Integrand size = 17, antiderivative size = 39 \[ \int \cosh ^3(a+b x) \sinh ^p(a+b x) \, dx=\frac {\sinh ^{1+p}(a+b x)}{b (1+p)}+\frac {\sinh ^{3+p}(a+b x)}{b (3+p)} \] Output:
sinh(b*x+a)^(p+1)/b/(p+1)+sinh(b*x+a)^(3+p)/b/(3+p)
Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \cosh ^3(a+b x) \sinh ^p(a+b x) \, dx=\frac {\sinh ^{1+p}(a+b x)}{b (1+p)}+\frac {\sinh ^{3+p}(a+b x)}{b (3+p)} \] Input:
Integrate[Cosh[a + b*x]^3*Sinh[a + b*x]^p,x]
Output:
Sinh[a + b*x]^(1 + p)/(b*(1 + p)) + Sinh[a + b*x]^(3 + p)/(b*(3 + p))
Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 3044, 244, 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.
\(\displaystyle \int \cosh ^3(a+b x) \sinh ^p(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (i a+i b x)^3 (-i \sin (i a+i b x))^pdx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \sinh ^p(a+b x) \left (\sinh ^2(a+b x)+1\right )d\sinh (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\sinh ^p(a+b x)+\sinh ^{p+2}(a+b x)\right )d\sinh (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\sinh ^{p+1}(a+b x)}{p+1}+\frac {\sinh ^{p+3}(a+b x)}{p+3}}{b}\) |
Input:
Int[Cosh[a + b*x]^3*Sinh[a + b*x]^p,x]
Output:
(Sinh[a + b*x]^(1 + p)/(1 + p) + Sinh[a + b*x]^(3 + p)/(3 + p))/b
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.38
\[\frac {\sinh \left (b x +a \right )^{3} {\mathrm e}^{p \ln \left (\sinh \left (b x +a \right )\right )}}{b \left (3+p \right )}+\frac {\sinh \left (b x +a \right ) {\mathrm e}^{p \ln \left (\sinh \left (b x +a \right )\right )}}{b \left (p +1\right )}\]
Input:
int(cosh(b*x+a)^3*sinh(b*x+a)^p,x)
Output:
1/b/(3+p)*sinh(b*x+a)^3*exp(p*ln(sinh(b*x+a)))+1/b/(p+1)*sinh(b*x+a)*exp(p *ln(sinh(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (39) = 78\).
Time = 0.08 (sec) , antiderivative size = 175, normalized size of antiderivative = 4.49 \[ \int \cosh ^3(a+b x) \sinh ^p(a+b x) \, dx=\frac {{\left ({\left (p + 1\right )} \sinh \left (b x + a\right )^{3} + {\left (3 \, {\left (p + 1\right )} \cosh \left (b x + a\right )^{2} + p + 9\right )} \sinh \left (b x + a\right )\right )} \cosh \left (p \log \left (\sinh \left (b x + a\right )\right )\right ) + {\left ({\left (p + 1\right )} \sinh \left (b x + a\right )^{3} + {\left (3 \, {\left (p + 1\right )} \cosh \left (b x + a\right )^{2} + p + 9\right )} \sinh \left (b x + a\right )\right )} \sinh \left (p \log \left (\sinh \left (b x + a\right )\right )\right )}{4 \, {\left ({\left (b p^{2} + 4 \, b p + 3 \, b\right )} \cosh \left (b x + a\right )^{4} - 2 \, {\left (b p^{2} + 4 \, b p + 3 \, b\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + {\left (b p^{2} + 4 \, b p + 3 \, b\right )} \sinh \left (b x + a\right )^{4}\right )}} \] Input:
integrate(cosh(b*x+a)^3*sinh(b*x+a)^p,x, algorithm="fricas")
Output:
1/4*(((p + 1)*sinh(b*x + a)^3 + (3*(p + 1)*cosh(b*x + a)^2 + p + 9)*sinh(b *x + a))*cosh(p*log(sinh(b*x + a))) + ((p + 1)*sinh(b*x + a)^3 + (3*(p + 1 )*cosh(b*x + a)^2 + p + 9)*sinh(b*x + a))*sinh(p*log(sinh(b*x + a))))/((b* p^2 + 4*b*p + 3*b)*cosh(b*x + a)^4 - 2*(b*p^2 + 4*b*p + 3*b)*cosh(b*x + a) ^2*sinh(b*x + a)^2 + (b*p^2 + 4*b*p + 3*b)*sinh(b*x + a)^4)
Leaf count of result is larger than twice the leaf count of optimal. 638 vs. \(2 (29) = 58\).
Time = 1.18 (sec) , antiderivative size = 638, normalized size of antiderivative = 16.36 \[ \int \cosh ^3(a+b x) \sinh ^p(a+b x) \, dx =\text {Too large to display} \] Input:
integrate(cosh(b*x+a)**3*sinh(b*x+a)**p,x)
Output:
Piecewise((x*sinh(a)**p*cosh(a)**3, Eq(b, 0)), (log(sinh(a + b*x))/b - cos h(a + b*x)**2/(2*b*sinh(a + b*x)**2), Eq(p, -3)), (b*x*tanh(a/2 + b*x/2)** 4/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) - 2*b*x*tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + b*x /(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) - 2*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + 4*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**2/(b*ta nh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) - 2*log(tanh(a/2 + b*x/ 2) + 1)/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + log(tanh (a/2 + b*x/2))*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) - 2*log(tanh(a/2 + b*x/2))*tanh(a/2 + b*x/2)**2/(b*tanh( a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + log(tanh(a/2 + b*x/2))/( b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + 2*tanh(a/2 + b*x/ 2)**2/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b), Eq(p, -1)), (p*sinh(a + b*x)*sinh(a + b*x)**p*cosh(a + b*x)**2/(b*p**2 + 4*b*p + 3*b) - 2*sinh(a + b*x)**3*sinh(a + b*x)**p/(b*p**2 + 4*b*p + 3*b) + 3*sinh(a + b*x)*sinh(a + b*x)**p*cosh(a + b*x)**2/(b*p**2 + 4*b*p + 3*b), True))
Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (39) = 78\).
Time = 0.19 (sec) , antiderivative size = 373, normalized size of antiderivative = 9.56 \[ \int \cosh ^3(a+b x) \sinh ^p(a+b x) \, dx=\frac {p e^{\left ({\left (b x + a\right )} p + 3 \, b x + p \log \left (e^{\left (-b x - a\right )} + 1\right ) + p \log \left (-e^{\left (-b x - a\right )} + 1\right ) + 3 \, a\right )}}{8 \, {\left (2^{p} p^{2} + 2^{p + 2} p + 3 \cdot 2^{p}\right )} b} + \frac {{\left (p + 9\right )} e^{\left ({\left (b x + a\right )} p + b x + p \log \left (e^{\left (-b x - a\right )} + 1\right ) + p \log \left (-e^{\left (-b x - a\right )} + 1\right ) + a\right )}}{8 \, {\left (2^{p} p^{2} + 2^{p + 2} p + 3 \cdot 2^{p}\right )} b} - \frac {{\left (p + 9\right )} e^{\left ({\left (b x + a\right )} p - b x + p \log \left (e^{\left (-b x - a\right )} + 1\right ) + p \log \left (-e^{\left (-b x - a\right )} + 1\right ) - a\right )}}{8 \, {\left (2^{p} p^{2} + 2^{p + 2} p + 3 \cdot 2^{p}\right )} b} - \frac {{\left (p + 1\right )} e^{\left ({\left (b x + a\right )} p - 3 \, b x + p \log \left (e^{\left (-b x - a\right )} + 1\right ) + p \log \left (-e^{\left (-b x - a\right )} + 1\right ) - 3 \, a\right )}}{8 \, {\left (2^{p} p^{2} + 2^{p + 2} p + 3 \cdot 2^{p}\right )} b} + \frac {e^{\left ({\left (b x + a\right )} p + 3 \, b x + p \log \left (e^{\left (-b x - a\right )} + 1\right ) + p \log \left (-e^{\left (-b x - a\right )} + 1\right ) + 3 \, a\right )}}{8 \, {\left (2^{p} p^{2} + 2^{p + 2} p + 3 \cdot 2^{p}\right )} b} \] Input:
integrate(cosh(b*x+a)^3*sinh(b*x+a)^p,x, algorithm="maxima")
Output:
1/8*p*e^((b*x + a)*p + 3*b*x + p*log(e^(-b*x - a) + 1) + p*log(-e^(-b*x - a) + 1) + 3*a)/((2^p*p^2 + 2^(p + 2)*p + 3*2^p)*b) + 1/8*(p + 9)*e^((b*x + a)*p + b*x + p*log(e^(-b*x - a) + 1) + p*log(-e^(-b*x - a) + 1) + a)/((2^ p*p^2 + 2^(p + 2)*p + 3*2^p)*b) - 1/8*(p + 9)*e^((b*x + a)*p - b*x + p*log (e^(-b*x - a) + 1) + p*log(-e^(-b*x - a) + 1) - a)/((2^p*p^2 + 2^(p + 2)*p + 3*2^p)*b) - 1/8*(p + 1)*e^((b*x + a)*p - 3*b*x + p*log(e^(-b*x - a) + 1 ) + p*log(-e^(-b*x - a) + 1) - 3*a)/((2^p*p^2 + 2^(p + 2)*p + 3*2^p)*b) + 1/8*e^((b*x + a)*p + 3*b*x + p*log(e^(-b*x - a) + 1) + p*log(-e^(-b*x - a) + 1) + 3*a)/((2^p*p^2 + 2^(p + 2)*p + 3*2^p)*b)
Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (39) = 78\).
Time = 0.18 (sec) , antiderivative size = 327, normalized size of antiderivative = 8.38 \[ \int \cosh ^3(a+b x) \sinh ^p(a+b x) \, dx=\frac {p e^{\left (-b p x - a p + 7 \, b x + p \log \left (\frac {1}{2} \, e^{\left (2 \, b x + 2 \, a\right )} - \frac {1}{2}\right ) + 7 \, a\right )} + p e^{\left (-b p x - a p + 5 \, b x + p \log \left (\frac {1}{2} \, e^{\left (2 \, b x + 2 \, a\right )} - \frac {1}{2}\right ) + 5 \, a\right )} - p e^{\left (-b p x - a p + 3 \, b x + p \log \left (\frac {1}{2} \, e^{\left (2 \, b x + 2 \, a\right )} - \frac {1}{2}\right ) + 3 \, a\right )} - p e^{\left (-b p x - a p + b x + p \log \left (\frac {1}{2} \, e^{\left (2 \, b x + 2 \, a\right )} - \frac {1}{2}\right ) + a\right )} + e^{\left (-b p x - a p + 7 \, b x + p \log \left (\frac {1}{2} \, e^{\left (2 \, b x + 2 \, a\right )} - \frac {1}{2}\right ) + 7 \, a\right )} + 9 \, e^{\left (-b p x - a p + 5 \, b x + p \log \left (\frac {1}{2} \, e^{\left (2 \, b x + 2 \, a\right )} - \frac {1}{2}\right ) + 5 \, a\right )} - 9 \, e^{\left (-b p x - a p + 3 \, b x + p \log \left (\frac {1}{2} \, e^{\left (2 \, b x + 2 \, a\right )} - \frac {1}{2}\right ) + 3 \, a\right )} - e^{\left (-b p x - a p + b x + p \log \left (\frac {1}{2} \, e^{\left (2 \, b x + 2 \, a\right )} - \frac {1}{2}\right ) + a\right )}}{8 \, {\left (b p^{2} e^{\left (4 \, b x + 4 \, a\right )} + 4 \, b p e^{\left (4 \, b x + 4 \, a\right )} + 3 \, b e^{\left (4 \, b x + 4 \, a\right )}\right )}} \] Input:
integrate(cosh(b*x+a)^3*sinh(b*x+a)^p,x, algorithm="giac")
Output:
1/8*(p*e^(-b*p*x - a*p + 7*b*x + p*log(1/2*e^(2*b*x + 2*a) - 1/2) + 7*a) + p*e^(-b*p*x - a*p + 5*b*x + p*log(1/2*e^(2*b*x + 2*a) - 1/2) + 5*a) - p*e ^(-b*p*x - a*p + 3*b*x + p*log(1/2*e^(2*b*x + 2*a) - 1/2) + 3*a) - p*e^(-b *p*x - a*p + b*x + p*log(1/2*e^(2*b*x + 2*a) - 1/2) + a) + e^(-b*p*x - a*p + 7*b*x + p*log(1/2*e^(2*b*x + 2*a) - 1/2) + 7*a) + 9*e^(-b*p*x - a*p + 5 *b*x + p*log(1/2*e^(2*b*x + 2*a) - 1/2) + 5*a) - 9*e^(-b*p*x - a*p + 3*b*x + p*log(1/2*e^(2*b*x + 2*a) - 1/2) + 3*a) - e^(-b*p*x - a*p + b*x + p*log (1/2*e^(2*b*x + 2*a) - 1/2) + a))/(b*p^2*e^(4*b*x + 4*a) + 4*b*p*e^(4*b*x + 4*a) + 3*b*e^(4*b*x + 4*a))
Time = 1.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 3.46 \[ \int \cosh ^3(a+b x) \sinh ^p(a+b x) \, dx=-{\left (\frac {1}{2}\right )}^p\,{\mathrm {e}}^{-3\,a-3\,b\,x}\,{\left ({\mathrm {e}}^{a+b\,x}-{\mathrm {e}}^{-a-b\,x}\right )}^p\,\left (\frac {\frac {p}{8}+\frac {1}{8}}{b\,\left (p^2+4\,p+3\right )}+\frac {{\mathrm {e}}^{2\,a+2\,b\,x}\,\left (p+9\right )}{8\,b\,\left (p^2+4\,p+3\right )}-\frac {{\mathrm {e}}^{6\,a+6\,b\,x}\,\left (p+1\right )}{8\,b\,\left (p^2+4\,p+3\right )}-\frac {{\mathrm {e}}^{4\,a+4\,b\,x}\,\left (p+9\right )}{8\,b\,\left (p^2+4\,p+3\right )}\right ) \] Input:
int(cosh(a + b*x)^3*sinh(a + b*x)^p,x)
Output:
-(1/2)^p*exp(- 3*a - 3*b*x)*(exp(a + b*x) - exp(- a - b*x))^p*((p/8 + 1/8) /(b*(4*p + p^2 + 3)) + (exp(2*a + 2*b*x)*(p + 9))/(8*b*(4*p + p^2 + 3)) - (exp(6*a + 6*b*x)*(p + 1))/(8*b*(4*p + p^2 + 3)) - (exp(4*a + 4*b*x)*(p + 9))/(8*b*(4*p + p^2 + 3)))
Time = 0.22 (sec) , antiderivative size = 129, normalized size of antiderivative = 3.31 \[ \int \cosh ^3(a+b x) \sinh ^p(a+b x) \, dx=\frac {\left (e^{2 b x +2 a}-1\right )^{p} \left (e^{6 b x +6 a} p +e^{6 b x +6 a}+e^{4 b x +4 a} p +9 e^{4 b x +4 a}-e^{2 b x +2 a} p -9 e^{2 b x +2 a}-p -1\right )}{8 e^{b p x +a p +3 b x +3 a} 2^{p} b \left (p^{2}+4 p +3\right )} \] Input:
int(cosh(b*x+a)^3*sinh(b*x+a)^p,x)
Output:
((e**(2*a + 2*b*x) - 1)**p*(e**(6*a + 6*b*x)*p + e**(6*a + 6*b*x) + e**(4* a + 4*b*x)*p + 9*e**(4*a + 4*b*x) - e**(2*a + 2*b*x)*p - 9*e**(2*a + 2*b*x ) - p - 1))/(8*e**(a*p + 3*a + b*p*x + 3*b*x)*2**p*b*(p**2 + 4*p + 3))