\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx\) [1290]

Optimal result

Integrand size = 43, antiderivative size = 250 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {3 (7 A-5 B+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(5 A-5 B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {3 (7 A-5 B+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {(5 A-5 B+3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac {(7 A-5 B+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A-B+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \]

[Out]

-1/3*(5*A-5*B+3*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d+1/5*(7*A-5*B+5*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d-(A-B+C)*s
ec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))+3/5*(7*A-5*B+5*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d-3/5*(7*A-5*B+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)/a/d-1/3*(5*A-5*B+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)/a/d

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4306, 3120, 2827, 2716, 2719, 2720} \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {(7 A-5 B+5 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 a d}-\frac {(5 A-5 B+3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a d}+\frac {3 (7 A-5 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}-\frac {(5 A-5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}-\frac {3 (7 A-5 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d} \]

[In]

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

[Out]

(-3*(7*A - 5*B + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a*d) - ((5*A - 5*B +
3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) + (3*(7*A - 5*B + 5*C)*Sqrt[Sec
[c + d*x]]*Sin[c + d*x])/(5*a*d) - ((5*A - 5*B + 3*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a*d) + ((7*A - 5*B +
5*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*a*d) - ((A - B + C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Cos[c
+ d*x]))

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 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 3120

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[(a*A - b*B + a*C)*Cos[e + f*x]*(a
+ b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Dist[1/(b*(b*c - a*d)*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c*(2*m + 1) - a*d*(m - n -
1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^
2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-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)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))} \, dx \\ & = -\frac {(A-B+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (7 A-5 B+5 C)-\frac {1}{2} a (5 A-5 B+3 C) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{a^2} \\ & = -\frac {(A-B+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {\left ((5 A-5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{2 a}+\frac {\left ((7 A-5 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{2 a} \\ & = -\frac {(5 A-5 B+3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac {(7 A-5 B+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A-B+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {\left ((5 A-5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a}+\frac {\left (3 (7 A-5 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{10 a} \\ & = -\frac {(5 A-5 B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {3 (7 A-5 B+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {(5 A-5 B+3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac {(7 A-5 B+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A-B+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {\left (3 (7 A-5 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{10 a} \\ & = -\frac {3 (7 A-5 B+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(5 A-5 B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {3 (7 A-5 B+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {(5 A-5 B+3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac {(7 A-5 B+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A-B+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.94 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.80 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (36 (7 A-5 B+5 C) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 (5 A-5 B+3 C) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-(100 A-40 B+60 C+(173 A-95 B+135 C) \cos (c+d x)+(76 A-40 B+60 C) \cos (2 (c+d x))+63 A \cos (3 (c+d x))-45 B \cos (3 (c+d x))+45 C \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{30 a d (1+\cos (c+d x))} \]

[In]

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

[Out]

-1/30*(Cos[(c + d*x)/2]^2*Sec[c + d*x]^(5/2)*(36*(7*A - 5*B + 5*C)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2
] + 20*(5*A - 5*B + 3*C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] - (100*A - 40*B + 60*C + (173*A - 95*B +
135*C)*Cos[c + d*x] + (76*A - 40*B + 60*C)*Cos[2*(c + d*x)] + 63*A*Cos[3*(c + d*x)] - 45*B*Cos[3*(c + d*x)] +
45*C*Cos[3*(c + d*x)])*Tan[(c + d*x)/2]))/(a*d*(1 + Cos[c + d*x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(784\) vs. \(2(276)=552\).

Time = 55.01 (sec) , antiderivative size = 785, normalized size of antiderivative = 3.14

method	result	size
default	\(\text {Expression too large to display}\)	\(785\)

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*((-A+B-C)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*
c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*
x+1/2*c)^2)^(1/2)+2/5*A/sin(1/2*d*x+1/2*c)^2/(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)*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+
12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2
*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(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+sin(1/2*d*x+1/2*c)^2)^(1/2)+(-2*A+2*
B)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+
1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+(2*A-2*B+2*C)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)
*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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)/(-
1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.48 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {5 \, {\left (\sqrt {2} {\left (-5 i \, A + 5 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-5 i \, A + 5 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (5 i \, A - 5 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (5 i \, A - 5 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (\sqrt {2} {\left (7 i \, A - 5 i \, B + 5 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (7 i \, A - 5 i \, B + 5 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (\sqrt {2} {\left (-7 i \, A + 5 i \, B - 5 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-7 i \, A + 5 i \, B - 5 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (9 \, {\left (7 \, A - 5 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (19 \, A - 10 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 6 \, A\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

-1/30*(5*(sqrt(2)*(-5*I*A + 5*I*B - 3*I*C)*cos(d*x + c)^3 + sqrt(2)*(-5*I*A + 5*I*B - 3*I*C)*cos(d*x + c)^2)*w
eierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(5*I*A - 5*I*B + 3*I*C)*cos(d*x + c)^3 +
sqrt(2)*(5*I*A - 5*I*B + 3*I*C)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9
*(sqrt(2)*(7*I*A - 5*I*B + 5*I*C)*cos(d*x + c)^3 + sqrt(2)*(7*I*A - 5*I*B + 5*I*C)*cos(d*x + c)^2)*weierstrass
Zeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 9*(sqrt(2)*(-7*I*A + 5*I*B - 5*I*C)*c
os(d*x + c)^3 + sqrt(2)*(-7*I*A + 5*I*B - 5*I*C)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4
, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(9*(7*A - 5*B + 5*C)*cos(d*x + c)^3 + 2*(19*A - 10*B + 15*C)*cos(d*x
+ c)^2 - 2*(2*A - 5*B)*cos(d*x + c) + 6*A)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x
+ c)^2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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