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

Optimal. Leaf size=108 \[ \frac{2 (a C+b B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 (5 a B+3 b C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (a C+b B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{2 b C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.219386, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {3029, 2968, 3023, 2748, 2639, 2635, 2641} \[ \frac{2 (a C+b B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 (5 a B+3 b C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (a C+b B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{2 b C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

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

Mathematica [A]  time = 0.421684, size = 86, normalized size = 0.8 \[ \frac{2 \left (5 (a C+b B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+3 (5 a B+3 b C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\sin (c+d x) \sqrt{\cos (c+d x)} (5 a C+5 b B+3 b C \cos (c+d x))\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.622, size = 371, normalized size = 3.4 \begin{align*} -{\frac{2}{15\,d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( -24\,Cb\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( 20\,bB+20\,aC+24\,Cb \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -10\,bB-10\,aC-6\,Cb \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +5\,bB\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -15\,B\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) a+5\,aC\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -9\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) b \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*C*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6
+(20*B*b+20*C*a+24*C*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*B*b-10*C*a-6*C*b)*sin(1/2*d*x+1/2*c)^2*co
s(1/2*d*x+1/2*c)+5*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))-15*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+5*a*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))-9*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))*
b)/(-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 \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )}{\left (b \cos \left (d x + c\right ) + a\right )}}{\sqrt{\cos \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))*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{2} + B a +{\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sqrt{\cos \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))*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((C*b*cos(d*x + c)^2 + B*a + (C*a + B*b)*cos(d*x + c))*sqrt(cos(d*x + c)), 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+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

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 )\right )}{\left (b \cos \left (d x + c\right ) + a\right )}}{\sqrt{\cos \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))*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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