3.279 \(\int \frac{\cos ^5(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=217 \[ \frac{2 (9 A+7 C) \sin (c+d x) (b \cos (c+d x))^{3/2}}{45 b^4 d}+\frac{2 (9 A+7 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{15 b^3 d \sqrt{\cos (c+d x)}}+\frac{2 B \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b^5 d}+\frac{10 B \sin (c+d x) \sqrt{b \cos (c+d x)}}{21 b^3 d}+\frac{10 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 b^2 d \sqrt{b \cos (c+d x)}}+\frac{2 C \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b^6 d} \]

[Out]

(2*(9*A + 7*C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*b^3*d*Sqrt[Cos[c + d*x]]) + (10*B*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*b^2*d*Sqrt[b*Cos[c + d*x]]) + (10*B*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]
)/(21*b^3*d) + (2*(9*A + 7*C)*(b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(45*b^4*d) + (2*B*(b*Cos[c + d*x])^(5/2)*Si
n[c + d*x])/(7*b^5*d) + (2*C*(b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*b^6*d)

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Rubi [A]  time = 0.225294, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.195, Rules used = {16, 3023, 2748, 2635, 2640, 2639, 2642, 2641} \[ \frac{2 (9 A+7 C) \sin (c+d x) (b \cos (c+d x))^{3/2}}{45 b^4 d}+\frac{2 (9 A+7 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{15 b^3 d \sqrt{\cos (c+d x)}}+\frac{2 B \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b^5 d}+\frac{10 B \sin (c+d x) \sqrt{b \cos (c+d x)}}{21 b^3 d}+\frac{10 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 b^2 d \sqrt{b \cos (c+d x)}}+\frac{2 C \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b^6 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(5/2),x]

[Out]

(2*(9*A + 7*C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*b^3*d*Sqrt[Cos[c + d*x]]) + (10*B*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*b^2*d*Sqrt[b*Cos[c + d*x]]) + (10*B*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]
)/(21*b^3*d) + (2*(9*A + 7*C)*(b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(45*b^4*d) + (2*B*(b*Cos[c + d*x])^(5/2)*Si
n[c + d*x])/(7*b^5*d) + (2*C*(b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*b^6*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 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 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 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]

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 ^5(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx &=\frac{\int (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{b^5}\\ &=\frac{2 C (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^6 d}+\frac{2 \int (b \cos (c+d x))^{5/2} \left (\frac{1}{2} b (9 A+7 C)+\frac{9}{2} b B \cos (c+d x)\right ) \, dx}{9 b^6}\\ &=\frac{2 C (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^6 d}+\frac{B \int (b \cos (c+d x))^{7/2} \, dx}{b^6}+\frac{(9 A+7 C) \int (b \cos (c+d x))^{5/2} \, dx}{9 b^5}\\ &=\frac{2 (9 A+7 C) (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 b^4 d}+\frac{2 B (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^5 d}+\frac{2 C (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^6 d}+\frac{(5 B) \int (b \cos (c+d x))^{3/2} \, dx}{7 b^4}+\frac{(9 A+7 C) \int \sqrt{b \cos (c+d x)} \, dx}{15 b^3}\\ &=\frac{10 B \sqrt{b \cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac{2 (9 A+7 C) (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 b^4 d}+\frac{2 B (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^5 d}+\frac{2 C (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^6 d}+\frac{(5 B) \int \frac{1}{\sqrt{b \cos (c+d x)}} \, dx}{21 b^2}+\frac{\left ((9 A+7 C) \sqrt{b \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{15 b^3 \sqrt{\cos (c+d x)}}\\ &=\frac{2 (9 A+7 C) \sqrt{b \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 b^3 d \sqrt{\cos (c+d x)}}+\frac{10 B \sqrt{b \cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac{2 (9 A+7 C) (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 b^4 d}+\frac{2 B (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^5 d}+\frac{2 C (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^6 d}+\frac{\left (5 B \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 b^2 \sqrt{b \cos (c+d x)}}\\ &=\frac{2 (9 A+7 C) \sqrt{b \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 b^3 d \sqrt{\cos (c+d x)}}+\frac{10 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 b^2 d \sqrt{b \cos (c+d x)}}+\frac{10 B \sqrt{b \cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac{2 (9 A+7 C) (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 b^4 d}+\frac{2 B (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^5 d}+\frac{2 C (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^6 d}\\ \end{align*}

Mathematica [A]  time = 0.687966, size = 130, normalized size = 0.6 \[ \frac{\sin (2 (c+d x)) (7 (36 A+43 C) \cos (c+d x)+5 (18 B \cos (2 (c+d x))+78 B+7 C \cos (3 (c+d x))))+168 (9 A+7 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+600 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{1260 b^2 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(5/2),x]

[Out]

(168*(9*A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 600*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2
, 2] + (7*(36*A + 43*C)*Cos[c + d*x] + 5*(78*B + 18*B*Cos[2*(c + d*x)] + 7*C*Cos[3*(c + d*x)]))*Sin[2*(c + d*x
)])/(1260*b^2*d*Sqrt[b*Cos[c + d*x]])

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Maple [A]  time = 3.701, size = 384, normalized size = 1.8 \begin{align*} -{\frac{2}{315\,{b}^{2}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 ( -1120\,C\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}+ \left ( 720\,B+2240\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{8}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -504\,A-1080\,B-2072\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( 504\,A+840\,B+952\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -126\,A-240\,B-168\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -189\,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 EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +75\,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 EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -147\,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 ) \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)^5*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x)

[Out]

-2/315*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/b^2*(-1120*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^10+(720*B+2240*C)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-504*A-1080*B-2072*C)*sin(1/2*d*x+1/2*c)^6*co
s(1/2*d*x+1/2*c)+(504*A+840*B+952*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-126*A-240*B-168*C)*sin(1/2*d*x+
1/2*c)^2*cos(1/2*d*x+1/2*c)-189*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))+75*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))-147*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)/(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} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{5}}{\left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^5/(b*cos(d*x + c))^(5/2), 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 )^{4} + B \cos \left (d x + c\right )^{3} + A \cos \left (d x + c\right )^{2}\right )} \sqrt{b \cos \left (d x + c\right )}}{b^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^4 + B*cos(d*x + c)^3 + A*cos(d*x + c)^2)*sqrt(b*cos(d*x + c))/b^3, 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)**5*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(5/2),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} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{5}}{\left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^5/(b*cos(d*x + c))^(5/2), x)