3.1163 \(\int (a+a \cos (c+d x)) (A+C \cos ^2(c+d x)) \sec ^{\frac{3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=135 \[ \frac{2 a (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 a (A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a A \sin (c+d x) \sqrt{\sec (c+d x)}}{d}+\frac{2 a C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.238345, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4221, 3032, 3023, 2748, 2641, 2639} \[ \frac{2 a (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 a (A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a A \sin (c+d x) \sqrt{\sec (c+d x)}}{d}+\frac{2 a C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 3032

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 + 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)*(a*C*(b*c - a*d) + A*b*(a*c - b*d)) - ((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, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[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 (a+a \cos (c+d x)) \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+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a A \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{\frac{a A}{2}-\frac{1}{2} a (A-C) \cos (c+d x)+\frac{1}{2} a C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 a A \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{3} \left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a (3 A+C)-\frac{3}{4} a (A-C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 a A \sqrt{\sec (c+d x)} \sin (c+d x)}{d}-\left (a (A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (a (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 a (A-C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{d}+\frac{2 a (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 a C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 a A \sqrt{\sec (c+d x)} \sin (c+d x)}{d}\\ \end{align*}

Mathematica [C]  time = 1.13892, size = 169, normalized size = 1.25 \[ \frac{a e^{-i d x} \sqrt{\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (2 i (A-C) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )+2 (3 A+C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+6 A \sin (c+d x)-6 i A \cos (c+d x)+C \sin (2 (c+d x))+6 i C \cos (c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*((-6*I)*A*Cos[c + d*x] + (6*I)*C*Cos[c + d*x] + 2*(3*A + C)*Sqrt
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (2*I)*(A - C)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hyperge
ometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + 6*A*Sin[c + d*x] + C*Sin[2*(c + d*x)]))/(3*d*E^(I*d*x))

________________________________________________________________________________________

Maple [B]  time = 1.201, size = 458, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a*(4*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)*sin(1/2*d*x+1/2*c)^4-2*(-2
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*A+C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+3*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)*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)+3*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))+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)*(-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))-3*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)*(-2*sin(1/2*d*x+1/2*c)^4+si
n(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
)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C a \cos \left (d x + c\right )^{3} + C a \cos \left (d x + c\right )^{2} + A a \cos \left (d x + c\right ) + A a\right )} \sec \left (d x + c\right )^{\frac{3}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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