\(\int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\) [261]

Optimal result

Integrand size = 41, antiderivative size = 188 \[ \int (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=-\frac {2 b^2 (3 A+5 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^3 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 b^3 (3 A+5 C) \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}} \]

[Out]

2/5*A*b^5*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+2/3*b^4*B*sin(d*x+c)/d/(b*cos(d*x+c))^(3/2)+2/5*b^3*(3*A+5*C)*sin(
d*x+c)/d/(b*cos(d*x+c))^(1/2)+2/3*b^3*B*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+
1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)-2/5*b^2*(3*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2
*d*x+1/2*c)*EllipticE(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 = 0.39 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {16, 3100, 2827, 2716, 2721, 2720, 2719} \[ \int (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^3 (3 A+5 C) \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}}-\frac {2 b^2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 b^3 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}} \]

[In]

Int[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(-2*b^2*(3*A + 5*C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*b^3*B*Sqrt[C
os[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*A*b^5*Sin[c + d*x])/(5*d*(b*Cos[c + d*
x])^(5/2)) + (2*b^4*B*Sin[c + d*x])/(3*d*(b*Cos[c + d*x])^(3/2)) + (2*b^3*(3*A + 5*C)*Sin[c + d*x])/(5*d*Sqrt[
b*Cos[c + d*x]])

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 2716

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

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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = b^6 \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{7/2}} \, dx \\ & = \frac {2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {1}{5} \left (2 b^3\right ) \int \frac {\frac {5 b^2 B}{2}+\frac {1}{2} b^2 (3 A+5 C) \cos (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx \\ & = \frac {2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\left (b^5 B\right ) \int \frac {1}{(b \cos (c+d x))^{5/2}} \, dx+\frac {1}{5} \left (b^4 (3 A+5 C)\right ) \int \frac {1}{(b \cos (c+d x))^{3/2}} \, dx \\ & = \frac {2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 b^3 (3 A+5 C) \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}}+\frac {1}{3} \left (b^3 B\right ) \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx-\frac {1}{5} \left (b^2 (3 A+5 C)\right ) \int \sqrt {b \cos (c+d x)} \, dx \\ & = \frac {2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 b^3 (3 A+5 C) \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}}+\frac {\left (b^3 B \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 \sqrt {b \cos (c+d x)}}-\frac {\left (b^2 (3 A+5 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 (3 A+5 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^3 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 b^3 (3 A+5 C) \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.64 \[ \int (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=-\frac {2 b^4 \left (3 (3 A+5 C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-5 B \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-5 B \sin (c+d x)-\frac {9}{2} A \sin (2 (c+d x))-\frac {15}{2} C \sin (2 (c+d x))-3 A \tan (c+d x)\right )}{15 d (b \cos (c+d x))^{3/2}} \]

[In]

Integrate[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(-2*b^4*(3*(3*A + 5*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] - 5*B*Cos[c + d*x]^(3/2)*EllipticF[(c + d*
x)/2, 2] - 5*B*Sin[c + d*x] - (9*A*Sin[2*(c + d*x)])/2 - (15*C*Sin[2*(c + d*x)])/2 - 3*A*Tan[c + d*x]))/(15*d*
(b*Cos[c + d*x])^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(807\) vs. \(2(216)=432\).

Time = 6.37 (sec) , antiderivative size = 808, normalized size of antiderivative = 4.30

\[\text {Expression too large to display}\]

[In]

int((cos(d*x+c)*b)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)

[Out]

-2/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2/sin(1/2*d*x+1/2*c)^3/(8*sin(1/2*d*x+1/2*
c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-36*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))*sin(1/2*d*x+1/
2*c)^4-20*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)*sin(1/2*d*x+1/2*c)^4+120*cos(1/2*d*x+1/2*c)*C*sin(1/2*d*x+1/2*c)^6-60*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))*sin(1/2*d*x+1/2*c)^4-72*A*cos(1/2*d*x+1/2*c)*s
in(1/2*d*x+1/2*c)^4+36*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))*sin(1/2*d*x+1/2*c)^2-20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+20*B*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-120*C*cos(1/2
*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+60*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))*sin(1/2*d*x+1/2*c)^2+24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-9*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))+10*B*cos(1/2*d*x+1/2*c
)*sin(1/2*d*x+1/2*c)^2-5*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x
+1/2*c),2^(1/2))+30*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-15*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)))*(-2*sin(1/2*d*x+1/2*c)^4*b+b*sin(1/2*d*x+1/2*c)^2)^(
1/2)/((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 = 0.10 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.22 \[ \int (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {-5 i \, \sqrt {2} B b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, {\left (3 \, A + 5 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 5 \, B b^{2} \cos \left (d x + c\right ) + 3 \, A b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/15*(-5*I*sqrt(2)*B*b^(5/2)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sq
rt(2)*B*b^(5/2)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(3*A +
5*C)*b^(5/2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)))
+ 3*I*sqrt(2)*(3*A + 5*C)*b^(5/2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c
) - I*sin(d*x + c))) + 2*(3*(3*A + 5*C)*b^2*cos(d*x + c)^2 + 5*B*b^2*cos(d*x + c) + 3*A*b^2)*sqrt(b*cos(d*x +
c))*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

Maxima [F]

\[ \int (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{6} \,d x } \]

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^6, x)

Giac [F]

\[ \int (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{6} \,d x } \]

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^6, x)

Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^6} \,d x \]

[In]

int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^6,x)

[Out]

int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^6, x)