3.1076 \(\int \frac{(a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=180 \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right )}{3 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A-C)-b^2 B\right )}{d}+\frac{2 a (3 a B+4 A b) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b^2 (A-C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]
[Out]
(-2*(a^2*B - b^2*B + 2*a*b*(A - C))*EllipticE[(c + d*x)/2, 2])/d + (2*(6*a*b*B + b^2*(3*A + C) + a^2*(A + 3*C)
)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*(4*A*b + 3*a*B)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]) - (2*b^2*(A -
C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) + (2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)
)
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Rubi [A] time = 0.500862, antiderivative size = 180, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used =
{3047, 3031, 3023, 2748, 2641, 2639} \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right )}{3 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A-C)-b^2 B\right )}{d}+\frac{2 a (3 a B+4 A b) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b^2 (A-C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]
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]^(5/2),x]
[Out]
(-2*(a^2*B - b^2*B + 2*a*b*(A - C))*EllipticE[(c + d*x)/2, 2])/d + (2*(6*a*b*B + b^2*(3*A + C) + a^2*(A + 3*C)
)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*(4*A*b + 3*a*B)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]) - (2*b^2*(A -
C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) + (2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)
)
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]
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 3023
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 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 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 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]
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{5}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{(a+b \cos (c+d x)) \left (\frac{1}{2} (4 A b+3 a B)+\frac{1}{2} (3 b B+a (A+3 C)) \cos (c+d x)-\frac{3}{2} b (A-C) \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a (4 A b+3 a B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{4}{3} \int \frac{\frac{1}{4} \left (-4 A b^2-6 a b B-a^2 (A+3 C)\right )+\frac{3}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) \cos (c+d x)+\frac{3}{4} b^2 (A-C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a (4 A b+3 a B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{8}{9} \int \frac{-\frac{3}{8} \left (6 a b B+b^2 (3 A+C)+a^2 (A+3 C)\right )+\frac{9}{8} \left (a^2 B-b^2 B+2 a b (A-C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a (4 A b+3 a B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\left (a^2 B-b^2 B+2 a b (A-C)\right ) \int \sqrt{\cos (c+d x)} \, dx-\frac{1}{3} \left (-6 a b B-b^2 (3 A+C)-a^2 (A+3 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 \left (6 a b B+b^2 (3 A+C)+a^2 (A+3 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a (4 A b+3 a B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 1.23892, size = 157, normalized size = 0.87 \[ \frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right )-6 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A-C)-b^2 B\right )+2 a^2 A \tan (c+d x)+6 a^2 B \sin (c+d x)+12 a A b \sin (c+d x)+b^2 C \sin (2 (c+d x))}{3 d \sqrt{\cos (c+d x)}} \]
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]^(5/2),x]
[Out]
(-6*(a^2*B - b^2*B + 2*a*b*(A - C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 2*(6*a*b*B + b^2*(3*A + C)
+ a^2*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 12*a*A*b*Sin[c + d*x] + 6*a^2*B*Sin[c + d*x] +
b^2*C*Sin[2*(c + d*x)] + 2*a^2*A*Tan[c + d*x])/(3*d*Sqrt[Cos[c + d*x]])
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Maple [B] time = 3.009, size = 1303, normalized size = 7.2 \begin{align*} \text{result too large to display} \end{align*}
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)^(5/2),x)
[Out]
2/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1
)/sin(1/2*d*x+1/2*c)^3*(-6*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))*a*b+6*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))*a*b-12*B*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+2*A*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^2+8*b^2*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-8*b^2*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*B*a^
2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+3*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-3*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-A*a^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*a^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+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))*b^2*sin(1/2*d*x+1/2*c)^2+2*A*(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))*a^2*sin(1/2*d*x+1/2*c)^
2+6*A*(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))*b^2*
sin(1/2*d*x+1/2*c)^2+6*B*(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))*a^2*sin(1/2*d*x+1/2*c)^2-6*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli
pticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2*sin(1/2*d*x+1/2*c)^2+6*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))*a^2*sin(1/2*d*x+1/2*c)^2-24*A*a*b*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^4+12*A*a*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-6*a*b*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))+12*A*(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))*a*b*sin(1/2*d*x+1/2*c)^2+12*B*(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))*a*b*sin(1/2*d*x+1/2*c)^2-12*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b*sin(1/2
*d*x+1/2*c)^2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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{5}{2}}}\,{d x} \end{align*}
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)^(5/2),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)^(5/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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{5}{2}}}, x\right ) \end{align*}
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)^(5/2),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)^(5/2), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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)**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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{5}{2}}}\,{d x} \end{align*}
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)^(5/2),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)^(5/2), x)