Integrand size = 13, antiderivative size = 100 \[ \int \sinh (a+b x) \tanh (c+d x) \, dx=\frac {\cosh (a+b x)}{b}-\frac {e^{-a-b x} \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{2 d},1-\frac {b}{2 d},-e^{2 (c+d x)}\right )}{b}-\frac {e^{a+b x} \operatorname {Hypergeometric2F1}\left (1,\frac {b}{2 d},1+\frac {b}{2 d},-e^{2 (c+d x)}\right )}{b} \] Output:
cosh(b*x+a)/b-exp(-b*x-a)*hypergeom([1, -1/2*b/d],[1-1/2*b/d],-exp(2*d*x+2 *c))/b-exp(b*x+a)*hypergeom([1, 1/2*b/d],[1+1/2*b/d],-exp(2*d*x+2*c))/b
Time = 0.68 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.03 \[ \int \sinh (a+b x) \tanh (c+d x) \, dx=\frac {e^{-a-b x} \left (1+e^{2 (a+b x)}-2 \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{2 d},1-\frac {b}{2 d},-e^{2 (c+d x)}\right )-2 e^{2 (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b}{2 d},1+\frac {b}{2 d},-e^{2 (c+d x)}\right )\right )}{2 b} \] Input:
Integrate[Sinh[a + b*x]*Tanh[c + d*x],x]
Output:
(E^(-a - b*x)*(1 + E^(2*(a + b*x)) - 2*Hypergeometric2F1[1, -1/2*b/d, 1 - b/(2*d), -E^(2*(c + d*x))] - 2*E^(2*(a + b*x))*Hypergeometric2F1[1, b/(2*d ), 1 + b/(2*d), -E^(2*(c + d*x))]))/(2*b)
Time = 0.35 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.21, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6135, 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.
\(\displaystyle \int \sinh (a+b x) \tanh (c+d x) \, dx\) |
\(\Big \downarrow \) 6135 |
\(\displaystyle \int \left (\frac {e^{-a-b x}}{e^{2 (c+d x)}+1}-\frac {e^{a+b x}}{e^{2 (c+d x)}+1}-\frac {1}{2} e^{-a-b x}+\frac {1}{2} e^{a+b x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{-a-b x} \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{2 d},1-\frac {b}{2 d},-e^{2 (c+d x)}\right )}{b}-\frac {e^{a+b x} \operatorname {Hypergeometric2F1}\left (1,\frac {b}{2 d},\frac {b}{2 d}+1,-e^{2 (c+d x)}\right )}{b}+\frac {e^{-a-b x}}{2 b}+\frac {e^{a+b x}}{2 b}\) |
Input:
Int[Sinh[a + b*x]*Tanh[c + d*x],x]
Output:
E^(-a - b*x)/(2*b) + E^(a + b*x)/(2*b) - (E^(-a - b*x)*Hypergeometric2F1[1 , -1/2*b/d, 1 - b/(2*d), -E^(2*(c + d*x))])/b - (E^(a + b*x)*Hypergeometri c2F1[1, b/(2*d), 1 + b/(2*d), -E^(2*(c + d*x))])/b
Int[Sinh[(a_.) + (b_.)*(x_)]*Tanh[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[-E^ (-(a + b*x))/2 + E^(a + b*x)/2 + 1/(E^(a + b*x)*(1 + E^(2*(c + d*x)))) - E^ (a + b*x)/(1 + E^(2*(c + d*x))), x] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - d^2, 0]
\[\int \sinh \left (b x +a \right ) \tanh \left (d x +c \right )d x\]
Input:
int(sinh(b*x+a)*tanh(d*x+c),x)
Output:
int(sinh(b*x+a)*tanh(d*x+c),x)
\[ \int \sinh (a+b x) \tanh (c+d x) \, dx=\int { \sinh \left (b x + a\right ) \tanh \left (d x + c\right ) \,d x } \] Input:
integrate(sinh(b*x+a)*tanh(d*x+c),x, algorithm="fricas")
Output:
integral(sinh(b*x + a)*tanh(d*x + c), x)
\[ \int \sinh (a+b x) \tanh (c+d x) \, dx=\int \sinh {\left (a + b x \right )} \tanh {\left (c + d x \right )}\, dx \] Input:
integrate(sinh(b*x+a)*tanh(d*x+c),x)
Output:
Integral(sinh(a + b*x)*tanh(c + d*x), x)
\[ \int \sinh (a+b x) \tanh (c+d x) \, dx=\int { \sinh \left (b x + a\right ) \tanh \left (d x + c\right ) \,d x } \] Input:
integrate(sinh(b*x+a)*tanh(d*x+c),x, algorithm="maxima")
Output:
1/2*(e^(2*b*x + 2*a) + 1)*e^(-b*x - a)/b - 1/2*integrate(2*(e^(2*b*x + 2*a ) - 1)/(e^(b*x + 2*d*x + a + 2*c) + e^(b*x + a)), x)
\[ \int \sinh (a+b x) \tanh (c+d x) \, dx=\int { \sinh \left (b x + a\right ) \tanh \left (d x + c\right ) \,d x } \] Input:
integrate(sinh(b*x+a)*tanh(d*x+c),x, algorithm="giac")
Output:
integrate(sinh(b*x + a)*tanh(d*x + c), x)
Timed out. \[ \int \sinh (a+b x) \tanh (c+d x) \, dx=\int \mathrm {sinh}\left (a+b\,x\right )\,\mathrm {tanh}\left (c+d\,x\right ) \,d x \] Input:
int(sinh(a + b*x)*tanh(c + d*x),x)
Output:
int(sinh(a + b*x)*tanh(c + d*x), x)
\[ \int \sinh (a+b x) \tanh (c+d x) \, dx=\frac {e^{2 b x +2 a}-2 e^{b x +2 a} \left (\int \frac {e^{b x}}{e^{2 d x +2 c}+1}d x \right ) b +2 e^{b x} \left (\int \frac {1}{e^{b x +2 d x +2 c}+e^{b x}}d x \right ) b +1}{2 e^{b x +a} b} \] Input:
int(sinh(b*x+a)*tanh(d*x+c),x)
Output:
(e**(2*a + 2*b*x) - 2*e**(2*a + b*x)*int(e**(b*x)/(e**(2*c + 2*d*x) + 1),x )*b + 2*e**(b*x)*int(1/(e**(b*x + 2*c + 2*d*x) + e**(b*x)),x)*b + 1)/(2*e* *(a + b*x)*b)