3.1371 \(\int (a+b \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=206 \[ -\frac {2 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {4 a b (3 A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {4 a b (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^2}{d}-\frac {2 b^2 (5 A-C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.53, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4221, 3048, 3033, 3023, 2748, 2641, 2639} \[ -\frac {2 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {4 a b (3 A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {4 a b (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^2}{d}-\frac {2 b^2 (5 A-C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3048

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*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*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 4221

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

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Mathematica [A]  time = 1.11, size = 139, normalized size = 0.67 \[ \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (12 \left (b^2 (5 A+3 C)-5 a^2 (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {2 \sin (c+d x) \left (30 a^2 A+20 a b C \cos (c+d x)+3 b^2 C \cos (2 (c+d x))+3 b^2 C\right )}{\sqrt {\cos (c+d x)}}+40 a b (3 A+C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(12*(-5*a^2*(A - C) + b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 40*a
*b*(3*A + C)*EllipticF[(c + d*x)/2, 2] + (2*(30*a^2*A + 3*b^2*C + 20*a*b*C*Cos[c + d*x] + 3*b^2*C*Cos[2*(c + d
*x)])*Sin[c + d*x])/Sqrt[Cos[c + d*x]]))/(30*d)

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + 2 \, C a b \cos \left (d x + c\right )^{3} + 2 \, A a b \cos \left (d x + c\right ) + A a^{2} + {\left (C a^{2} + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sec \left (d x + c\right )^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.99, size = 694, normalized size = 3.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(-24*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+
8*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(5*a+3*b)*sin(1/2*d*x+1/2*c)^4*cos(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)*(15*A*a^2+10*C*a*b+3*C*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2
*d*x+1/2*c)+30*A*a*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+15*A*(-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)
)*a^2-15*A*(-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))*b^2+10*C*a*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-
15*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)^(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^2-9*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)^(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)/(-2*si
n(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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2),x)

[Out]

Timed out

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