3.1.79 \(\int \cos ^3(c+d x) (b \cos (c+d x))^{3/2} \, dx\) [79]

Optimal. Leaf size=95 \[ \frac {14 b \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {14 (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}+\frac {2 (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^2 d} \]

[Out]

14/45*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/9*(b*cos(d*x+c))^(7/2)*sin(d*x+c)/b^2/d+14/15*b*(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 = 0.04, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 2715, 2721, 2719} \begin {gather*} \frac {2 \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b^2 d}+\frac {14 \sin (c+d x) (b \cos (c+d x))^{3/2}}{45 d}+\frac {14 b 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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(b*Cos[c + d*x])^(3/2),x]

[Out]

(14*b*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*Sqrt[Cos[c + d*x]]) + (14*(b*Cos[c + d*x])^(3/2)*S
in[c + d*x])/(45*d) + (2*(b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*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 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]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (b \cos (c+d x))^{3/2} \, dx &=\frac {\int (b \cos (c+d x))^{9/2} \, dx}{b^3}\\ &=\frac {2 (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^2 d}+\frac {7 \int (b \cos (c+d x))^{5/2} \, dx}{9 b}\\ &=\frac {14 (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}+\frac {2 (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^2 d}+\frac {1}{15} (7 b) \int \sqrt {b \cos (c+d x)} \, dx\\ &=\frac {14 (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}+\frac {2 (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^2 d}+\frac {\left (7 b \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 \sqrt {\cos (c+d x)}}\\ &=\frac {14 b \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 {14 (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}+\frac {2 (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 75, normalized size = 0.79 \begin {gather*} \frac {(b \cos (c+d x))^{3/2} \left (168 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sqrt {\cos (c+d x)} (38 \sin (2 (c+d x))+5 \sin (4 (c+d x)))\right )}{180 d \cos ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(b*Cos[c + d*x])^(3/2),x]

[Out]

((b*Cos[c + d*x])^(3/2)*(168*EllipticE[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(38*Sin[2*(c + d*x)] + 5*Sin[4*(c
+ d*x)])))/(180*d*Cos[c + d*x]^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(222\) vs. \(2(107)=214\).
time = 0.08, size = 223, normalized size = 2.35

method	result	size
default	\(-\frac {2 \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{2} \left (160 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-480 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+616 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-432 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+160 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-21 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{45 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}\)	\(223\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(b*cos(d*x+c))^(3/2),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^2*(160*cos(1/2*d*x+1/2*c)^11-480*cos(1/2*d*x
+1/2*c)^9+616*cos(1/2*d*x+1/2*c)^7-432*cos(1/2*d*x+1/2*c)^5+160*cos(1/2*d*x+1/2*c)^3-21*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-24*cos(1/2*d*x+1/2*c))/(-b*(2*s
in(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(cos(d*x+c)^3*(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c))^(3/2)*cos(d*x + c)^3, x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 103, normalized size = 1.08 \begin {gather*} \frac {21 i \, \sqrt {2} b^{\frac {3}{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 ) - 21 i \, \sqrt {2} b^{\frac {3}{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 \, b \cos \left (d x + c\right )^{3} + 7 \, b \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(cos(d*x+c)^3*(b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/45*(21*I*sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) -
21*I*sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(5
*b*cos(d*x + c)^3 + 7*b*cos(d*x + c))*sqrt(b*cos(d*x + c))*sin(d*x + c))/d

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(b*cos(d*x+c))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4848 deep

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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(cos(d*x+c)^3*(b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c))^(3/2)*cos(d*x + c)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^3\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(b*cos(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^3*(b*cos(c + d*x))^(3/2), x)

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