3.28 \(\int \frac {x^2 \cosh (c+d x)}{(a+b x)^2} \, dx\)

Optimal. Leaf size=147 \[ \frac {a^2 d \sinh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (x d+\frac {a d}{b}\right )}{b^4}+\frac {a^2 d \cosh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (x d+\frac {a d}{b}\right )}{b^4}-\frac {a^2 \cosh (c+d x)}{b^3 (a+b x)}-\frac {2 a \cosh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (x d+\frac {a d}{b}\right )}{b^3}-\frac {2 a \sinh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (x d+\frac {a d}{b}\right )}{b^3}+\frac {\sinh (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.37, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {6742, 2637, 3297, 3303, 3298, 3301} \[ \frac {a^2 d \sinh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (x d+\frac {a d}{b}\right )}{b^4}+\frac {a^2 d \cosh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (x d+\frac {a d}{b}\right )}{b^4}-\frac {a^2 \cosh (c+d x)}{b^3 (a+b x)}-\frac {2 a \cosh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (x d+\frac {a d}{b}\right )}{b^3}-\frac {2 a \sinh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (x d+\frac {a d}{b}\right )}{b^3}+\frac {\sinh (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*Cosh[c + d*x])/(b^3*(a + b*x))) - (2*a*Cosh[c - (a*d)/b]*CoshIntegral[(a*d)/b + d*x])/b^3 + (a^2*d*Cosh
Integral[(a*d)/b + d*x]*Sinh[c - (a*d)/b])/b^4 + Sinh[c + d*x]/(b^2*d) + (a^2*d*Cosh[c - (a*d)/b]*SinhIntegral
[(a*d)/b + d*x])/b^4 - (2*a*Sinh[c - (a*d)/b]*SinhIntegral[(a*d)/b + d*x])/b^3

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.71, size = 115, normalized size = 0.78 \[ \frac {b \left (\frac {b \sinh (c+d x)}{d}-\frac {a^2 \cosh (c+d x)}{a+b x}\right )+a \text {Chi}\left (d \left (\frac {a}{b}+x\right )\right ) \left (a d \sinh \left (c-\frac {a d}{b}\right )-2 b \cosh \left (c-\frac {a d}{b}\right )\right )+a \text {Shi}\left (d \left (\frac {a}{b}+x\right )\right ) \left (a d \cosh \left (c-\frac {a d}{b}\right )-2 b \sinh \left (c-\frac {a d}{b}\right )\right )}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*CoshIntegral[d*(a/b + x)]*(-2*b*Cosh[c - (a*d)/b] + a*d*Sinh[c - (a*d)/b]) + b*(-((a^2*Cosh[c + d*x])/(a +
b*x)) + (b*Sinh[c + d*x])/d) + a*(a*d*Cosh[c - (a*d)/b] - 2*b*Sinh[c - (a*d)/b])*SinhIntegral[d*(a/b + x)])/b^
4

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fricas [A]  time = 0.47, size = 274, normalized size = 1.86 \[ -\frac {2 \, a^{2} b d \cosh \left (d x + c\right ) - {\left ({\left (a^{3} d^{2} - 2 \, a^{2} b d + {\left (a^{2} b d^{2} - 2 \, a b^{2} d\right )} x\right )} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) - {\left (a^{3} d^{2} + 2 \, a^{2} b d + {\left (a^{2} b d^{2} + 2 \, a b^{2} d\right )} x\right )} {\rm Ei}\left (-\frac {b d x + a d}{b}\right )\right )} \cosh \left (-\frac {b c - a d}{b}\right ) - 2 \, {\left (b^{3} x + a b^{2}\right )} \sinh \left (d x + c\right ) + {\left ({\left (a^{3} d^{2} - 2 \, a^{2} b d + {\left (a^{2} b d^{2} - 2 \, a b^{2} d\right )} x\right )} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) + {\left (a^{3} d^{2} + 2 \, a^{2} b d + {\left (a^{2} b d^{2} + 2 \, a b^{2} d\right )} x\right )} {\rm Ei}\left (-\frac {b d x + a d}{b}\right )\right )} \sinh \left (-\frac {b c - a d}{b}\right )}{2 \, {\left (b^{5} d x + a b^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^2*b*d*cosh(d*x + c) - ((a^3*d^2 - 2*a^2*b*d + (a^2*b*d^2 - 2*a*b^2*d)*x)*Ei((b*d*x + a*d)/b) - (a^3*
d^2 + 2*a^2*b*d + (a^2*b*d^2 + 2*a*b^2*d)*x)*Ei(-(b*d*x + a*d)/b))*cosh(-(b*c - a*d)/b) - 2*(b^3*x + a*b^2)*si
nh(d*x + c) + ((a^3*d^2 - 2*a^2*b*d + (a^2*b*d^2 - 2*a*b^2*d)*x)*Ei((b*d*x + a*d)/b) + (a^3*d^2 + 2*a^2*b*d +
(a^2*b*d^2 + 2*a*b^2*d)*x)*Ei(-(b*d*x + a*d)/b))*sinh(-(b*c - a*d)/b))/(b^5*d*x + a*b^4*d)

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giac [B]  time = 0.18, size = 1308, normalized size = 8.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b*x + a)*a^2*(b*c/(b*x + a) - a*d/(b*x + a) + d)*d^2*Ei(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) -
 b*c + a*d)/b)*e^((b*c - a*d)/b) - a^2*b*c*d^2*Ei(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/
b)*e^((b*c - a*d)/b) + a^3*d^3*Ei(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b)*e^((b*c - a*d
)/b) - (b*x + a)*a^2*(b*c/(b*x + a) - a*d/(b*x + a) + d)*d^2*Ei(-((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d
) - b*c + a*d)/b)*e^(-(b*c - a*d)/b) + a^2*b*c*d^2*Ei(-((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c +
a*d)/b)*e^(-(b*c - a*d)/b) - a^3*d^3*Ei(-((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b)*e^(-(b
*c - a*d)/b) - 2*(b*x + a)*a*b*(b*c/(b*x + a) - a*d/(b*x + a) + d)*d*Ei(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x +
 a) + d) - b*c + a*d)/b)*e^((b*c - a*d)/b) + 2*a*b^2*c*d*Ei(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b
*c + a*d)/b)*e^((b*c - a*d)/b) - 2*a^2*b*d^2*Ei(((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*c + a*d)/b)
*e^((b*c - a*d)/b) - 2*(b*x + a)*a*b*(b*c/(b*x + a) - a*d/(b*x + a) + d)*d*Ei(-((b*x + a)*(b*c/(b*x + a) - a*d
/(b*x + a) + d) - b*c + a*d)/b)*e^(-(b*c - a*d)/b) + 2*a*b^2*c*d*Ei(-((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a)
 + d) - b*c + a*d)/b)*e^(-(b*c - a*d)/b) - 2*a^2*b*d^2*Ei(-((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d) - b*
c + a*d)/b)*e^(-(b*c - a*d)/b) - a^2*b*d^2*e^((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) - a^2*b*d^2*e^(
-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) + (b*x + a)*b^2*(b*c/(b*x + a) - a*d/(b*x + a) + d)*e^((b*x
+ a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) - b^3*c*e^((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) + a*b^
2*d*e^((b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) - (b*x + a)*b^2*(b*c/(b*x + a) - a*d/(b*x + a) + d)*e^
(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b) + b^3*c*e^(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b
) - a*b^2*d*e^(-(b*x + a)*(b*c/(b*x + a) - a*d/(b*x + a) + d)/b))*b^2/(((b*x + a)*b^6*(b*c/(b*x + a) - a*d/(b*
x + a) + d) - b^7*c + a*b^6*d)*d)

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maple [A]  time = 0.12, size = 254, normalized size = 1.73 \[ -\frac {{\mathrm e}^{-d x -c}}{2 d \,b^{2}}-\frac {d \,{\mathrm e}^{-d x -c} a^{2}}{2 b^{3} \left (b d x +d a \right )}+\frac {d \,{\mathrm e}^{\frac {d a -c b}{b}} \Ei \left (1, d x +c +\frac {d a -c b}{b}\right ) a^{2}}{2 b^{4}}+\frac {{\mathrm e}^{\frac {d a -c b}{b}} \Ei \left (1, d x +c +\frac {d a -c b}{b}\right ) a}{b^{3}}+\frac {{\mathrm e}^{d x +c}}{2 d \,b^{2}}+\frac {{\mathrm e}^{-\frac {d a -c b}{b}} \Ei \left (1, -d x -c -\frac {d a -c b}{b}\right ) a}{b^{3}}-\frac {d \,{\mathrm e}^{d x +c} a^{2}}{2 b^{4} \left (\frac {a d}{b}+d x \right )}-\frac {d \,{\mathrm e}^{-\frac {d a -c b}{b}} \Ei \left (1, -d x -c -\frac {d a -c b}{b}\right ) a^{2}}{2 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*exp(-d*x-c)/b^2-1/2*d*exp(-d*x-c)/b^3/(b*d*x+a*d)*a^2+1/2*d/b^4*exp((a*d-b*c)/b)*Ei(1,d*x+c+(a*d-b*c)/b
)*a^2+1/b^3*exp((a*d-b*c)/b)*Ei(1,d*x+c+(a*d-b*c)/b)*a+1/2/d/b^2*exp(d*x+c)+1/b^3*exp(-(a*d-b*c)/b)*Ei(1,-d*x-
c-(a*d-b*c)/b)*a-1/2*d/b^4*exp(d*x+c)/(a*d/b+d*x)*a^2-1/2*d/b^4*exp(-(a*d-b*c)/b)*Ei(1,-d*x-c-(a*d-b*c)/b)*a^2

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maxima [A]  time = 0.41, size = 236, normalized size = 1.61 \[ \frac {1}{2} \, {\left (a^{2} {\left (\frac {e^{\left (-c + \frac {a d}{b}\right )} E_{1}\left (\frac {{\left (b x + a\right )} d}{b}\right )}{b^{4}} - \frac {e^{\left (c - \frac {a d}{b}\right )} E_{1}\left (-\frac {{\left (b x + a\right )} d}{b}\right )}{b^{4}}\right )} + \frac {2 \, a {\left (\frac {e^{\left (-c + \frac {a d}{b}\right )} E_{1}\left (\frac {{\left (b x + a\right )} d}{b}\right )}{b} + \frac {e^{\left (c - \frac {a d}{b}\right )} E_{1}\left (-\frac {{\left (b x + a\right )} d}{b}\right )}{b}\right )}}{b^{2} d} - \frac {\frac {{\left (d x e^{c} - e^{c}\right )} e^{\left (d x\right )}}{d^{2}} + \frac {{\left (d x + 1\right )} e^{\left (-d x - c\right )}}{d^{2}}}{b^{2}} + \frac {4 \, a \cosh \left (d x + c\right ) \log \left (b x + a\right )}{b^{3} d}\right )} d - {\left (\frac {a^{2}}{b^{4} x + a b^{3}} - \frac {x}{b^{2}} + \frac {2 \, a \log \left (b x + a\right )}{b^{3}}\right )} \cosh \left (d x + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*(e^(-c + a*d/b)*exp_integral_e(1, (b*x + a)*d/b)/b^4 - e^(c - a*d/b)*exp_integral_e(1, -(b*x + a)*d/b
)/b^4) + 2*a*(e^(-c + a*d/b)*exp_integral_e(1, (b*x + a)*d/b)/b + e^(c - a*d/b)*exp_integral_e(1, -(b*x + a)*d
/b)/b)/(b^2*d) - ((d*x*e^c - e^c)*e^(d*x)/d^2 + (d*x + 1)*e^(-d*x - c)/d^2)/b^2 + 4*a*cosh(d*x + c)*log(b*x +
a)/(b^3*d))*d - (a^2/(b^4*x + a*b^3) - x/b^2 + 2*a*log(b*x + a)/b^3)*cosh(d*x + c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\mathrm {cosh}\left (c+d\,x\right )}{{\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \cosh {\left (c + d x \right )}}{\left (a + b x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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