\(\int (c+d x)^m \cosh ^2(a+b x) \, dx\) [77]

Optimal result

Integrand size = 16, antiderivative size = 144 \[ \int (c+d x)^m \cosh ^2(a+b x) \, dx=\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {2^{-3-m} e^{2 a-\frac {2 b c}{d}} (c+d x)^m \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-3-m} e^{-2 a+\frac {2 b c}{d}} (c+d x)^m \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 b (c+d x)}{d}\right )}{b} \] Output:

1/2*(d*x+c)^(1+m)/d/(1+m)+2^(-3-m)*exp(2*a-2*b*c/d)*(d*x+c)^m*GAMMA(1+m,-2 
*b*(d*x+c)/d)/b/((-b*(d*x+c)/d)^m)-2^(-3-m)*exp(-2*a+2*b*c/d)*(d*x+c)^m*GA 
MMA(1+m,2*b*(d*x+c)/d)/b/((b*(d*x+c)/d)^m)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.92 \[ \int (c+d x)^m \cosh ^2(a+b x) \, dx=\frac {1}{8} (c+d x)^m \left (\frac {4 c+4 d x}{d+d m}+\frac {2^{-m} e^{2 a-\frac {2 b c}{d}} \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-m} e^{-2 a+\frac {2 b c}{d}} \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 b (c+d x)}{d}\right )}{b}\right ) \] Input:

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

Output:

((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
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3042, 3793, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (i a+i b x+\frac {\pi }{2}\right )^2 (c+d x)^mdx\)

\(\Big \downarrow \) 3793

\(\displaystyle \int \left (\frac {1}{2} \cosh (2 a+2 b x) (c+d x)^m+\frac {1}{2} (c+d x)^m\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2^{-m-3} e^{2 a-\frac {2 b c}{d}} (c+d x)^m \left (-\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 b (c+d x)}{d}\right )}{b}-\frac {2^{-m-3} e^{\frac {2 b c}{d}-2 a} (c+d x)^m \left (\frac {b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 b (c+d x)}{d}\right )}{b}+\frac {(c+d x)^{m+1}}{2 d (m+1)}\)

Input:

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

Output:

(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)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[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]))
 

Maple [F]

\[\int \left (d x +c \right )^{m} \cosh \left (b x +a \right )^{2}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.67 \[ \int (c+d x)^m \cosh ^2(a+b x) \, dx=-\frac {{\left (d m + d\right )} \cosh \left (\frac {d m \log \left (\frac {2 \, b}{d}\right ) - 2 \, b c + 2 \, a d}{d}\right ) \Gamma \left (m + 1, \frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (d m + d\right )} \cosh \left (\frac {d m \log \left (-\frac {2 \, b}{d}\right ) + 2 \, b c - 2 \, a d}{d}\right ) \Gamma \left (m + 1, -\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (d m + d\right )} \Gamma \left (m + 1, \frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (\frac {2 \, b}{d}\right ) - 2 \, b c + 2 \, a d}{d}\right ) + {\left (d m + d\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (-\frac {2 \, b}{d}\right ) + 2 \, b c - 2 \, a d}{d}\right ) - 4 \, {\left (b d x + b c\right )} \cosh \left (m \log \left (d x + c\right )\right ) - 4 \, {\left (b d x + b c\right )} \sinh \left (m \log \left (d x + c\right )\right )}{8 \, {\left (b d m + b d\right )}} \] Input:

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

Output:

-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)*si 
nh((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)
 

Sympy [F]

\[ \int (c+d x)^m \cosh ^2(a+b x) \, dx=\int \left (c + d x\right )^{m} \cosh ^{2}{\left (a + b x \right )}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.71 \[ \int (c+d x)^m \cosh ^2(a+b x) \, dx=-\frac {{\left (d x + c\right )}^{m + 1} e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )} E_{-m}\left (\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, d} - \frac {{\left (d x + c\right )}^{m + 1} e^{\left (2 \, a - \frac {2 \, b c}{d}\right )} E_{-m}\left (-\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, d} + \frac {{\left (d x + c\right )}^{m + 1}}{2 \, d {\left (m + 1\right )}} \] Input:

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

Output:

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

Giac [F]

\[ \int (c+d x)^m \cosh ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cosh \left (b x + a\right )^{2} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^m \cosh ^2(a+b x) \, dx=\int {\mathrm {cosh}\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^m \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (c+d x)^m \cosh ^2(a+b x) \, dx=\frac {e^{4 b x +4 a} \left (d x +c \right )^{m} d m +e^{4 b x +4 a} \left (d x +c \right )^{m} d +4 e^{2 b x +2 a} \left (d x +c \right )^{m} b c +4 e^{2 b x +2 a} \left (d x +c \right )^{m} b d x -\left (d x +c \right )^{m} d m -\left (d x +c \right )^{m} d -e^{2 b x +4 a} \left (\int \frac {e^{2 b x} \left (d x +c \right )^{m}}{d x +c}d x \right ) d^{2} m^{2}-e^{2 b x +4 a} \left (\int \frac {e^{2 b x} \left (d x +c \right )^{m}}{d x +c}d x \right ) d^{2} m +e^{2 b x +2 a} \left (\int \frac {\left (d x +c \right )^{m}}{e^{2 b x +2 a} c +e^{2 b x +2 a} d x}d x \right ) d^{2} m^{2}+e^{2 b x +2 a} \left (\int \frac {\left (d x +c \right )^{m}}{e^{2 b x +2 a} c +e^{2 b x +2 a} d x}d x \right ) d^{2} m}{8 e^{2 b x +2 a} b d \left (m +1\right )} \] Input:

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

Output:

(e**(4*a + 4*b*x)*(c + d*x)**m*d*m + e**(4*a + 4*b*x)*(c + d*x)**m*d + 4*e 
**(2*a + 2*b*x)*(c + d*x)**m*b*c + 4*e**(2*a + 2*b*x)*(c + d*x)**m*b*d*x - 
 (c + d*x)**m*d*m - (c + d*x)**m*d - e**(4*a + 2*b*x)*int((e**(2*b*x)*(c + 
 d*x)**m)/(c + d*x),x)*d**2*m**2 - e**(4*a + 2*b*x)*int((e**(2*b*x)*(c + d 
*x)**m)/(c + d*x),x)*d**2*m + e**(2*a + 2*b*x)*int((c + d*x)**m/(e**(2*a + 
 2*b*x)*c + e**(2*a + 2*b*x)*d*x),x)*d**2*m**2 + e**(2*a + 2*b*x)*int((c + 
 d*x)**m/(e**(2*a + 2*b*x)*c + e**(2*a + 2*b*x)*d*x),x)*d**2*m)/(8*e**(2*a 
 + 2*b*x)*b*d*(m + 1))