3.61 \(\int \frac{\cos ^4(c+d x) (A+C \cos ^2(c+d x))}{\sqrt{b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=147 \[ \frac{2 (11 A+9 C) \sin (c+d x) (b \cos (c+d x))^{5/2}}{77 b^3 d}+\frac{10 (11 A+9 C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{231 b d}+\frac{10 (11 A+9 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d \sqrt{b \cos (c+d x)}}+\frac{2 C \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 b^5 d} \]

[Out]

(10*(11*A + 9*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(231*d*Sqrt[b*Cos[c + d*x]]) + (10*(11*A + 9*C)
*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(231*b*d) + (2*(11*A + 9*C)*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(77*b^3*d
) + (2*C*(b*Cos[c + d*x])^(9/2)*Sin[c + d*x])/(11*b^5*d)

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Rubi [A]  time = 0.126337, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {16, 3014, 2635, 2642, 2641} \[ \frac{2 (11 A+9 C) \sin (c+d x) (b \cos (c+d x))^{5/2}}{77 b^3 d}+\frac{10 (11 A+9 C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{231 b d}+\frac{10 (11 A+9 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d \sqrt{b \cos (c+d x)}}+\frac{2 C \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 b^5 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*(A + C*Cos[c + d*x]^2))/Sqrt[b*Cos[c + d*x]],x]

[Out]

(10*(11*A + 9*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(231*d*Sqrt[b*Cos[c + d*x]]) + (10*(11*A + 9*C)
*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(231*b*d) + (2*(11*A + 9*C)*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(77*b^3*d
) + (2*C*(b*Cos[c + d*x])^(9/2)*Sin[c + d*x])/(11*b^5*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 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

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]

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt{b \cos (c+d x)}} \, dx &=\frac{\int (b \cos (c+d x))^{7/2} \left (A+C \cos ^2(c+d x)\right ) \, dx}{b^4}\\ &=\frac{2 C (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^5 d}+\frac{(11 A+9 C) \int (b \cos (c+d x))^{7/2} \, dx}{11 b^4}\\ &=\frac{2 (11 A+9 C) (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^3 d}+\frac{2 C (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^5 d}+\frac{(5 (11 A+9 C)) \int (b \cos (c+d x))^{3/2} \, dx}{77 b^2}\\ &=\frac{10 (11 A+9 C) \sqrt{b \cos (c+d x)} \sin (c+d x)}{231 b d}+\frac{2 (11 A+9 C) (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^3 d}+\frac{2 C (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^5 d}+\frac{1}{231} (5 (11 A+9 C)) \int \frac{1}{\sqrt{b \cos (c+d x)}} \, dx\\ &=\frac{10 (11 A+9 C) \sqrt{b \cos (c+d x)} \sin (c+d x)}{231 b d}+\frac{2 (11 A+9 C) (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^3 d}+\frac{2 C (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^5 d}+\frac{\left (5 (11 A+9 C) \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{231 \sqrt{b \cos (c+d x)}}\\ &=\frac{10 (11 A+9 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d \sqrt{b \cos (c+d x)}}+\frac{10 (11 A+9 C) \sqrt{b \cos (c+d x)} \sin (c+d x)}{231 b d}+\frac{2 (11 A+9 C) (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^3 d}+\frac{2 C (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^5 d}\\ \end{align*}

Mathematica [A]  time = 0.375043, size = 94, normalized size = 0.64 \[ \frac{\sin (2 (c+d x)) (12 (11 A+16 C) \cos (2 (c+d x))+572 A+21 C \cos (4 (c+d x))+531 C)+80 (11 A+9 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{1848 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*(A + C*Cos[c + d*x]^2))/Sqrt[b*Cos[c + d*x]],x]

[Out]

(80*(11*A + 9*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (572*A + 531*C + 12*(11*A + 16*C)*Cos[2*(c + d
*x)] + 21*C*Cos[4*(c + d*x)])*Sin[2*(c + d*x)])/(1848*d*Sqrt[b*Cos[c + d*x]])

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Maple [B]  time = 3.71, size = 349, normalized size = 2.4 \begin{align*} -{\frac{2}{231\,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 ( 1344\,C\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{12}-3360\,C\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}+ \left ( 528\,A+3792\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{8}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -792\,A-2328\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( 616\,A+924\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -176\,A-186\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +55\,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 ) +45\,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 EllipticF} \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)^4*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/2),x)

[Out]

-2/231*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(1344*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^12-3360*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+(528*A+3792*C)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-7
92*A-2328*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(616*A+924*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-1
76*A-186*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+55*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))+45*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)))/(-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 (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{4}}{\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)^4*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^4/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 (C \cos \left (d x + c\right )^{5} + A \cos \left (d x + c\right )^{3}\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)^4*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^5 + A*cos(d*x + c)^3)*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)**4*(A+C*cos(d*x+c)**2)/(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 (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{4}}{\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)^4*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^4/sqrt(b*cos(d*x + c)), x)