\(\int (b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [55]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 33, antiderivative size = 78 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 b^2 (5 A+3 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b C (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d} \]

[Out]

2/5*b*C*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/5*b^2*(5*A+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {16, 3093, 2721, 2719} \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 b^2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d} \]

[In]

Int[(b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]

[Out]

(2*b^2*(5*A + 3*C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*b*C*(b*Cos[c
+ d*x])^(3/2)*Sin[c + d*x])/(5*d)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 3093

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos
[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e +
f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = b^2 \int \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx \\ & = \frac {2 b C (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{5} \left (b^2 (5 A+3 C)\right ) \int \sqrt {b \cos (c+d x)} \, dx \\ & = \frac {2 b C (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {\left (b^2 (5 A+3 C) \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)}} \\ & = \frac {2 b^2 (5 A+3 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b C (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {b^2 \sqrt {b \cos (c+d x)} \left (2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+C \sqrt {\cos (c+d x)} \sin (2 (c+d x))\right )}{5 d \sqrt {\cos (c+d x)}} \]

[In]

Integrate[(b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]

[Out]

(b^2*Sqrt[b*Cos[c + d*x]]*(2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2] + C*Sqrt[Cos[c + d*x]]*Sin[2*(c + d*x)]))/(
5*d*Sqrt[Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(262\) vs. \(2(94)=188\).

Time = 18.82 (sec) , antiderivative size = 263, normalized size of antiderivative = 3.37

method result size
default \(\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \left (8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(263\)
parts \(\frac {2 A \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}-\frac {2 C \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(359\)

[In]

int((cos(d*x+c)*b)^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

2/5*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*(8*cos(1/2*d*x+1/2*c)*C*sin(1/2*d*x+1/2*c)^6
-8*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+5*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+3*C*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(
1/2*d*x+1/2*c)^2))^(1/2)/sin(1/2*d*x+1/2*c)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.36 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \, \sqrt {b \cos \left (d x + c\right )} C b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{5 \, d} \]

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

1/5*(2*sqrt(b*cos(d*x + c))*C*b^2*cos(d*x + c)*sin(d*x + c) + I*sqrt(2)*(5*A + 3*C)*b^(5/2)*weierstrassZeta(-4
, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - I*sqrt(2)*(5*A + 3*C)*b^(5/2)*weierstrassZet
a(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d

Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

Timed out

Maxima [F]

\[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^2, x)

Giac [F]

\[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^2, x)

Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \]

[In]

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^2,x)

[Out]

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^2, x)