3.1.16 \(\int (b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \, dx\) [16]

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

[Out]

2/45*b*(9*A+7*C)*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/9*C*(b*cos(d*x+c))^(7/2)*sin(d*x+c)/b/d+2/15*b^2*(9*A+7*C
)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(b*cos(d*x+c))^(1/2)/d
/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3093, 2715, 2721, 2719} \begin {gather*} \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} \end {gather*}

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 2715

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] && Integ
erQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 3093

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]

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*}

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Mathematica [A]
time = 0.44, size = 88, normalized size = 0.78 \begin {gather*} \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 \sqrt {\cos (c+d x)} (18 A+19 C+5 C \cos (2 (c+d x))) \sin (2 (c+d x))\right )}{180 d \cos ^{\frac {5}{2}}(c+d x)} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(323\) vs. \(2(125)=250\).
time = 0.40, size = 324, normalized size = 2.87

method	result	size
default	\(-\frac {2 \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \left (-160 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+320 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-72 A -296 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (72 A +136 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-18 A -24 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-27 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-21 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{45 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}\)	\(324\)

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,method=_RETURNVERBOSE)

[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*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+(-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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 128, normalized size = 1.13 \begin {gather*} \frac {3 i \, \sqrt {2} {\left (9 \, A + 7 \, C\right )} b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (9 \, A + 7 \, C\right )} b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (5 \, C b^{2} \cos \left (d x + c\right )^{3} + {\left (9 \, A + 7 \, C\right )} b^{2} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{45 \, d} \end {gather*}

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]

1/45*(3*I*sqrt(2)*(9*A + 7*C)*b^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d
*x + c))) - 3*I*sqrt(2)*(9*A + 7*C)*b^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I
*sin(d*x + c))) + 2*(5*C*b^2*cos(d*x + c)^3 + (9*A + 7*C)*b^2*cos(d*x + c))*sqrt(b*cos(d*x + c))*sin(d*x + c))
/d

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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