3.90 \(\int \cos ^3(c+d x) (b \cos (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=125 \[ \frac{2 \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 b^2 d}+\frac{30 b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{77 d}+\frac{30 b^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{77 d \sqrt{b \cos (c+d x)}}+\frac{18 \sin (c+d x) (b \cos (c+d x))^{5/2}}{77 d} \]

[Out]

(30*b^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(77*d*Sqrt[b*Cos[c + d*x]]) + (30*b^2*Sqrt[b*Cos[c + d*x
]]*Sin[c + d*x])/(77*d) + (18*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(77*d) + (2*(b*Cos[c + d*x])^(9/2)*Sin[c +
d*x])/(11*b^2*d)

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Rubi [A]  time = 0.0804959, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {16, 2635, 2642, 2641} \[ \frac{2 \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 b^2 d}+\frac{30 b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{77 d}+\frac{30 b^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{77 d \sqrt{b \cos (c+d x)}}+\frac{18 \sin (c+d x) (b \cos (c+d x))^{5/2}}{77 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(b*Cos[c + d*x])^(5/2),x]

[Out]

(30*b^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(77*d*Sqrt[b*Cos[c + d*x]]) + (30*b^2*Sqrt[b*Cos[c + d*x
]]*Sin[c + d*x])/(77*d) + (18*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(77*d) + (2*(b*Cos[c + d*x])^(9/2)*Sin[c +
d*x])/(11*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 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 \cos ^3(c+d x) (b \cos (c+d x))^{5/2} \, dx &=\frac{\int (b \cos (c+d x))^{11/2} \, dx}{b^3}\\ &=\frac{2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^2 d}+\frac{9 \int (b \cos (c+d x))^{7/2} \, dx}{11 b}\\ &=\frac{18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}+\frac{2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^2 d}+\frac{1}{77} (45 b) \int (b \cos (c+d x))^{3/2} \, dx\\ &=\frac{30 b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{77 d}+\frac{18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}+\frac{2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^2 d}+\frac{1}{77} \left (15 b^3\right ) \int \frac{1}{\sqrt{b \cos (c+d x)}} \, dx\\ &=\frac{30 b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{77 d}+\frac{18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}+\frac{2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^2 d}+\frac{\left (15 b^3 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{77 \sqrt{b \cos (c+d x)}}\\ &=\frac{30 b^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{77 d \sqrt{b \cos (c+d x)}}+\frac{30 b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{77 d}+\frac{18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}+\frac{2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^2 d}\\ \end{align*}

Mathematica [A]  time = 0.195242, size = 83, normalized size = 0.66 \[ \frac{(b \cos (c+d x))^{5/2} \left (240 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+(290 \sin (c+d x)+57 \sin (3 (c+d x))+7 \sin (5 (c+d x))) \sqrt{\cos (c+d x)}\right )}{616 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(b*Cos[c + d*x])^(5/2),x]

[Out]

((b*Cos[c + d*x])^(5/2)*(240*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(290*Sin[c + d*x] + 57*Sin[3*(c +
d*x)] + 7*Sin[5*(c + d*x)])))/(616*d*Cos[c + d*x]^(5/2))

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Maple [A]  time = 2.044, size = 236, normalized size = 1.9 \begin{align*} -{\frac{2\,{b}^{3}}{77\,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 ( 448\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{13}-1568\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{11}+2384\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}-2040\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}+1084\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-370\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+15\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +62\,\cos \left ( 1/2\,dx+c/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)^3*(b*cos(d*x+c))^(5/2),x)

[Out]

-2/77*(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*(448*cos(1/2*d*x+1/2*c)^13-1568*cos(1/2*d*
x+1/2*c)^11+2384*cos(1/2*d*x+1/2*c)^9-2040*cos(1/2*d*x+1/2*c)^7+1084*cos(1/2*d*x+1/2*c)^5-370*cos(1/2*d*x+1/2*
c)^3+15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6
2*cos(1/2*d*x+1/2*c))/(-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 (b \cos \left (d x + c\right )\right )^{\frac{5}{2}} \cos \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c))^(5/2)*cos(d*x + c)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \cos \left (d x + c\right )} b^{2} \cos \left (d x + c\right )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c))^(5/2)*cos(d*x + c)^3, x)