\(\int \frac {\cosh ^2(a+b x)}{(c+d x)^{5/2}} \, dx\) [53]

Optimal result

Integrand size = 18, antiderivative size = 174 \[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{5/2}} \, dx=-\frac {2 \cosh ^2(a+b x)}{3 d (c+d x)^{3/2}}+\frac {2 b^{3/2} e^{-2 a+\frac {2 b c}{d}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {2 b^{3/2} e^{2 a-\frac {2 b c}{d}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{3 d^2 \sqrt {c+d x}} \] Output:

-2/3*cosh(b*x+a)^2/d/(d*x+c)^(3/2)+2/3*b^(3/2)*exp(-2*a+2*b*c/d)*2^(1/2)*P 
i^(1/2)*erf(2^(1/2)*b^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^(5/2)+2/3*b^(3/2)*exp 
(2*a-2*b*c/d)*2^(1/2)*Pi^(1/2)*erfi(2^(1/2)*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)) 
/d^(5/2)-8/3*b*cosh(b*x+a)*sinh(b*x+a)/d^2/(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.90 \[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{5/2}} \, dx=-\frac {2 e^{-2 \left (a+\frac {b c}{d}\right )} \left (\sqrt {2} d e^{4 a} \left (-\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 b (c+d x)}{d}\right )+\sqrt {2} d e^{\frac {4 b c}{d}} \left (\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {2 b (c+d x)}{d}\right )+e^{2 \left (a+\frac {b c}{d}\right )} \left (d \cosh ^2(a+b x)+2 b (c+d x) \sinh (2 (a+b x))\right )\right )}{3 d^2 (c+d x)^{3/2}} \] Input:

Integrate[Cosh[a + b*x]^2/(c + d*x)^(5/2),x]
 

Output:

(-2*(Sqrt[2]*d*E^(4*a)*(-((b*(c + d*x))/d))^(3/2)*Gamma[1/2, (-2*b*(c + d* 
x))/d] + Sqrt[2]*d*E^((4*b*c)/d)*((b*(c + d*x))/d)^(3/2)*Gamma[1/2, (2*b*( 
c + d*x))/d] + E^(2*(a + (b*c)/d))*(d*Cosh[a + b*x]^2 + 2*b*(c + d*x)*Sinh 
[2*(a + b*x)])))/(3*d^2*E^(2*(a + (b*c)/d))*(c + d*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.27, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3795, 17, 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.

\(\displaystyle \int \frac {\cosh ^2(a+b x)}{(c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (i a+i b x+\frac {\pi }{2}\right )^2}{(c+d x)^{5/2}}dx\)

\(\Big \downarrow \) 3795

\(\displaystyle \frac {16 b^2 \int \frac {\cosh ^2(a+b x)}{\sqrt {c+d x}}dx}{3 d^2}-\frac {8 b^2 \int \frac {1}{\sqrt {c+d x}}dx}{3 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \cosh ^2(a+b x)}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 17

\(\displaystyle \frac {16 b^2 \int \frac {\cosh ^2(a+b x)}{\sqrt {c+d x}}dx}{3 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \cosh ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac {16 b^2 \sqrt {c+d x}}{3 d^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {16 b^2 \int \frac {\sin \left (i a+i b x+\frac {\pi }{2}\right )^2}{\sqrt {c+d x}}dx}{3 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \cosh ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac {16 b^2 \sqrt {c+d x}}{3 d^3}\)

\(\Big \downarrow \) 3793

\(\displaystyle \frac {16 b^2 \int \left (\frac {\cosh (2 a+2 b x)}{2 \sqrt {c+d x}}+\frac {1}{2 \sqrt {c+d x}}\right )dx}{3 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \cosh ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac {16 b^2 \sqrt {c+d x}}{3 d^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {16 b^2 \left (\frac {\sqrt {\frac {\pi }{2}} e^{\frac {2 b c}{d}-2 a} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 \sqrt {b} \sqrt {d}}+\frac {\sqrt {\frac {\pi }{2}} e^{2 a-\frac {2 b c}{d}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 \sqrt {b} \sqrt {d}}+\frac {\sqrt {c+d x}}{d}\right )}{3 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \cosh ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac {16 b^2 \sqrt {c+d x}}{3 d^3}\)

Input:

Int[Cosh[a + b*x]^2/(c + d*x)^(5/2),x]
 

Output:

(-16*b^2*Sqrt[c + d*x])/(3*d^3) - (2*Cosh[a + b*x]^2)/(3*d*(c + d*x)^(3/2) 
) + (16*b^2*(Sqrt[c + d*x]/d + (E^(-2*a + (2*b*c)/d)*Sqrt[Pi/2]*Erf[(Sqrt[ 
2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(4*Sqrt[b]*Sqrt[d]) + (E^(2*a - (2*b*c 
)/d)*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(4*Sqrt[b]* 
Sqrt[d])))/(3*d^2) - (8*b*Cosh[a + b*x]*Sinh[a + b*x])/(3*d^2*Sqrt[c + d*x 
])
 

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

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]))
 

rule 3795

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo 
l] :> Simp[(c + d*x)^(m + 1)*((b*Sin[e + f*x])^n/(d*(m + 1))), x] + (-Simp[ 
b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(d^2*(m + 1) 
*(m + 2))), x] + Simp[b^2*f^2*n*((n - 1)/(d^2*(m + 1)*(m + 2)))   Int[(c + 
d*x)^(m + 2)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[f^2*(n^2/(d^2*(m + 1)* 
(m + 2)))   Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x], x]) /; FreeQ[{b, 
c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]
 

Maple [F]

\[\int \frac {\cosh \left (b x +a \right )^{2}}{\left (d x +c \right )^{\frac {5}{2}}}d x\]

Input:

int(cosh(b*x+a)^2/(d*x+c)^(5/2),x)
 

Output:

int(cosh(b*x+a)^2/(d*x+c)^(5/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 861 vs. \(2 (134) = 268\).

Time = 0.15 (sec) , antiderivative size = 861, normalized size of antiderivative = 4.95 \[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(cosh(b*x+a)^2/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

1/6*(4*sqrt(2)*sqrt(pi)*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)^2*c 
osh(-2*(b*c - a*d)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)^2*si 
nh(-2*(b*c - a*d)/d) + ((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(-2*(b*c - a*d 
)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*sinh(-2*(b*c - a*d)/d))*sinh(b*x + 
a)^2 + 2*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*cosh(-2*(b*c - a*d 
)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*sinh(-2*(b*c - a*d)/d 
))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(2)*sqrt(d*x + c)*sqrt(b/d)) - 4*sqrt( 
2)*sqrt(pi)*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)^2*cosh(-2*(b*c 
- a*d)/d) + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)^2*sinh(-2*(b*c - 
 a*d)/d) + ((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(-2*(b*c - a*d)/d) + (b*d^ 
2*x^2 + 2*b*c*d*x + b*c^2)*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*((b 
*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*cosh(-2*(b*c - a*d)/d) + (b*d^ 
2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*sinh(-2*(b*c - a*d)/d))*sinh(b*x 
+ a))*sqrt(-b/d)*erf(sqrt(2)*sqrt(d*x + c)*sqrt(-b/d)) - ((4*b*d*x + 4*b*c 
 + d)*cosh(b*x + a)^4 + 4*(4*b*d*x + 4*b*c + d)*cosh(b*x + a)*sinh(b*x + a 
)^3 + (4*b*d*x + 4*b*c + d)*sinh(b*x + a)^4 - 4*b*d*x + 2*d*cosh(b*x + a)^ 
2 + 2*(3*(4*b*d*x + 4*b*c + d)*cosh(b*x + a)^2 + d)*sinh(b*x + a)^2 - 4*b* 
c + 4*((4*b*d*x + 4*b*c + d)*cosh(b*x + a)^3 + d*cosh(b*x + a))*sinh(b*x + 
 a) + d)*sqrt(d*x + c))/((d^4*x^2 + 2*c*d^3*x + c^2*d^2)*cosh(b*x + a)^2 + 
 2*(d^4*x^2 + 2*c*d^3*x + c^2*d^2)*cosh(b*x + a)*sinh(b*x + a) + (d^4*x...
 

Sympy [F]

\[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{5/2}} \, dx=\int \frac {\cosh ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(cosh(b*x+a)**2/(d*x+c)**(5/2),x)
 

Output:

Integral(cosh(a + b*x)**2/(c + d*x)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.68 \[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{5/2}} \, dx=-\frac {\frac {3 \, \sqrt {2} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {3}{2}, \frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}} + \frac {3 \, \sqrt {2} \left (-\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {3}{2}, -\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}} + \frac {2}{{\left (d x + c\right )}^{\frac {3}{2}}}}{6 \, d} \] Input:

integrate(cosh(b*x+a)^2/(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

-1/6*(3*sqrt(2)*((d*x + c)*b/d)^(3/2)*e^(2*(b*c - a*d)/d)*gamma(-3/2, 2*(d 
*x + c)*b/d)/(d*x + c)^(3/2) + 3*sqrt(2)*(-(d*x + c)*b/d)^(3/2)*e^(-2*(b*c 
 - a*d)/d)*gamma(-3/2, -2*(d*x + c)*b/d)/(d*x + c)^(3/2) + 2/(d*x + c)^(3/ 
2))/d
 

Giac [F]

\[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{5/2}} \, dx=\int { \frac {\cosh \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(cosh(b*x+a)^2/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

integrate(cosh(b*x + a)^2/(d*x + c)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{5/2}} \, dx=\int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(cosh(a + b*x)^2/(c + d*x)^(5/2),x)
 

Output:

int(cosh(a + b*x)^2/(c + d*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{5/2}} \, dx=\int \frac {\cosh \left (b x +a \right )^{2}}{\sqrt {d x +c}\, c^{2}+2 \sqrt {d x +c}\, c d x +\sqrt {d x +c}\, d^{2} x^{2}}d x \] Input:

int(cosh(b*x+a)^2/(d*x+c)^(5/2),x)
 

Output:

int(cosh(a + b*x)**2/(sqrt(c + d*x)*c**2 + 2*sqrt(c + d*x)*c*d*x + sqrt(c 
+ d*x)*d**2*x**2),x)