3.94 \(\int x^{-3+m} \cosh ^2(a+b x) \, dx\)

Optimal. Leaf size=84 \[ -e^{2 a} b^2 2^{-m} x^m (-b x)^{-m} \Gamma (m-2,-2 b x)-e^{-2 a} b^2 2^{-m} x^m (b x)^{-m} \Gamma (m-2,2 b x)-\frac {x^{m-2}}{2 (2-m)} \]

[Out]

-1/2*x^(-2+m)/(2-m)-b^2*exp(2*a)*x^m*GAMMA(-2+m,-2*b*x)/(2^m)/((-b*x)^m)-b^2*x^m*GAMMA(-2+m,2*b*x)/(2^m)/exp(2
*a)/((b*x)^m)

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Rubi [A]  time = 0.14, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3312, 3307, 2181} \[ -e^{2 a} b^2 2^{-m} x^m (-b x)^{-m} \text {Gamma}(m-2,-2 b x)-e^{-2 a} b^2 2^{-m} x^m (b x)^{-m} \text {Gamma}(m-2,2 b x)-\frac {x^{m-2}}{2 (2-m)} \]

Antiderivative was successfully verified.

[In]

Int[x^(-3 + m)*Cosh[a + b*x]^2,x]

[Out]

-x^(-2 + m)/(2*(2 - m)) - (b^2*E^(2*a)*x^m*Gamma[-2 + m, -2*b*x])/(2^m*(-(b*x))^m) - (b^2*x^m*Gamma[-2 + m, 2*
b*x])/(2^m*E^(2*a)*(b*x)^m)

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3307

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps

\begin {align*} \int x^{-3+m} \cosh ^2(a+b x) \, dx &=\int \left (\frac {x^{-3+m}}{2}+\frac {1}{2} x^{-3+m} \cosh (2 a+2 b x)\right ) \, dx\\ &=-\frac {x^{-2+m}}{2 (2-m)}+\frac {1}{2} \int x^{-3+m} \cosh (2 a+2 b x) \, dx\\ &=-\frac {x^{-2+m}}{2 (2-m)}+\frac {1}{4} \int e^{-i (2 i a+2 i b x)} x^{-3+m} \, dx+\frac {1}{4} \int e^{i (2 i a+2 i b x)} x^{-3+m} \, dx\\ &=-\frac {x^{-2+m}}{2 (2-m)}-2^{-m} b^2 e^{2 a} x^m (-b x)^{-m} \Gamma (-2+m,-2 b x)-2^{-m} b^2 e^{-2 a} x^m (b x)^{-m} \Gamma (-2+m,2 b x)\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 84, normalized size = 1.00 \[ -e^{2 a} b^2 2^{-m} x^m (-b x)^{-m} \Gamma (m-2,-2 b x)-e^{-2 a} b^2 2^{-m} x^m (b x)^{-m} \Gamma (m-2,2 b x)-\frac {x^{m-2}}{2 (2-m)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-3 + m)*Cosh[a + b*x]^2,x]

[Out]

-1/2*x^(-2 + m)/(2 - m) - (b^2*E^(2*a)*x^m*Gamma[-2 + m, -2*b*x])/(2^m*(-(b*x))^m) - (b^2*x^m*Gamma[-2 + m, 2*
b*x])/(2^m*E^(2*a)*(b*x)^m)

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fricas [A]  time = 0.44, size = 136, normalized size = 1.62 \[ \frac {4 \, b x \cosh \left ({\left (m - 3\right )} \log \relax (x)\right ) - {\left (m - 2\right )} \cosh \left ({\left (m - 3\right )} \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m - 2, 2 \, b x\right ) + {\left (m - 2\right )} \cosh \left ({\left (m - 3\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m - 2, -2 \, b x\right ) + {\left (m - 2\right )} \Gamma \left (m - 2, 2 \, b x\right ) \sinh \left ({\left (m - 3\right )} \log \left (2 \, b\right ) + 2 \, a\right ) - {\left (m - 2\right )} \Gamma \left (m - 2, -2 \, b x\right ) \sinh \left ({\left (m - 3\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left ({\left (m - 3\right )} \log \relax (x)\right )}{8 \, {\left (b m - 2 \, b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-3+m)*cosh(b*x+a)^2,x, algorithm="fricas")

[Out]

1/8*(4*b*x*cosh((m - 3)*log(x)) - (m - 2)*cosh((m - 3)*log(2*b) + 2*a)*gamma(m - 2, 2*b*x) + (m - 2)*cosh((m -
3)*log(-2*b) - 2*a)*gamma(m - 2, -2*b*x) + (m - 2)*gamma(m - 2, 2*b*x)*sinh((m - 3)*log(2*b) + 2*a) - (m - 2)
*gamma(m - 2, -2*b*x)*sinh((m - 3)*log(-2*b) - 2*a) + 4*b*x*sinh((m - 3)*log(x)))/(b*m - 2*b)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m - 3} \cosh \left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-3+m)*cosh(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^(m - 3)*cosh(b*x + a)^2, x)

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int x^{-3+m} \left (\cosh ^{2}\left (b x +a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-3+m)*cosh(b*x+a)^2,x)

[Out]

int(x^(-3+m)*cosh(b*x+a)^2,x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-3+m)*cosh(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(m-3>0)', see `assume?` for mor
e details)Is m-3 equal to -1?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{m-3}\,{\mathrm {cosh}\left (a+b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(m - 3)*cosh(a + b*x)^2,x)

[Out]

int(x^(m - 3)*cosh(a + b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m - 3} \cosh ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-3+m)*cosh(b*x+a)**2,x)

[Out]

Integral(x**(m - 3)*cosh(a + b*x)**2, x)

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