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

Optimal. Leaf size=144 \[ \frac{2^{-m-3} e^{2 a-\frac{2 b c}{d}} (c+d x)^m \left (-\frac{b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 b (c+d x)}{d}\right )}{b}-\frac{2^{-m-3} e^{\frac{2 b c}{d}-2 a} (c+d x)^m \left (\frac{b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 b (c+d x)}{d}\right )}{b}+\frac{(c+d x)^{m+1}}{2 d (m+1)} \]

[Out]

(c + d*x)^(1 + m)/(2*d*(1 + m)) + (2^(-3 - m)*E^(2*a - (2*b*c)/d)*(c + d*x)^m*Gamma[1 + m, (-2*b*(c + d*x))/d]
)/(b*(-((b*(c + d*x))/d))^m) - (2^(-3 - m)*E^(-2*a + (2*b*c)/d)*(c + d*x)^m*Gamma[1 + m, (2*b*(c + d*x))/d])/(
b*((b*(c + d*x))/d)^m)

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Rubi [A]  time = 0.184035, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3312, 3307, 2181} \[ \frac{2^{-m-3} e^{2 a-\frac{2 b c}{d}} (c+d x)^m \left (-\frac{b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 b (c+d x)}{d}\right )}{b}-\frac{2^{-m-3} e^{\frac{2 b c}{d}-2 a} (c+d x)^m \left (\frac{b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 b (c+d x)}{d}\right )}{b}+\frac{(c+d x)^{m+1}}{2 d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^m*Cosh[a + b*x]^2,x]

[Out]

(c + d*x)^(1 + m)/(2*d*(1 + m)) + (2^(-3 - m)*E^(2*a - (2*b*c)/d)*(c + d*x)^m*Gamma[1 + m, (-2*b*(c + d*x))/d]
)/(b*(-((b*(c + d*x))/d))^m) - (2^(-3 - m)*E^(-2*a + (2*b*c)/d)*(c + d*x)^m*Gamma[1 + m, (2*b*(c + d*x))/d])/(
b*((b*(c + d*x))/d)^m)

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])
)

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 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]

Rubi steps

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

Mathematica [A]  time = 0.201685, size = 132, normalized size = 0.92 \[ \frac{1}{8} (c+d x)^m \left (\frac{2^{-m} e^{2 a-\frac{2 b c}{d}} \left (-\frac{b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 b (c+d x)}{d}\right )}{b}-\frac{2^{-m} e^{\frac{2 b c}{d}-2 a} \left (\frac{b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 b (c+d x)}{d}\right )}{b}+\frac{4 c+4 d x}{d m+d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c + d*x)^m*((4*c + 4*d*x)/(d + d*m) + (E^(2*a - (2*b*c)/d)*Gamma[1 + m, (-2*b*(c + d*x))/d])/(2^m*b*(-((b*(c
+ d*x))/d))^m) - (E^(-2*a + (2*b*c)/d)*Gamma[1 + m, (2*b*(c + d*x))/d])/(2^m*b*((b*(c + d*x))/d)^m)))/8

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Maple [F]  time = 0.091, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{m} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.79576, size = 597, normalized size = 4.15 \begin{align*} -\frac{{\left (d m + d\right )} \cosh \left (\frac{d m \log \left (\frac{2 \, b}{d}\right ) - 2 \, b c + 2 \, a d}{d}\right ) \Gamma \left (m + 1, \frac{2 \,{\left (b d x + b c\right )}}{d}\right ) -{\left (d m + d\right )} \cosh \left (\frac{d m \log \left (-\frac{2 \, b}{d}\right ) + 2 \, b c - 2 \, a d}{d}\right ) \Gamma \left (m + 1, -\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) -{\left (d m + d\right )} \Gamma \left (m + 1, \frac{2 \,{\left (b d x + b c\right )}}{d}\right ) \sinh \left (\frac{d m \log \left (\frac{2 \, b}{d}\right ) - 2 \, b c + 2 \, a d}{d}\right ) +{\left (d m + d\right )} \Gamma \left (m + 1, -\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) \sinh \left (\frac{d m \log \left (-\frac{2 \, b}{d}\right ) + 2 \, b c - 2 \, a d}{d}\right ) - 4 \,{\left (b d x + b c\right )} \cosh \left (m \log \left (d x + c\right )\right ) - 4 \,{\left (b d x + b c\right )} \sinh \left (m \log \left (d x + c\right )\right )}{8 \,{\left (b d m + b d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{m} \cosh ^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{m} \cosh \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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