3.1456 \(\int \frac{(a+b \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=194 \[ \frac{2 \sin (c+d x) (7 a B+7 A b+5 b C)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (7 a B+7 A b+5 b C)}{21 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (5 a A+3 a C+3 b B)}{5 d}+\frac{2 (a C+b B) \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b C \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(2*(5*a*A + 3*b*B + 3*a*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(7*A*b
+ 7*a*B + 5*b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*b*C*Sin[c + d*x]
)/(7*d*Sec[c + d*x]^(5/2)) + (2*(b*B + a*C)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*(7*A*b + 7*a*B + 5*b*C
)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.312874, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4221, 3033, 3023, 2748, 2639, 2635, 2641} \[ \frac{2 \sin (c+d x) (7 a B+7 A b+5 b C)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (7 a B+7 A b+5 b C)}{21 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (5 a A+3 a C+3 b B)}{5 d}+\frac{2 (a C+b B) \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b C \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*a*A + 3*b*B + 3*a*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(7*A*b
+ 7*a*B + 5*b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*b*C*Sin[c + d*x]
)/(7*d*Sec[c + d*x]^(5/2)) + (2*(b*B + a*C)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*(7*A*b + 7*a*B + 5*b*C
)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

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 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 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 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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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]

Rubi steps

\begin{align*} \int \frac{(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac{2 b C \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{7} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \left (\frac{7 a A}{2}+\frac{1}{2} (7 A b+7 a B+5 b C) \cos (c+d x)+\frac{7}{2} (b B+a C) \cos ^2(c+d x)\right ) \, dx\\ &=\frac{2 b C \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B+a C) \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{35} \left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \left (\frac{7}{4} (5 a A+3 b B+3 a C)+\frac{5}{4} (7 A b+7 a B+5 b C) \cos (c+d x)\right ) \, dx\\ &=\frac{2 b C \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B+a C) \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} \left ((5 a A+3 b B+3 a C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{7} \left ((7 A b+7 a B+5 b C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx\\ &=\frac{2 (5 a A+3 b B+3 a C) \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{2 b C \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B+a C) \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (7 A b+7 a B+5 b C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{1}{21} \left ((7 A b+7 a B+5 b C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 (5 a A+3 b B+3 a C) \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{2 (7 A b+7 a B+5 b C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{2 b C \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B+a C) \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (7 A b+7 a B+5 b C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.11189, size = 139, normalized size = 0.72 \[ \frac{\sqrt{\sec (c+d x)} \left (\sin (2 (c+d x)) (42 (a C+b B) \cos (c+d x)+70 a B+70 A b+15 b C \cos (2 (c+d x))+65 b C)+20 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (7 a B+7 A b+5 b C)+84 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (5 a A+3 a C+3 b B)\right )}{210 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(84*(5*a*A + 3*b*B + 3*a*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 20*(7*A*b + 7*a
*B + 5*b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (70*A*b + 70*a*B + 65*b*C + 42*(b*B + a*C)*Cos[c +
d*x] + 15*b*C*Cos[2*(c + d*x)])*Sin[2*(c + d*x)]))/(210*d)

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Maple [B]  time = 1.309, size = 515, normalized size = 2.7 \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))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x)

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*C*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8+(-168*B*b-168*C*a-360*C*b)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A*b+140*B*a+168*B*b+168*C*a+280*C*b)
*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-70*A*b-70*B*a-42*B*b-42*C*a-80*C*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*
x+1/2*c)-105*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+35*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))-63*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
+35*B*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))-63
*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+25*C*
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)/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., 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 )}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/sec(d*x+c)**(1/2),x)

[Out]

Integral((a + b*cos(c + d*x))*(A + B*cos(c + d*x) + C*cos(c + d*x)**2)/sqrt(sec(c + d*x)), x)

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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 )}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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