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

Optimal. Leaf size=144 \[ \frac{2 A \sin (c+d x) \sqrt{b \cos (c+d x)}}{3 b d}+\frac{2 A \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{b \cos (c+d x)}}+\frac{2 B \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b^2 d}+\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{5 b d \sqrt{\cos (c+d x)}} \]

[Out]

(6*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*b*d*Sqrt[Cos[c + d*x]]) + (2*A*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*A*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(3*b*d) + (2*B*(b*
Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*b^2*d)

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Rubi [A]  time = 0.113366, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {16, 2748, 2635, 2642, 2641, 2640, 2639} \[ \frac{2 A \sin (c+d x) \sqrt{b \cos (c+d x)}}{3 b d}+\frac{2 A \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{b \cos (c+d x)}}+\frac{2 B \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b^2 d}+\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{5 b d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*b*d*Sqrt[Cos[c + d*x]]) + (2*A*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*A*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(3*b*d) + (2*B*(b*
Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*b^2*d)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, 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 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, 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{\cos ^2(c+d x) (A+B \cos (c+d x))}{\sqrt{b \cos (c+d x)}} \, dx &=\frac{\int (b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx}{b^2}\\ &=\frac{A \int (b \cos (c+d x))^{3/2} \, dx}{b^2}+\frac{B \int (b \cos (c+d x))^{5/2} \, dx}{b^3}\\ &=\frac{2 A \sqrt{b \cos (c+d x)} \sin (c+d x)}{3 b d}+\frac{2 B (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b^2 d}+\frac{1}{3} A \int \frac{1}{\sqrt{b \cos (c+d x)}} \, dx+\frac{(3 B) \int \sqrt{b \cos (c+d x)} \, dx}{5 b}\\ &=\frac{2 A \sqrt{b \cos (c+d x)} \sin (c+d x)}{3 b d}+\frac{2 B (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b^2 d}+\frac{\left (A \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 \sqrt{b \cos (c+d x)}}+\frac{\left (3 B \sqrt{b \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 b \sqrt{\cos (c+d x)}}\\ &=\frac{6 B \sqrt{b \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b d \sqrt{\cos (c+d x)}}+\frac{2 A \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{b \cos (c+d x)}}+\frac{2 A \sqrt{b \cos (c+d x)} \sin (c+d x)}{3 b d}+\frac{2 B (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b^2 d}\\ \end{align*}

Mathematica [A]  time = 0.145038, size = 88, normalized size = 0.61 \[ \frac{2 \sqrt{\cos (c+d x)} \left (\sin (c+d x) \sqrt{\cos (c+d x)} (5 A+3 B \cos (c+d x))+5 A F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+9 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{15 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Cos[c + d*x]]*(9*B*EllipticE[(c + d*x)/2, 2] + 5*A*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(5*A
+ 3*B*Cos[c + d*x])*Sin[c + d*x]))/(15*d*Sqrt[b*Cos[c + d*x]])

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Maple [A]  time = 3.006, size = 270, normalized size = 1.9 \begin{align*} -{\frac{2}{15\,d}\sqrt{b \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\,B\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( 20\,A+24\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -10\,A-6\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +5\,A\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\,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 ) \right ){\frac{1}{\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) }}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6
+(20*A+24*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*A-6*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*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))-9*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*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)/(b*(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 (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\sqrt{b \cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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