (a+b \sqrt {c+d x}\right ) \, dx &=\frac {\operatorname {Subst}\left (\int (-c+x) \cosh \left (a+b \sqrt {x}\right ) \, dx,x,c+d x\right )}{d^2}\\ &=\frac {2 \operatorname {Subst}\left (\int x \left (-c+x^2\right ) \cosh (a+b x) \, dx,x,\sqrt {c+d x}\right )}{d^2}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (-c x \cosh (a+b x)+x^3 \cosh (a+b x)\right ) \, dx,x,\sqrt {c+d x}\right )}{d^2}\\ &=\frac {2 \operatorname {Subst}\left (\int x^3 \cosh (a+b x) \, dx,x,\sqrt {c+d x}\right )}{d^2}-\frac {(2 c) \operatorname {Subst}\left (\int x \cosh (a+b x) \, dx,x,\sqrt {c+d x}\right )}{d^2}\\ &=-\frac {2 c \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2}+\frac {2 (c+d x)^{3/2} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {6 \operatorname {Subst}\left (\int x^2 \sinh (a+b x) \, dx,x,\sqrt {c+d x}\right )}{b d^2}+\frac {(2 c) \operatorname {Subst}\left (\int \sinh (a+b x) \, dx,x,\sqrt {c+d x}\right )}{b d^2}\\ &=\frac {2 c \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}-\frac {6 (c+d x) \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}-\frac {2 c \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2}+\frac {2 (c+d x)^{3/2} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2}+\frac {12 \operatorname {Subst}\left (\int x \cosh (a+b x) \, dx,x,\sqrt {c+d x}\right )}{b^2 d^2}\\ &=\frac {2 c \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}-\frac {6 (c+d x) \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {12 \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b^3 d^2}-\frac {2 c \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2}+\frac {2 (c+d x)^{3/2} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {12 \operatorname {Subst}\left (\int \sinh (a+b x) \, dx,x,\sqrt {c+d x}\right )}{b^3 d^2}\\ &=-\frac {12 \cosh \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}+\frac {2 c \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}-\frac {6 (c+d x) \cosh \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {12 \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b^3 d^2}-\frac {2 c \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2}+\frac {2 (c+d x)^{3/2} \sinh \left (a+b \sqrt {c+d x}\right )}{b d^2}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 72, normalized size = 0.43 \[ \frac {2 b \left (b^2 d x+6\right ) \sqrt {c+d x} \sinh \left (a+b \sqrt {c+d x}\right )-2 \left (b^2 (2 c+3 d x)+6\right ) \cosh \left (a+b \sqrt {c+d x}\right )}{b^4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(6 + b^2*(2*c + 3*d*x))*Cosh[a + b*Sqrt[c + d*x]] + 2*b*Sqrt[c + d*x]*(6 + b^2*d*x)*Sinh[a + b*Sqrt[c + d*
x]])/(b^4*d^2)

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fricas [A]  time = 0.55, size = 68, normalized size = 0.41 \[ \frac {2 \, {\left ({\left (b^{3} d x + 6 \, b\right )} \sqrt {d x + c} \sinh \left (\sqrt {d x + c} b + a\right ) - {\left (3 \, b^{2} d x + 2 \, b^{2} c + 6\right )} \cosh \left (\sqrt {d x + c} b + a\right )\right )}}{b^{4} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.13, size = 300, normalized size = 1.80 \[ -\frac {\frac {{\left ({\left (\sqrt {d x + c} b + a\right )} b^{2} c - a b^{2} c - {\left (\sqrt {d x + c} b + a\right )}^{3} + 3 \, {\left (\sqrt {d x + c} b + a\right )}^{2} a - 3 \, {\left (\sqrt {d x + c} b + a\right )} a^{2} + a^{3} - b^{2} c + 3 \, {\left (\sqrt {d x + c} b + a\right )}^{2} - 6 \, {\left (\sqrt {d x + c} b + a\right )} a + 3 \, a^{2} - 6 \, \sqrt {d x + c} b + 6\right )} e^{\left (\sqrt {d x + c} b + a\right )}}{b^{3} d} - \frac {{\left ({\left (\sqrt {d x + c} b + a\right )} b^{2} c - a b^{2} c - {\left (\sqrt {d x + c} b + a\right )}^{3} + 3 \, {\left (\sqrt {d x + c} b + a\right )}^{2} a - 3 \, {\left (\sqrt {d x + c} b + a\right )} a^{2} + a^{3} + b^{2} c - 3 \, {\left (\sqrt {d x + c} b + a\right )}^{2} + 6 \, {\left (\sqrt {d x + c} b + a\right )} a - 3 \, a^{2} - 6 \, \sqrt {d x + c} b - 6\right )} e^{\left (-\sqrt {d x + c} b - a\right )}}{b^{3} d}}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.08, size = 303, normalized size = 1.81 \[ \frac {\frac {2 \left (\left (a +b \sqrt {d x +c}\right )^{3} \sinh \left (a +b \sqrt {d x +c}\right )-3 \left (a +b \sqrt {d x +c}\right )^{2} \cosh \left (a +b \sqrt {d x +c}\right )+6 \sinh \left (a +b \sqrt {d x +c}\right ) \left (a +b \sqrt {d x +c}\right )-6 \cosh \left (a +b \sqrt {d x +c}\right )\right )}{b^{2}}-\frac {6 a \left (\sinh \left (a +b \sqrt {d x +c}\right ) \left (a +b \sqrt {d x +c}\right )^{2}-2 \left (a +b \sqrt {d x +c}\right ) \cosh \left (a +b \sqrt {d x +c}\right )+2 \sinh \left (a +b \sqrt {d x +c}\right )\right )}{b^{2}}+\frac {6 a^{2} \left (\sinh \left (a +b \sqrt {d x +c}\right ) \left (a +b \sqrt {d x +c}\right )-\cosh \left (a +b \sqrt {d x +c}\right )\right )}{b^{2}}-\frac {2 a^{3} \sinh \left (a +b \sqrt {d x +c}\right )}{b^{2}}-2 c \left (\sinh \left (a +b \sqrt {d x +c}\right ) \left (a +b \sqrt {d x +c}\right )-\cosh \left (a +b \sqrt {d x +c}\right )\right )+2 a c \sinh \left (a +b \sqrt {d x +c}\right )}{d^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 291, normalized size = 1.74 \[ \frac {2 \, d^{2} x^{2} \cosh \left (\sqrt {d x + c} b + a\right ) - {\left (\frac {c^{2} e^{\left (\sqrt {d x + c} b + a\right )}}{b} + \frac {c^{2} e^{\left (-\sqrt {d x + c} b - a\right )}}{b} - \frac {2 \, {\left ({\left (d x + c\right )} b^{2} e^{a} - 2 \, \sqrt {d x + c} b e^{a} + 2 \, e^{a}\right )} c e^{\left (\sqrt {d x + c} b\right )}}{b^{3}} - \frac {2 \, {\left ({\left (d x + c\right )} b^{2} + 2 \, \sqrt {d x + c} b + 2\right )} c e^{\left (-\sqrt {d x + c} b - a\right )}}{b^{3}} + \frac {{\left ({\left (d x + c\right )}^{2} b^{4} e^{a} - 4 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} e^{a} + 12 \, {\left (d x + c\right )} b^{2} e^{a} - 24 \, \sqrt {d x + c} b e^{a} + 24 \, e^{a}\right )} e^{\left (\sqrt {d x + c} b\right )}}{b^{5}} + \frac {{\left ({\left (d x + c\right )}^{2} b^{4} + 4 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} + 12 \, {\left (d x + c\right )} b^{2} + 24 \, \sqrt {d x + c} b + 24\right )} e^{\left (-\sqrt {d x + c} b - a\right )}}{b^{5}}\right )} b}{4 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*d^2*x^2*cosh(sqrt(d*x + c)*b + a) - (c^2*e^(sqrt(d*x + c)*b + a)/b + c^2*e^(-sqrt(d*x + c)*b - a)/b - 2
*((d*x + c)*b^2*e^a - 2*sqrt(d*x + c)*b*e^a + 2*e^a)*c*e^(sqrt(d*x + c)*b)/b^3 - 2*((d*x + c)*b^2 + 2*sqrt(d*x
+ c)*b + 2)*c*e^(-sqrt(d*x + c)*b - a)/b^3 + ((d*x + c)^2*b^4*e^a - 4*(d*x + c)^(3/2)*b^3*e^a + 12*(d*x + c)*
b^2*e^a - 24*sqrt(d*x + c)*b*e^a + 24*e^a)*e^(sqrt(d*x + c)*b)/b^5 + ((d*x + c)^2*b^4 + 4*(d*x + c)^(3/2)*b^3
+ 12*(d*x + c)*b^2 + 24*sqrt(d*x + c)*b + 24)*e^(-sqrt(d*x + c)*b - a)/b^5)*b)/d^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {cosh}\left (a+b\,\sqrt {c+d\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(a + b*(c + d*x)^(1/2)),x)

[Out]

int(x*cosh(a + b*(c + d*x)^(1/2)), x)

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sympy [A]  time = 0.59, size = 151, normalized size = 0.90 \[ \begin {cases} \frac {x^{2} \cosh {\relax (a )}}{2} & \text {for}\: b = 0 \wedge \left (b = 0 \vee d = 0\right ) \\\frac {x^{2} \cosh {\left (a + b \sqrt {c} \right )}}{2} & \text {for}\: d = 0 \\\frac {2 x \sqrt {c + d x} \sinh {\left (a + b \sqrt {c + d x} \right )}}{b d} - \frac {4 c \cosh {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d^{2}} - \frac {6 x \cosh {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d} + \frac {12 \sqrt {c + d x} \sinh {\left (a + b \sqrt {c + d x} \right )}}{b^{3} d^{2}} - \frac {12 \cosh {\left (a + b \sqrt {c + d x} \right )}}{b^{4} d^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*cosh(a)/2, Eq(b, 0) & (Eq(b, 0) | Eq(d, 0))), (x**2*cosh(a + b*sqrt(c))/2, Eq(d, 0)), (2*x*sqr
t(c + d*x)*sinh(a + b*sqrt(c + d*x))/(b*d) - 4*c*cosh(a + b*sqrt(c + d*x))/(b**2*d**2) - 6*x*cosh(a + b*sqrt(c
+ d*x))/(b**2*d) + 12*sqrt(c + d*x)*sinh(a + b*sqrt(c + d*x))/(b**3*d**2) - 12*cosh(a + b*sqrt(c + d*x))/(b**
4*d**2), True))

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