\(\int (c+d x)^{5/2} \cosh ^2(a+b x) \, dx\) [48]

Optimal result

Integrand size = 18, antiderivative size = 239 \[ \int (c+d x)^{5/2} \cosh ^2(a+b x) \, dx=\frac {5 d (c+d x)^{3/2}}{16 b^2}+\frac {(c+d x)^{7/2}}{7 d}-\frac {5 d (c+d x)^{3/2} \cosh ^2(a+b x)}{8 b^2}+\frac {15 d^{5/2} e^{-2 a+\frac {2 b c}{d}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{256 b^{7/2}}-\frac {15 d^{5/2} e^{2 a-\frac {2 b c}{d}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{256 b^{7/2}}+\frac {(c+d x)^{5/2} \cosh (a+b x) \sinh (a+b x)}{2 b}+\frac {15 d^2 \sqrt {c+d x} \sinh (2 a+2 b x)}{64 b^3} \] Output:

5/16*d*(d*x+c)^(3/2)/b^2+1/7*(d*x+c)^(7/2)/d-5/8*d*(d*x+c)^(3/2)*cosh(b*x+ 
a)^2/b^2+15/512*d^(5/2)*exp(-2*a+2*b*c/d)*2^(1/2)*Pi^(1/2)*erf(2^(1/2)*b^( 
1/2)*(d*x+c)^(1/2)/d^(1/2))/b^(7/2)-15/512*d^(5/2)*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))/b^(7/2)+1/2*(d*x+c) 
^(5/2)*cosh(b*x+a)*sinh(b*x+a)/b+15/64*d^2*(d*x+c)^(1/2)*sinh(2*b*x+2*a)/b 
^3
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.57 \[ \int (c+d x)^{5/2} \cosh ^2(a+b x) \, dx=\frac {\frac {64 (c+d x)^4}{d}-\frac {7 \sqrt {2} d^3 e^{2 a-\frac {2 b c}{d}} \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {7}{2},-\frac {2 b (c+d x)}{d}\right )}{b^4}-\frac {7 \sqrt {2} d^3 e^{-2 a+\frac {2 b c}{d}} \sqrt {\frac {b (c+d x)}{d}} \Gamma \left (\frac {7}{2},\frac {2 b (c+d x)}{d}\right )}{b^4}}{448 \sqrt {c+d x}} \] Input:

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

Output:

((64*(c + d*x)^4)/d - (7*Sqrt[2]*d^3*E^(2*a - (2*b*c)/d)*Sqrt[-((b*(c + d* 
x))/d)]*Gamma[7/2, (-2*b*(c + d*x))/d])/b^4 - (7*Sqrt[2]*d^3*E^(-2*a + (2* 
b*c)/d)*Sqrt[(b*(c + d*x))/d]*Gamma[7/2, (2*b*(c + d*x))/d])/b^4)/(448*Sqr 
t[c + d*x])
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3792, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3792

\(\displaystyle \frac {15 d^2 \int \sqrt {c+d x} \cosh ^2(a+b x)dx}{16 b^2}+\frac {1}{2} \int (c+d x)^{5/2}dx-\frac {5 d (c+d x)^{3/2} \cosh ^2(a+b x)}{8 b^2}+\frac {(c+d x)^{5/2} \sinh (a+b x) \cosh (a+b x)}{2 b}\)

\(\Big \downarrow \) 17

\(\displaystyle \frac {15 d^2 \int \sqrt {c+d x} \cosh ^2(a+b x)dx}{16 b^2}-\frac {5 d (c+d x)^{3/2} \cosh ^2(a+b x)}{8 b^2}+\frac {(c+d x)^{5/2} \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {(c+d x)^{7/2}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {15 d^2 \int \sqrt {c+d x} \sin \left (i a+i b x+\frac {\pi }{2}\right )^2dx}{16 b^2}-\frac {5 d (c+d x)^{3/2} \cosh ^2(a+b x)}{8 b^2}+\frac {(c+d x)^{5/2} \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {(c+d x)^{7/2}}{7 d}\)

\(\Big \downarrow \) 3793

\(\displaystyle \frac {15 d^2 \int \left (\frac {1}{2} \sqrt {c+d x} \cosh (2 a+2 b x)+\frac {1}{2} \sqrt {c+d x}\right )dx}{16 b^2}-\frac {5 d (c+d x)^{3/2} \cosh ^2(a+b x)}{8 b^2}+\frac {(c+d x)^{5/2} \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {(c+d x)^{7/2}}{7 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {5 d (c+d x)^{3/2} \cosh ^2(a+b x)}{8 b^2}+\frac {15 d^2 \left (\frac {\sqrt {\frac {\pi }{2}} \sqrt {d} e^{\frac {2 b c}{d}-2 a} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {d} e^{2 a-\frac {2 b c}{d}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{3/2}}+\frac {\sqrt {c+d x} \sinh (2 a+2 b x)}{4 b}+\frac {(c+d x)^{3/2}}{3 d}\right )}{16 b^2}+\frac {(c+d x)^{5/2} \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {(c+d x)^{7/2}}{7 d}\)

Input:

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

Output:

(c + d*x)^(7/2)/(7*d) - (5*d*(c + d*x)^(3/2)*Cosh[a + b*x]^2)/(8*b^2) + (( 
c + d*x)^(5/2)*Cosh[a + b*x]*Sinh[a + b*x])/(2*b) + (15*d^2*((c + d*x)^(3/ 
2)/(3*d) + (Sqrt[d]*E^(-2*a + (2*b*c)/d)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[b]*S 
qrt[c + d*x])/Sqrt[d]])/(16*b^(3/2)) - (Sqrt[d]*E^(2*a - (2*b*c)/d)*Sqrt[P 
i/2]*Erfi[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(16*b^(3/2)) + (Sqrt[c 
 + d*x]*Sinh[2*a + 2*b*x])/(4*b)))/(16*b^2)
 

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 3792

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

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]

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1001 vs. \(2 (183) = 366\).

Time = 0.11 (sec) , antiderivative size = 1001, normalized size of antiderivative = 4.19 \[ \int (c+d x)^{5/2} \cosh ^2(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/3584*(105*sqrt(2)*sqrt(pi)*(d^4*cosh(b*x + a)^2*cosh(-2*(b*c - a*d)/d) - 
 d^4*cosh(b*x + a)^2*sinh(-2*(b*c - a*d)/d) + (d^4*cosh(-2*(b*c - a*d)/d) 
- d^4*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*(d^4*cosh(b*x + a)*cosh( 
-2*(b*c - a*d)/d) - d^4*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)) + 105*sqrt(2)*sqrt(pi)*( 
d^4*cosh(b*x + a)^2*cosh(-2*(b*c - a*d)/d) + d^4*cosh(b*x + a)^2*sinh(-2*( 
b*c - a*d)/d) + (d^4*cosh(-2*(b*c - a*d)/d) + d^4*sinh(-2*(b*c - a*d)/d))* 
sinh(b*x + a)^2 + 2*(d^4*cosh(b*x + a)*cosh(-2*(b*c - a*d)/d) + d^4*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*(112*b^3*d^3*x^2 + 112*b^3*c^2*d + 140*b^2*c*d^2 
 - 7*(16*b^3*d^3*x^2 + 16*b^3*c^2*d - 20*b^2*c*d^2 + 15*b*d^3 + 4*(8*b^3*c 
*d^2 - 5*b^2*d^3)*x)*cosh(b*x + a)^4 - 28*(16*b^3*d^3*x^2 + 16*b^3*c^2*d - 
 20*b^2*c*d^2 + 15*b*d^3 + 4*(8*b^3*c*d^2 - 5*b^2*d^3)*x)*cosh(b*x + a)*si 
nh(b*x + a)^3 - 7*(16*b^3*d^3*x^2 + 16*b^3*c^2*d - 20*b^2*c*d^2 + 15*b*d^3 
 + 4*(8*b^3*c*d^2 - 5*b^2*d^3)*x)*sinh(b*x + a)^4 + 105*b*d^3 - 128*(b^4*d 
^3*x^3 + 3*b^4*c*d^2*x^2 + 3*b^4*c^2*d*x + b^4*c^3)*cosh(b*x + a)^2 - 2*(6 
4*b^4*d^3*x^3 + 192*b^4*c*d^2*x^2 + 192*b^4*c^2*d*x + 64*b^4*c^3 + 21*(16* 
b^3*d^3*x^2 + 16*b^3*c^2*d - 20*b^2*c*d^2 + 15*b*d^3 + 4*(8*b^3*c*d^2 - 5* 
b^2*d^3)*x)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 28*(8*b^3*c*d^2 + 5*b^2*d^3 
)*x - 4*(7*(16*b^3*d^3*x^2 + 16*b^3*c^2*d - 20*b^2*c*d^2 + 15*b*d^3 + 4...
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.18 \[ \int (c+d x)^{5/2} \cosh ^2(a+b x) \, dx=\frac {512 \, {\left (d x + c\right )}^{\frac {7}{2}} - \frac {105 \, \sqrt {2} \sqrt {\pi } d^{3} \operatorname {erf}\left (\sqrt {2} \sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) e^{\left (2 \, a - \frac {2 \, b c}{d}\right )}}{b^{3} \sqrt {-\frac {b}{d}}} + \frac {105 \, \sqrt {2} \sqrt {\pi } d^{3} \operatorname {erf}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )}}{b^{3} \sqrt {\frac {b}{d}}} - \frac {28 \, {\left (16 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d e^{\left (\frac {2 \, b c}{d}\right )} + 20 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} e^{\left (\frac {2 \, b c}{d}\right )} + 15 \, \sqrt {d x + c} d^{3} e^{\left (\frac {2 \, b c}{d}\right )}\right )} e^{\left (-2 \, a - \frac {2 \, {\left (d x + c\right )} b}{d}\right )}}{b^{3}} + \frac {28 \, {\left (16 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d e^{\left (2 \, a\right )} - 20 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} e^{\left (2 \, a\right )} + 15 \, \sqrt {d x + c} d^{3} e^{\left (2 \, a\right )}\right )} e^{\left (\frac {2 \, {\left (d x + c\right )} b}{d} - \frac {2 \, b c}{d}\right )}}{b^{3}}}{3584 \, d} \] Input:

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

Output:

1/3584*(512*(d*x + c)^(7/2) - 105*sqrt(2)*sqrt(pi)*d^3*erf(sqrt(2)*sqrt(d* 
x + c)*sqrt(-b/d))*e^(2*a - 2*b*c/d)/(b^3*sqrt(-b/d)) + 105*sqrt(2)*sqrt(p 
i)*d^3*erf(sqrt(2)*sqrt(d*x + c)*sqrt(b/d))*e^(-2*a + 2*b*c/d)/(b^3*sqrt(b 
/d)) - 28*(16*(d*x + c)^(5/2)*b^2*d*e^(2*b*c/d) + 20*(d*x + c)^(3/2)*b*d^2 
*e^(2*b*c/d) + 15*sqrt(d*x + c)*d^3*e^(2*b*c/d))*e^(-2*a - 2*(d*x + c)*b/d 
)/b^3 + 28*(16*(d*x + c)^(5/2)*b^2*d*e^(2*a) - 20*(d*x + c)^(3/2)*b*d^2*e^ 
(2*a) + 15*sqrt(d*x + c)*d^3*e^(2*a))*e^(2*(d*x + c)*b/d - 2*b*c/d)/b^3)/d
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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