\(\int (a+b \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx\) [1377]

Optimal result

Integrand size = 35, antiderivative size = 269 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 a \left (15 b^2 (A-C)+a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 b \left (b^2 (3 A+C)+3 a^2 (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 b^3 (9 A-5 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 a \left (8 A b^2+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 A b (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \]

[Out]

4/5*A*b*(a+b*cos(d*x+c))^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*A*(a+b*cos(d*x+c))^3*sec(d*x+c)^(5/2)*sin(d*x+c)/
d-2/15*b^3*(9*A-5*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/5*a*(8*A*b^2+a^2*(3*A+5*C))*sin(d*x+c)*sec(d*x+c)^(1/2)/d
-2/5*a*(15*b^2*(A-C)+a^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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/3*b*(b^2*(3*A+C)+3*a^2*(A+3*C))*(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)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4306, 3127, 3126, 3110, 3102, 2827, 2720, 2719} \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 a \left (a^2 (3 A+5 C)+8 A b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 b \left (3 a^2 (A+3 C)+b^2 (3 A+C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {2 a \left (a^2 (3 A+5 C)+15 b^2 (A-C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 A b \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2}{5 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^3}{5 d}-\frac {2 b^3 (9 A-5 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}} \]

[In]

Int[(a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]

[Out]

(-2*a*(15*b^2*(A - C) + a^2*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d
) + (2*b*(b^2*(3*A + C) + 3*a^2*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3
*d) - (2*b^3*(9*A - 5*C)*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]) + (2*a*(8*A*b^2 + a^2*(3*A + 5*C))*Sqrt[Sec[c
+ d*x]]*Sin[c + d*x])/(5*d) + (4*A*b*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*A*(a
+ b*Cos[c + d*x])^3*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d)

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 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 3102

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

Rule 3110

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

Rule 3126

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

Rule 3127

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Si
n[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n
+ 2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2
*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 4306

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{5} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^2 \left (3 A b+\frac {1}{2} a (3 A+5 C) \cos (c+d x)-\frac {1}{2} b (3 A-5 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {4 A b (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{15} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x)) \left (\frac {3}{4} \left (8 A b^2+a^2 (3 A+5 C)\right )+\frac {3}{2} a b (A+5 C) \cos (c+d x)-\frac {3}{4} b^2 (9 A-5 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a \left (8 A b^2+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 A b (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac {1}{15} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3}{8} b \left (8 A b^2+5 a^2 (A+3 C)\right )+\frac {3}{8} a \left (15 b^2 (A-C)+a^2 (3 A+5 C)\right ) \cos (c+d x)+\frac {3}{8} b^3 (9 A-5 C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {2 b^3 (9 A-5 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 a \left (8 A b^2+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 A b (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac {1}{45} \left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {15}{16} b \left (b^2 (3 A+C)+3 a^2 (A+3 C)\right )+\frac {9}{16} a \left (15 b^2 (A-C)+a^2 (3 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {2 b^3 (9 A-5 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 a \left (8 A b^2+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 A b (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{3} \left (b \left (b^2 (3 A+C)+3 a^2 (A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (a \left (15 b^2 (A-C)+a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {2 a \left (15 b^2 (A-C)+a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 b \left (b^2 (3 A+C)+3 a^2 (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 b^3 (9 A-5 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 a \left (8 A b^2+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {4 A b (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.90 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.80 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {\sec ^{\frac {5}{2}}(c+d x) \left (-6 a \left (15 b^2 (A-C)+a^2 (3 A+5 C)\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 b \left (b^2 (3 A+C)+3 a^2 (A+3 C)\right ) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+6 a^3 A \sin (c+d x)+18 a^3 A \cos ^2(c+d x) \sin (c+d x)+90 a A b^2 \cos ^2(c+d x) \sin (c+d x)+30 a^3 C \cos ^2(c+d x) \sin (c+d x)+10 b^3 C \cos ^3(c+d x) \sin (c+d x)+15 a^2 A b \sin (2 (c+d x))\right )}{15 d} \]

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]

[Out]

(Sec[c + d*x]^(5/2)*(-6*a*(15*b^2*(A - C) + a^2*(3*A + 5*C))*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 10
*b*(b^2*(3*A + C) + 3*a^2*(A + 3*C))*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 6*a^3*A*Sin[c + d*x] + 18*
a^3*A*Cos[c + d*x]^2*Sin[c + d*x] + 90*a*A*b^2*Cos[c + d*x]^2*Sin[c + d*x] + 30*a^3*C*Cos[c + d*x]^2*Sin[c + d
*x] + 10*b^3*C*Cos[c + d*x]^3*Sin[c + d*x] + 15*a^2*A*b*Sin[2*(c + d*x)]))/(15*d)

Maple [F(-1)]

Timed out.

hanged

[In]

int((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x)

[Out]

int((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x)

Fricas [C] (verification not implemented)

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

Time = 0.13 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.17 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, {\left (A + 3 \, C\right )} a^{2} b + i \, {\left (3 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, {\left (A + 3 \, C\right )} a^{2} b - i \, {\left (3 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a^{3} + 15 i \, {\left (A - C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a^{3} - 15 i \, {\left (A - C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (5 \, C b^{3} \cos \left (d x + c\right )^{3} + 15 \, A a^{2} b \cos \left (d x + c\right ) + 3 \, A a^{3} + 3 \, {\left ({\left (3 \, A + 5 \, C\right )} a^{3} + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))**3*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2), x)

Giac [F]

\[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3 \,d x \]

[In]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^3,x)

[Out]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^3, x)