3.1077 \(\int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=200 \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right )}{3 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right )}{5 d}+\frac {2 \sin (c+d x) \left (a^2 (3 A+5 C)+10 a b B+4 A b^2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a (5 a B+4 A b) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]

[Out]

-2/5*(10*a*b*B+5*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))/d+2/3*(a^2*B+3*b^2*B+2*a*b*(A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
F(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/15*a*(4*A*b+5*B*a)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*A*(a+b*cos(d*x+c))^2*si
n(d*x+c)/d/cos(d*x+c)^(5/2)+2/5*(4*A*b^2+10*a*b*B+a^2*(3*A+5*C))*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.54, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3047, 3031, 3021, 2748, 2641, 2639} \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right )}{3 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right )}{5 d}+\frac {2 \sin (c+d x) \left (a^2 (3 A+5 C)+10 a b B+4 A b^2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a (5 a B+4 A b) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(10*a*b*B + 5*b^2*(A - C) + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(a^2*B + 3*b^2*B + 2*a*
b*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*(4*A*b + 5*a*B)*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)) +
(2*(4*A*b^2 + 10*a*b*B + a^2*(3*A + 5*C))*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*A*(a + b*Cos[c + d*x])^
2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))

Rule 2639

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

Rule 2641

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

Rule 2748

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 3021

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]

Rule 3031

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] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3047

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]

Rubi steps

\begin {align*} \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{2} (4 A b+5 a B)+\frac {1}{2} (3 a A+5 b B+5 a C) \cos (c+d x)-\frac {1}{2} b (A-5 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a (4 A b+5 a B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {4}{15} \int \frac {-\frac {3}{4} \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right )-\frac {5}{4} \left (a^2 B+3 b^2 B+2 a b (A+3 C)\right ) \cos (c+d x)+\frac {3}{4} b^2 (A-5 C) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a (4 A b+5 a B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {8}{15} \int \frac {-\frac {5}{8} \left (a^2 B+3 b^2 B+2 a b (A+3 C)\right )+\frac {3}{8} \left (10 a b B+5 b^2 (A-C)+a^2 (3 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a (4 A b+5 a B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {1}{3} \left (-a^2 B-3 b^2 B-2 a b (A+3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (10 a b B+5 b^2 (A-C)+a^2 (3 A+5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (10 a b B+5 b^2 (A-C)+a^2 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (a^2 B+3 b^2 B+2 a b (A+3 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a (4 A b+5 a B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 1.63, size = 202, normalized size = 1.01 \[ \frac {10 \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right )-6 \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right )+9 a^2 A \sin (2 (c+d x))+6 a^2 A \tan (c+d x)+10 a^2 B \sin (c+d x)+15 a^2 C \sin (2 (c+d x))+20 a A b \sin (c+d x)+30 a b B \sin (2 (c+d x))+15 A b^2 \sin (2 (c+d x))}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(10*a*b*B + 5*b^2*(A - C) + a^2*(3*A + 5*C))*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*(a^2*B + 3*
b^2*B + 2*a*b*(A + 3*C))*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 20*a*A*b*Sin[c + d*x] + 10*a^2*B*Sin[c
+ d*x] + 9*a^2*A*Sin[2*(c + d*x)] + 15*A*b^2*Sin[2*(c + d*x)] + 30*a*b*B*Sin[2*(c + d*x)] + 15*a^2*C*Sin[2*(c
+ d*x)] + 6*a^2*A*Tan[c + d*x])/(15*d*Cos[c + d*x]^(3/2))

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{2} \cos \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + A a^{2} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b^2*cos(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + c)^3 + A*a^2 + (C*a^2 + 2*B*a*b + A*b^2)*cos(d*x
+ c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c))/cos(d*x + c)^(7/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 8.76, size = 1000, normalized size = 5.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b^2*C*(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
))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*b^2*B*(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))+4*C*a*b*(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)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-2*b^2*C*(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*A*b+B*a)*(-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)/(-1/2+cos(1/2*d*x+1/2*c)^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*b^2+2*B*a*b+C*a^2)*(-(-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)^(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*(-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)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/
2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)-2/5*a^2*A/(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)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*EllipticE(cos(1/2*d*
x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/
2*d*x+1/2*c)^4*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(co
s(1/2*d*x+1/2*c),2^(1/2))-8*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))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.00, size = 310, normalized size = 1.55 \[ \frac {6\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,A\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,B\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,B\,a\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*A*a^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 30*A*b^2*cos(c + d*x)^2*sin(c + d*x)*hype
rgeom([-1/4, 1/2], 3/4, cos(c + d*x)^2) + 20*A*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c
+ d*x)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (2*B*b^2*ellipticF(c/2 + (d*x)/2, 2))/d + (
2*C*b^2*ellipticE(c/2 + (d*x)/2, 2))/d + (4*C*a*b*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*a^2*sin(c + d*x)*hyper
geom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*C*a^2*sin(c + d*x
)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (4*B*a*b*sin(c
+ d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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