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

Optimal. Leaf size=113 \[ \frac{2 b^2 (9 A+7 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{15 d \sqrt{\cos (c+d x)}}+\frac{2 b (9 A+7 C) \sin (c+d x) (b \cos (c+d x))^{3/2}}{45 d}+\frac{2 C \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b d} \]

[Out]

(2*b^2*(9*A + 7*C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*Sqrt[Cos[c + d*x]]) + (2*b*(9*A + 7*C
)*(b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(45*d) + (2*C*(b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*b*d)

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Rubi [A]  time = 0.0804051, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3014, 2635, 2640, 2639} \[ \frac{2 b^2 (9 A+7 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{15 d \sqrt{\cos (c+d x)}}+\frac{2 b (9 A+7 C) \sin (c+d x) (b \cos (c+d x))^{3/2}}{45 d}+\frac{2 C \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*(9*A + 7*C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*Sqrt[Cos[c + d*x]]) + (2*b*(9*A + 7*C
)*(b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(45*d) + (2*C*(b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*b*d)

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

Mathematica [A]  time = 0.0631868, size = 88, normalized size = 0.78 \[ \frac{(b \cos (c+d x))^{5/2} \left (24 (9 A+7 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 \sin (2 (c+d x)) \sqrt{\cos (c+d x)} (18 A+5 C \cos (2 (c+d x))+19 C)\right )}{180 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*(24*(9*A + 7*C)*EllipticE[(c + d*x)/2, 2] + 2*Sqrt[Cos[c + d*x]]*(18*A + 19*C + 5*C*Co
s[2*(c + d*x)])*Sin[2*(c + d*x)]))/(180*d*Cos[c + d*x]^(5/2))

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Maple [B]  time = 3.396, size = 324, normalized size = 2.9 \begin{align*} -{\frac{2\,{b}^{3}}{45\,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 ( -160\,C\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}+320\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}\cos \left ( 1/2\,dx+c/2 \right ) + \left ( -72\,A-296\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( 72\,A+136\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -18\,A-24\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -27\,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 ) -21\,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((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x)

[Out]

-2/45*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*(-160*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^10+320*C*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-72*A-296*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(72
*A+136*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-18*A-24*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-27*A*(s
in(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))-21*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{\left (C \cos \left (d x + c\right )^{2} + A\right )} \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((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b^2*cos(d*x + c)^4 + A*b^2*cos(d*x + c)^2)*sqrt(b*cos(d*x + c)), 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((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\right )} \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((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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