3.51 \(\int (a+b x^2)^2 \cosh (c+d x) \, dx\)

Optimal. Leaf size=136 \[ \frac{a^2 \sinh (c+d x)}{d}+\frac{4 a b \sinh (c+d x)}{d^3}-\frac{4 a b x \cosh (c+d x)}{d^2}+\frac{2 a b x^2 \sinh (c+d x)}{d}+\frac{12 b^2 x^2 \sinh (c+d x)}{d^3}-\frac{4 b^2 x^3 \cosh (c+d x)}{d^2}+\frac{24 b^2 \sinh (c+d x)}{d^5}-\frac{24 b^2 x \cosh (c+d x)}{d^4}+\frac{b^2 x^4 \sinh (c+d x)}{d} \]

[Out]

(-24*b^2*x*Cosh[c + d*x])/d^4 - (4*a*b*x*Cosh[c + d*x])/d^2 - (4*b^2*x^3*Cosh[c + d*x])/d^2 + (24*b^2*Sinh[c +
d*x])/d^5 + (4*a*b*Sinh[c + d*x])/d^3 + (a^2*Sinh[c + d*x])/d + (12*b^2*x^2*Sinh[c + d*x])/d^3 + (2*a*b*x^2*S
inh[c + d*x])/d + (b^2*x^4*Sinh[c + d*x])/d

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Rubi [A]  time = 0.18139, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {5277, 2637, 3296} \[ \frac{a^2 \sinh (c+d x)}{d}+\frac{4 a b \sinh (c+d x)}{d^3}-\frac{4 a b x \cosh (c+d x)}{d^2}+\frac{2 a b x^2 \sinh (c+d x)}{d}+\frac{12 b^2 x^2 \sinh (c+d x)}{d^3}-\frac{4 b^2 x^3 \cosh (c+d x)}{d^2}+\frac{24 b^2 \sinh (c+d x)}{d^5}-\frac{24 b^2 x \cosh (c+d x)}{d^4}+\frac{b^2 x^4 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2*Cosh[c + d*x],x]

[Out]

(-24*b^2*x*Cosh[c + d*x])/d^4 - (4*a*b*x*Cosh[c + d*x])/d^2 - (4*b^2*x^3*Cosh[c + d*x])/d^2 + (24*b^2*Sinh[c +
d*x])/d^5 + (4*a*b*Sinh[c + d*x])/d^3 + (a^2*Sinh[c + d*x])/d + (12*b^2*x^2*Sinh[c + d*x])/d^3 + (2*a*b*x^2*S
inh[c + d*x])/d + (b^2*x^4*Sinh[c + d*x])/d

Rule 5277

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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]

Rubi steps

\begin{align*} \int \left (a+b x^2\right )^2 \cosh (c+d x) \, dx &=\int \left (a^2 \cosh (c+d x)+2 a b x^2 \cosh (c+d x)+b^2 x^4 \cosh (c+d x)\right ) \, dx\\ &=a^2 \int \cosh (c+d x) \, dx+(2 a b) \int x^2 \cosh (c+d x) \, dx+b^2 \int x^4 \cosh (c+d x) \, dx\\ &=\frac{a^2 \sinh (c+d x)}{d}+\frac{2 a b x^2 \sinh (c+d x)}{d}+\frac{b^2 x^4 \sinh (c+d x)}{d}-\frac{(4 a b) \int x \sinh (c+d x) \, dx}{d}-\frac{\left (4 b^2\right ) \int x^3 \sinh (c+d x) \, dx}{d}\\ &=-\frac{4 a b x \cosh (c+d x)}{d^2}-\frac{4 b^2 x^3 \cosh (c+d x)}{d^2}+\frac{a^2 \sinh (c+d x)}{d}+\frac{2 a b x^2 \sinh (c+d x)}{d}+\frac{b^2 x^4 \sinh (c+d x)}{d}+\frac{(4 a b) \int \cosh (c+d x) \, dx}{d^2}+\frac{\left (12 b^2\right ) \int x^2 \cosh (c+d x) \, dx}{d^2}\\ &=-\frac{4 a b x \cosh (c+d x)}{d^2}-\frac{4 b^2 x^3 \cosh (c+d x)}{d^2}+\frac{4 a b \sinh (c+d x)}{d^3}+\frac{a^2 \sinh (c+d x)}{d}+\frac{12 b^2 x^2 \sinh (c+d x)}{d^3}+\frac{2 a b x^2 \sinh (c+d x)}{d}+\frac{b^2 x^4 \sinh (c+d x)}{d}-\frac{\left (24 b^2\right ) \int x \sinh (c+d x) \, dx}{d^3}\\ &=-\frac{24 b^2 x \cosh (c+d x)}{d^4}-\frac{4 a b x \cosh (c+d x)}{d^2}-\frac{4 b^2 x^3 \cosh (c+d x)}{d^2}+\frac{4 a b \sinh (c+d x)}{d^3}+\frac{a^2 \sinh (c+d x)}{d}+\frac{12 b^2 x^2 \sinh (c+d x)}{d^3}+\frac{2 a b x^2 \sinh (c+d x)}{d}+\frac{b^2 x^4 \sinh (c+d x)}{d}+\frac{\left (24 b^2\right ) \int \cosh (c+d x) \, dx}{d^4}\\ &=-\frac{24 b^2 x \cosh (c+d x)}{d^4}-\frac{4 a b x \cosh (c+d x)}{d^2}-\frac{4 b^2 x^3 \cosh (c+d x)}{d^2}+\frac{24 b^2 \sinh (c+d x)}{d^5}+\frac{4 a b \sinh (c+d x)}{d^3}+\frac{a^2 \sinh (c+d x)}{d}+\frac{12 b^2 x^2 \sinh (c+d x)}{d^3}+\frac{2 a b x^2 \sinh (c+d x)}{d}+\frac{b^2 x^4 \sinh (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.17899, size = 85, normalized size = 0.62 \[ \frac{\left (a^2 d^4+2 a b d^2 \left (d^2 x^2+2\right )+b^2 \left (d^4 x^4+12 d^2 x^2+24\right )\right ) \sinh (c+d x)-4 b d x \left (a d^2+b \left (d^2 x^2+6\right )\right ) \cosh (c+d x)}{d^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2*Cosh[c + d*x],x]

[Out]

(-4*b*d*x*(a*d^2 + b*(6 + d^2*x^2))*Cosh[c + d*x] + (a^2*d^4 + 2*a*b*d^2*(2 + d^2*x^2) + b^2*(24 + 12*d^2*x^2
+ d^4*x^4))*Sinh[c + d*x])/d^5

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Maple [B]  time = 0.008, size = 332, normalized size = 2.4 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{2} \left ( \left ( dx+c \right ) ^{4}\sinh \left ( dx+c \right ) -4\, \left ( dx+c \right ) ^{3}\cosh \left ( dx+c \right ) +12\, \left ( dx+c \right ) ^{2}\sinh \left ( dx+c \right ) -24\, \left ( dx+c \right ) \cosh \left ( dx+c \right ) +24\,\sinh \left ( dx+c \right ) \right ) }{{d}^{4}}}-4\,{\frac{c{b}^{2} \left ( \left ( dx+c \right ) ^{3}\sinh \left ( dx+c \right ) -3\, \left ( dx+c \right ) ^{2}\cosh \left ( dx+c \right ) +6\, \left ( dx+c \right ) \sinh \left ( dx+c \right ) -6\,\cosh \left ( dx+c \right ) \right ) }{{d}^{4}}}+6\,{\frac{{c}^{2}{b}^{2} \left ( \left ( dx+c \right ) ^{2}\sinh \left ( dx+c \right ) -2\, \left ( dx+c \right ) \cosh \left ( dx+c \right ) +2\,\sinh \left ( dx+c \right ) \right ) }{{d}^{4}}}+2\,{\frac{ab \left ( \left ( dx+c \right ) ^{2}\sinh \left ( dx+c \right ) -2\, \left ( dx+c \right ) \cosh \left ( dx+c \right ) +2\,\sinh \left ( dx+c \right ) \right ) }{{d}^{2}}}-4\,{\frac{{b}^{2}{c}^{3} \left ( \left ( dx+c \right ) \sinh \left ( dx+c \right ) -\cosh \left ( dx+c \right ) \right ) }{{d}^{4}}}-4\,{\frac{cba \left ( \left ( dx+c \right ) \sinh \left ( dx+c \right ) -\cosh \left ( dx+c \right ) \right ) }{{d}^{2}}}+{\frac{{b}^{2}{c}^{4}\sinh \left ( dx+c \right ) }{{d}^{4}}}+2\,{\frac{b{c}^{2}a\sinh \left ( dx+c \right ) }{{d}^{2}}}+{a}^{2}\sinh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/d^4*b^2*((d*x+c)^4*sinh(d*x+c)-4*(d*x+c)^3*cosh(d*x+c)+12*(d*x+c)^2*sinh(d*x+c)-24*(d*x+c)*cosh(d*x+c)+
24*sinh(d*x+c))-4/d^4*b^2*c*((d*x+c)^3*sinh(d*x+c)-3*(d*x+c)^2*cosh(d*x+c)+6*(d*x+c)*sinh(d*x+c)-6*cosh(d*x+c)
)+6/d^4*b^2*c^2*((d*x+c)^2*sinh(d*x+c)-2*(d*x+c)*cosh(d*x+c)+2*sinh(d*x+c))+2/d^2*b*a*((d*x+c)^2*sinh(d*x+c)-2
*(d*x+c)*cosh(d*x+c)+2*sinh(d*x+c))-4/d^4*b^2*c^3*((d*x+c)*sinh(d*x+c)-cosh(d*x+c))-4/d^2*b*c*a*((d*x+c)*sinh(
d*x+c)-cosh(d*x+c))+1/d^4*b^2*c^4*sinh(d*x+c)+2/d^2*b*c^2*a*sinh(d*x+c)+a^2*sinh(d*x+c))

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Maxima [A]  time = 1.03875, size = 255, normalized size = 1.88 \begin{align*} \frac{a^{2} e^{\left (d x + c\right )}}{2 \, d} - \frac{a^{2} e^{\left (-d x - c\right )}}{2 \, d} + \frac{{\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} a b e^{\left (d x\right )}}{d^{3}} - \frac{{\left (d^{2} x^{2} + 2 \, d x + 2\right )} a b e^{\left (-d x - c\right )}}{d^{3}} + \frac{{\left (d^{4} x^{4} e^{c} - 4 \, d^{3} x^{3} e^{c} + 12 \, d^{2} x^{2} e^{c} - 24 \, d x e^{c} + 24 \, e^{c}\right )} b^{2} e^{\left (d x\right )}}{2 \, d^{5}} - \frac{{\left (d^{4} x^{4} + 4 \, d^{3} x^{3} + 12 \, d^{2} x^{2} + 24 \, d x + 24\right )} b^{2} e^{\left (-d x - c\right )}}{2 \, d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.18069, size = 207, normalized size = 1.52 \begin{align*} -\frac{4 \,{\left (b^{2} d^{3} x^{3} +{\left (a b d^{3} + 6 \, b^{2} d\right )} x\right )} \cosh \left (d x + c\right ) -{\left (b^{2} d^{4} x^{4} + a^{2} d^{4} + 4 \, a b d^{2} + 2 \,{\left (a b d^{4} + 6 \, b^{2} d^{2}\right )} x^{2} + 24 \, b^{2}\right )} \sinh \left (d x + c\right )}{d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*(b^2*d^3*x^3 + (a*b*d^3 + 6*b^2*d)*x)*cosh(d*x + c) - (b^2*d^4*x^4 + a^2*d^4 + 4*a*b*d^2 + 2*(a*b*d^4 + 6*
b^2*d^2)*x^2 + 24*b^2)*sinh(d*x + c))/d^5

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Sympy [A]  time = 2.95396, size = 172, normalized size = 1.26 \begin{align*} \begin{cases} \frac{a^{2} \sinh{\left (c + d x \right )}}{d} + \frac{2 a b x^{2} \sinh{\left (c + d x \right )}}{d} - \frac{4 a b x \cosh{\left (c + d x \right )}}{d^{2}} + \frac{4 a b \sinh{\left (c + d x \right )}}{d^{3}} + \frac{b^{2} x^{4} \sinh{\left (c + d x \right )}}{d} - \frac{4 b^{2} x^{3} \cosh{\left (c + d x \right )}}{d^{2}} + \frac{12 b^{2} x^{2} \sinh{\left (c + d x \right )}}{d^{3}} - \frac{24 b^{2} x \cosh{\left (c + d x \right )}}{d^{4}} + \frac{24 b^{2} \sinh{\left (c + d x \right )}}{d^{5}} & \text{for}\: d \neq 0 \\\left (a^{2} x + \frac{2 a b x^{3}}{3} + \frac{b^{2} x^{5}}{5}\right ) \cosh{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*sinh(c + d*x)/d + 2*a*b*x**2*sinh(c + d*x)/d - 4*a*b*x*cosh(c + d*x)/d**2 + 4*a*b*sinh(c + d*x
)/d**3 + b**2*x**4*sinh(c + d*x)/d - 4*b**2*x**3*cosh(c + d*x)/d**2 + 12*b**2*x**2*sinh(c + d*x)/d**3 - 24*b**
2*x*cosh(c + d*x)/d**4 + 24*b**2*sinh(c + d*x)/d**5, Ne(d, 0)), ((a**2*x + 2*a*b*x**3/3 + b**2*x**5/5)*cosh(c)
, True))

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Giac [A]  time = 1.17539, size = 243, normalized size = 1.79 \begin{align*} \frac{{\left (b^{2} d^{4} x^{4} + 2 \, a b d^{4} x^{2} - 4 \, b^{2} d^{3} x^{3} + a^{2} d^{4} - 4 \, a b d^{3} x + 12 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} - 24 \, b^{2} d x + 24 \, b^{2}\right )} e^{\left (d x + c\right )}}{2 \, d^{5}} - \frac{{\left (b^{2} d^{4} x^{4} + 2 \, a b d^{4} x^{2} + 4 \, b^{2} d^{3} x^{3} + a^{2} d^{4} + 4 \, a b d^{3} x + 12 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} + 24 \, b^{2} d x + 24 \, b^{2}\right )} e^{\left (-d x - c\right )}}{2 \, d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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