### 3.14 $$\int \cosh ^m(a+b x) \sinh ^5(a+b x) \, dx$$

Optimal. Leaf size=59 $\frac{\cosh ^{m+1}(a+b x)}{b (m+1)}-\frac{2 \cosh ^{m+3}(a+b x)}{b (m+3)}+\frac{\cosh ^{m+5}(a+b x)}{b (m+5)}$

[Out]

Cosh[a + b*x]^(1 + m)/(b*(1 + m)) - (2*Cosh[a + b*x]^(3 + m))/(b*(3 + m)) + Cosh[a + b*x]^(5 + m)/(b*(5 + m))

________________________________________________________________________________________

Rubi [A]  time = 0.0607817, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {2565, 270} $\frac{\cosh ^{m+1}(a+b x)}{b (m+1)}-\frac{2 \cosh ^{m+3}(a+b x)}{b (m+3)}+\frac{\cosh ^{m+5}(a+b x)}{b (m+5)}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[a + b*x]^m*Sinh[a + b*x]^5,x]

[Out]

Cosh[a + b*x]^(1 + m)/(b*(1 + m)) - (2*Cosh[a + b*x]^(3 + m))/(b*(3 + m)) + Cosh[a + b*x]^(5 + m)/(b*(5 + m))

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \cosh ^m(a+b x) \sinh ^5(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^m \left (1-x^2\right )^2 \, dx,x,\cosh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^m-2 x^{2+m}+x^{4+m}\right ) \, dx,x,\cosh (a+b x)\right )}{b}\\ &=\frac{\cosh ^{1+m}(a+b x)}{b (1+m)}-\frac{2 \cosh ^{3+m}(a+b x)}{b (3+m)}+\frac{\cosh ^{5+m}(a+b x)}{b (5+m)}\\ \end{align*}

Mathematica [A]  time = 0.217858, size = 77, normalized size = 1.31 $\frac{\cosh ^{m+1}(a+b x) \left (-4 \left (m^2+8 m+7\right ) \cosh (2 (a+b x))+\left (m^2+4 m+3\right ) \cosh (4 (a+b x))+3 m^2+28 m+89\right )}{8 b (m+1) (m+3) (m+5)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cosh[a + b*x]^m*Sinh[a + b*x]^5,x]

[Out]

(Cosh[a + b*x]^(1 + m)*(89 + 28*m + 3*m^2 - 4*(7 + 8*m + m^2)*Cosh[2*(a + b*x)] + (3 + 4*m + m^2)*Cosh[4*(a +
b*x)]))/(8*b*(1 + m)*(3 + m)*(5 + m))

________________________________________________________________________________________

Maple [F]  time = 0.345, size = 0, normalized size = 0. \begin{align*} \int \left ( \cosh \left ( bx+a \right ) \right ) ^{m} \left ( \sinh \left ( bx+a \right ) \right ) ^{5}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^m*sinh(b*x+a)^5,x)

[Out]

int(cosh(b*x+a)^m*sinh(b*x+a)^5,x)

________________________________________________________________________________________

Maxima [B]  time = 1.69795, size = 753, normalized size = 12.76 \begin{align*} \frac{m^{2} e^{\left ({\left (b x + a\right )} m + 5 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) + 5 \, a\right )}}{32 \,{\left (2^{m} m^{3} + 9 \cdot 2^{m} m^{2} + 23 \cdot 2^{m} m + 15 \cdot 2^{m}\right )} b} + \frac{m e^{\left ({\left (b x + a\right )} m + 5 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) + 5 \, a\right )}}{8 \,{\left (2^{m} m^{3} + 9 \cdot 2^{m} m^{2} + 23 \cdot 2^{m} m + 15 \cdot 2^{m}\right )} b} - \frac{{\left (3 \, m^{2} + 28 \, m + 25\right )} e^{\left ({\left (b x + a\right )} m + 3 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) + 3 \, a\right )}}{32 \,{\left (2^{m} m^{3} + 9 \cdot 2^{m} m^{2} + 23 \cdot 2^{m} m + 15 \cdot 2^{m}\right )} b} + \frac{{\left (m^{2} + 12 \, m + 75\right )} e^{\left ({\left (b x + a\right )} m + b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) + a\right )}}{16 \,{\left (2^{m} m^{3} + 9 \cdot 2^{m} m^{2} + 23 \cdot 2^{m} m + 15 \cdot 2^{m}\right )} b} + \frac{{\left (m^{2} + 12 \, m + 75\right )} e^{\left ({\left (b x + a\right )} m - b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) - a\right )}}{16 \,{\left (2^{m} m^{3} + 9 \cdot 2^{m} m^{2} + 23 \cdot 2^{m} m + 15 \cdot 2^{m}\right )} b} - \frac{{\left (3 \, m^{2} + 28 \, m + 25\right )} e^{\left ({\left (b x + a\right )} m - 3 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) - 3 \, a\right )}}{32 \,{\left (2^{m} m^{3} + 9 \cdot 2^{m} m^{2} + 23 \cdot 2^{m} m + 15 \cdot 2^{m}\right )} b} + \frac{{\left (m^{2} + 4 \, m + 3\right )} e^{\left ({\left (b x + a\right )} m - 5 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) - 5 \, a\right )}}{32 \,{\left (2^{m} m^{3} + 9 \cdot 2^{m} m^{2} + 23 \cdot 2^{m} m + 15 \cdot 2^{m}\right )} b} + \frac{3 \, e^{\left ({\left (b x + a\right )} m + 5 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) + 5 \, a\right )}}{32 \,{\left (2^{m} m^{3} + 9 \cdot 2^{m} m^{2} + 23 \cdot 2^{m} m + 15 \cdot 2^{m}\right )} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^m*sinh(b*x+a)^5,x, algorithm="maxima")

[Out]

1/32*m^2*e^((b*x + a)*m + 5*b*x + m*log(e^(-2*b*x - 2*a) + 1) + 5*a)/((2^m*m^3 + 9*2^m*m^2 + 23*2^m*m + 15*2^m
)*b) + 1/8*m*e^((b*x + a)*m + 5*b*x + m*log(e^(-2*b*x - 2*a) + 1) + 5*a)/((2^m*m^3 + 9*2^m*m^2 + 23*2^m*m + 15
*2^m)*b) - 1/32*(3*m^2 + 28*m + 25)*e^((b*x + a)*m + 3*b*x + m*log(e^(-2*b*x - 2*a) + 1) + 3*a)/((2^m*m^3 + 9*
2^m*m^2 + 23*2^m*m + 15*2^m)*b) + 1/16*(m^2 + 12*m + 75)*e^((b*x + a)*m + b*x + m*log(e^(-2*b*x - 2*a) + 1) +
a)/((2^m*m^3 + 9*2^m*m^2 + 23*2^m*m + 15*2^m)*b) + 1/16*(m^2 + 12*m + 75)*e^((b*x + a)*m - b*x + m*log(e^(-2*b
*x - 2*a) + 1) - a)/((2^m*m^3 + 9*2^m*m^2 + 23*2^m*m + 15*2^m)*b) - 1/32*(3*m^2 + 28*m + 25)*e^((b*x + a)*m -
3*b*x + m*log(e^(-2*b*x - 2*a) + 1) - 3*a)/((2^m*m^3 + 9*2^m*m^2 + 23*2^m*m + 15*2^m)*b) + 1/32*(m^2 + 4*m + 3
)*e^((b*x + a)*m - 5*b*x + m*log(e^(-2*b*x - 2*a) + 1) - 5*a)/((2^m*m^3 + 9*2^m*m^2 + 23*2^m*m + 15*2^m)*b) +
3/32*e^((b*x + a)*m + 5*b*x + m*log(e^(-2*b*x - 2*a) + 1) + 5*a)/((2^m*m^3 + 9*2^m*m^2 + 23*2^m*m + 15*2^m)*b)

________________________________________________________________________________________

Fricas [B]  time = 2.20771, size = 1094, normalized size = 18.54 \begin{align*} \frac{{\left ({\left (m^{2} + 4 \, m + 3\right )} \cosh \left (b x + a\right )^{5} + 5 \,{\left (m^{2} + 4 \, m + 3\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} -{\left (3 \, m^{2} + 28 \, m + 25\right )} \cosh \left (b x + a\right )^{3} +{\left (10 \,{\left (m^{2} + 4 \, m + 3\right )} \cosh \left (b x + a\right )^{3} - 3 \,{\left (3 \, m^{2} + 28 \, m + 25\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 2 \,{\left (m^{2} + 12 \, m + 75\right )} \cosh \left (b x + a\right )\right )} \cosh \left (m \log \left (\cosh \left (b x + a\right )\right )\right ) +{\left ({\left (m^{2} + 4 \, m + 3\right )} \cosh \left (b x + a\right )^{5} + 5 \,{\left (m^{2} + 4 \, m + 3\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} -{\left (3 \, m^{2} + 28 \, m + 25\right )} \cosh \left (b x + a\right )^{3} +{\left (10 \,{\left (m^{2} + 4 \, m + 3\right )} \cosh \left (b x + a\right )^{3} - 3 \,{\left (3 \, m^{2} + 28 \, m + 25\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 2 \,{\left (m^{2} + 12 \, m + 75\right )} \cosh \left (b x + a\right )\right )} \sinh \left (m \log \left (\cosh \left (b x + a\right )\right )\right )}{16 \,{\left ({\left (b m^{3} + 9 \, b m^{2} + 23 \, b m + 15 \, b\right )} \cosh \left (b x + a\right )^{6} - 3 \,{\left (b m^{3} + 9 \, b m^{2} + 23 \, b m + 15 \, b\right )} \cosh \left (b x + a\right )^{4} \sinh \left (b x + a\right )^{2} + 3 \,{\left (b m^{3} + 9 \, b m^{2} + 23 \, b m + 15 \, b\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} -{\left (b m^{3} + 9 \, b m^{2} + 23 \, b m + 15 \, b\right )} \sinh \left (b x + a\right )^{6}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^m*sinh(b*x+a)^5,x, algorithm="fricas")

[Out]

1/16*(((m^2 + 4*m + 3)*cosh(b*x + a)^5 + 5*(m^2 + 4*m + 3)*cosh(b*x + a)*sinh(b*x + a)^4 - (3*m^2 + 28*m + 25)
*cosh(b*x + a)^3 + (10*(m^2 + 4*m + 3)*cosh(b*x + a)^3 - 3*(3*m^2 + 28*m + 25)*cosh(b*x + a))*sinh(b*x + a)^2
+ 2*(m^2 + 12*m + 75)*cosh(b*x + a))*cosh(m*log(cosh(b*x + a))) + ((m^2 + 4*m + 3)*cosh(b*x + a)^5 + 5*(m^2 +
4*m + 3)*cosh(b*x + a)*sinh(b*x + a)^4 - (3*m^2 + 28*m + 25)*cosh(b*x + a)^3 + (10*(m^2 + 4*m + 3)*cosh(b*x +
a)^3 - 3*(3*m^2 + 28*m + 25)*cosh(b*x + a))*sinh(b*x + a)^2 + 2*(m^2 + 12*m + 75)*cosh(b*x + a))*sinh(m*log(co
sh(b*x + a))))/((b*m^3 + 9*b*m^2 + 23*b*m + 15*b)*cosh(b*x + a)^6 - 3*(b*m^3 + 9*b*m^2 + 23*b*m + 15*b)*cosh(b
*x + a)^4*sinh(b*x + a)^2 + 3*(b*m^3 + 9*b*m^2 + 23*b*m + 15*b)*cosh(b*x + a)^2*sinh(b*x + a)^4 - (b*m^3 + 9*b
*m^2 + 23*b*m + 15*b)*sinh(b*x + a)^6)

________________________________________________________________________________________

Sympy [A]  time = 78.7123, size = 2351, normalized size = 39.85 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**m*sinh(b*x+a)**5,x)

[Out]

Piecewise((x*sinh(a)**5*cosh(a)**m, Eq(b, 0)), (log(cosh(a + b*x))/b - sinh(a + b*x)**4/(4*b*cosh(a + b*x)**4)
- sinh(a + b*x)**2/(2*b*cosh(a + b*x)**2), Eq(m, -5)), (-2*b*x*tanh(a/2 + b*x/2)**8/(b*tanh(a/2 + b*x/2)**8 -
2*b*tanh(a/2 + b*x/2)**4 + b) + 4*b*x*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**8 - 2*b*tanh(a/2 + b*x/2)**4
+ b) - 2*b*x/(b*tanh(a/2 + b*x/2)**8 - 2*b*tanh(a/2 + b*x/2)**4 + b) + 4*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2
+ b*x/2)**8/(b*tanh(a/2 + b*x/2)**8 - 2*b*tanh(a/2 + b*x/2)**4 + b) - 8*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 +
b*x/2)**4/(b*tanh(a/2 + b*x/2)**8 - 2*b*tanh(a/2 + b*x/2)**4 + b) + 4*log(tanh(a/2 + b*x/2) + 1)/(b*tanh(a/2 +
b*x/2)**8 - 2*b*tanh(a/2 + b*x/2)**4 + b) - 2*log(tanh(a/2 + b*x/2)**2 + 1)*tanh(a/2 + b*x/2)**8/(b*tanh(a/2
+ b*x/2)**8 - 2*b*tanh(a/2 + b*x/2)**4 + b) + 4*log(tanh(a/2 + b*x/2)**2 + 1)*tanh(a/2 + b*x/2)**4/(b*tanh(a/2
+ b*x/2)**8 - 2*b*tanh(a/2 + b*x/2)**4 + b) - 2*log(tanh(a/2 + b*x/2)**2 + 1)/(b*tanh(a/2 + b*x/2)**8 - 2*b*t
anh(a/2 + b*x/2)**4 + b) + 4*tanh(a/2 + b*x/2)**6/(b*tanh(a/2 + b*x/2)**8 - 2*b*tanh(a/2 + b*x/2)**4 + b) + 4*
tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**8 - 2*b*tanh(a/2 + b*x/2)**4 + b), Eq(m, -3)), (b*x*tanh(a/2 + b*x/
2)**8/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2
+ b) - 4*b*x*tanh(a/2 + b*x/2)**6/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)*
*4 - 4*b*tanh(a/2 + b*x/2)**2 + b) + 6*b*x*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2
)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 4*b*x*tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x
/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) + b*x/(b*tanh(a/2
+ b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 2*log(tan
h(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**8/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b
*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) + 8*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**6/(b*tanh(a/2 + b*x
/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 12*log(tanh(a/2
+ b*x/2) + 1)*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)
**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) + 8*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**
8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*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)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)
**2 + b) + log(tanh(a/2 + b*x/2)**2 + 1)*tanh(a/2 + b*x/2)**8/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)*
*6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 4*log(tanh(a/2 + b*x/2)**2 + 1)*tanh(a/2 + b*x
/2)**6/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**
2 + b) + 6*log(tanh(a/2 + b*x/2)**2 + 1)*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)*
*6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 4*log(tanh(a/2 + b*x/2)**2 + 1)*tanh(a/2 + b*x
/2)**2/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**
2 + b) + log(tanh(a/2 + b*x/2)**2 + 1)/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x
/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 2*tanh(a/2 + b*x/2)**6/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2
)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) + 8*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)*
*8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 2*tanh(a/2 + b*x/2)
**2/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 +
b), Eq(m, -1)), (m**2*sinh(a + b*x)**4*cosh(a + b*x)*cosh(a + b*x)**m/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) + 8
*m*sinh(a + b*x)**4*cosh(a + b*x)*cosh(a + b*x)**m/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) - 4*m*sinh(a + b*x)**2*
cosh(a + b*x)**3*cosh(a + b*x)**m/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) + 15*sinh(a + b*x)**4*cosh(a + b*x)*cosh
(a + b*x)**m/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) - 20*sinh(a + b*x)**2*cosh(a + b*x)**3*cosh(a + b*x)**m/(b*m*
*3 + 9*b*m**2 + 23*b*m + 15*b) + 8*cosh(a + b*x)**5*cosh(a + b*x)**m/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b), True
))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh \left (b x + a\right )^{m} \sinh \left (b x + a\right )^{5}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^m*sinh(b*x+a)^5,x, algorithm="giac")

[Out]

integrate(cosh(b*x + a)^m*sinh(b*x + a)^5, x)