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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3392, 32, 3393, 3377, 3389, 2211, 2235, 2236} \[ \int (c+d x)^{5/2} \cosh ^2(a+b x) \, dx=\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} e^{\frac {2 b c}{d}-2 a} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{256 b^{7/2}}-\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} e^{2 a-\frac {2 b c}{d}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{256 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \sinh (2 a+2 b x)}{64 b^3}-\frac {5 d (c+d x)^{3/2} \cosh ^2(a+b x)}{8 b^2}+\frac {(c+d x)^{5/2} \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {5 d (c+d x)^{3/2}}{16 b^2}+\frac {(c+d x)^{7/2}}{7 d} \]

[In]

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

[Out]

(5*d*(c + d*x)^(3/2))/(16*b^2) + (c + d*x)^(7/2)/(7*d) - (5*d*(c + d*x)^(3/2)*Cosh[a + b*x]^2)/(8*b^2) + (15*d
^(5/2)*E^(-2*a + (2*b*c)/d)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(256*b^(7/2)) - (15*d^(5/
2)*E^(2*a - (2*b*c)/d)*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(256*b^(7/2)) + ((c + d*x)^(5
/2)*Cosh[a + b*x]*Sinh[a + b*x])/(2*b) + (15*d^2*Sqrt[c + d*x]*Sinh[2*a + 2*b*x])/(64*b^3)

Rule 32

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

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3389

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 3392

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

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[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])
)

Rubi steps \begin{align*} \text {integral}& = -\frac {5 d (c+d x)^{3/2} \cosh ^2(a+b x)}{8 b^2}+\frac {(c+d x)^{5/2} \cosh (a+b x) \sinh (a+b x)}{2 b}+\frac {1}{2} \int (c+d x)^{5/2} \, dx+\frac {\left (15 d^2\right ) \int \sqrt {c+d x} \cosh ^2(a+b x) \, dx}{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 {(c+d x)^{5/2} \cosh (a+b x) \sinh (a+b x)}{2 b}+\frac {\left (15 d^2\right ) \int \left (\frac {1}{2} \sqrt {c+d x}+\frac {1}{2} \sqrt {c+d x} \cosh (2 a+2 b x)\right ) \, dx}{16 b^2} \\ & = \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 {(c+d x)^{5/2} \cosh (a+b x) \sinh (a+b x)}{2 b}+\frac {\left (15 d^2\right ) \int \sqrt {c+d x} \cosh (2 a+2 b x) \, dx}{32 b^2} \\ & = \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 {(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}-\frac {\left (15 d^3\right ) \int \frac {\sinh (2 a+2 b x)}{\sqrt {c+d x}} \, dx}{128 b^3} \\ & = \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 {(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}-\frac {\left (15 d^3\right ) \int \frac {e^{-i (2 i a+2 i b x)}}{\sqrt {c+d x}} \, dx}{256 b^3}+\frac {\left (15 d^3\right ) \int \frac {e^{i (2 i a+2 i b x)}}{\sqrt {c+d x}} \, dx}{256 b^3} \\ & = \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 {(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}+\frac {\left (15 d^2\right ) \text {Subst}\left (\int e^{i \left (2 i a-\frac {2 i b c}{d}\right )-\frac {2 b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{128 b^3}-\frac {\left (15 d^2\right ) \text {Subst}\left (\int e^{-i \left (2 i a-\frac {2 i b c}{d}\right )+\frac {2 b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{128 b^3} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.57 \[ \int (c+d x)^{5/2} \cosh ^2(a+b x) \, dx=\frac {\frac {64 (c+d x)^4}{d}-\frac {7 \sqrt {2} d^3 e^{2 a-\frac {2 b c}{d}} \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {7}{2},-\frac {2 b (c+d x)}{d}\right )}{b^4}-\frac {7 \sqrt {2} d^3 e^{-2 a+\frac {2 b c}{d}} \sqrt {\frac {b (c+d x)}{d}} \Gamma \left (\frac {7}{2},\frac {2 b (c+d x)}{d}\right )}{b^4}}{448 \sqrt {c+d x}} \]

[In]

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

[Out]

((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*S
qrt[c + d*x])

Maple [F]

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

[In]

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

[Out]

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 = 0.29 (sec) , antiderivative size = 1001, normalized size of antiderivative = 4.19 \[ \int (c+d x)^{5/2} \cosh ^2(a+b x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

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)*cos
h(-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)*sinh(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*(64*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*(8*b^3*c*d^2 - 5*b^2*d^3)*x)*cosh(b*x + a)^3 + 64*(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))*sinh(b*x + a))*sqrt(d*x + c))/(b^4*d*cosh(b*x + a
)^2 + 2*b^4*d*cosh(b*x + a)*sinh(b*x + a) + b^4*d*sinh(b*x + a)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.18 \[ \int (c+d x)^{5/2} \cosh ^2(a+b x) \, dx=\frac {512 \, {\left (d x + c\right )}^{\frac {7}{2}} - \frac {105 \, \sqrt {2} \sqrt {\pi } d^{3} \operatorname {erf}\left (\sqrt {2} \sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) e^{\left (2 \, a - \frac {2 \, b c}{d}\right )}}{b^{3} \sqrt {-\frac {b}{d}}} + \frac {105 \, \sqrt {2} \sqrt {\pi } d^{3} \operatorname {erf}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )}}{b^{3} \sqrt {\frac {b}{d}}} - \frac {28 \, {\left (16 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d e^{\left (\frac {2 \, b c}{d}\right )} + 20 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} e^{\left (\frac {2 \, b c}{d}\right )} + 15 \, \sqrt {d x + c} d^{3} e^{\left (\frac {2 \, b c}{d}\right )}\right )} e^{\left (-2 \, a - \frac {2 \, {\left (d x + c\right )} b}{d}\right )}}{b^{3}} + \frac {28 \, {\left (16 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d e^{\left (2 \, a\right )} - 20 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} e^{\left (2 \, a\right )} + 15 \, \sqrt {d x + c} d^{3} e^{\left (2 \, a\right )}\right )} e^{\left (\frac {2 \, {\left (d x + c\right )} b}{d} - \frac {2 \, b c}{d}\right )}}{b^{3}}}{3584 \, d} \]

[In]

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

[Out]

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(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)) - 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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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