3.102 \(\int (b \cos (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=98 \[ \frac{10 b^3 \sin (c+d x) \sqrt{b \cos (c+d x)}}{21 d}+\frac{10 b^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{b \cos (c+d x)}}+\frac{2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d} \]

[Out]

(10*b^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (10*b^3*Sqrt[b*Cos[c + d*x
]]*Sin[c + d*x])/(21*d) + (2*b*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d)

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Rubi [A]  time = 0.0505346, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2635, 2642, 2641} \[ \frac{10 b^3 \sin (c+d x) \sqrt{b \cos (c+d x)}}{21 d}+\frac{10 b^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{b \cos (c+d x)}}+\frac{2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(7/2),x]

[Out]

(10*b^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (10*b^3*Sqrt[b*Cos[c + d*x
]]*Sin[c + d*x])/(21*d) + (2*b*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d)

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

Mathematica [A]  time = 0.0209971, size = 76, normalized size = 0.78 \[ \frac{b^3 \sqrt{b \cos (c+d x)} \left (20 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+(23 \sin (c+d x)+3 \sin (3 (c+d x))) \sqrt{\cos (c+d x)}\right )}{42 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(7/2),x]

[Out]

(b^3*Sqrt[b*Cos[c + d*x]]*(20*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(23*Sin[c + d*x] + 3*Sin[3*(c + d
*x)])))/(42*d*Sqrt[Cos[c + d*x]])

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Maple [A]  time = 1.878, size = 210, normalized size = 2.1 \begin{align*} -{\frac{2\,{b}^{4}}{21\,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 ( 48\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}-120\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}+128\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-72\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+5\,\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 ) +16\,\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((b*cos(d*x+c))^(7/2),x)

[Out]

-2/21*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^4*(48*cos(1/2*d*x+1/2*c)^9-120*cos(1/2*d*x+1
/2*c)^7+128*cos(1/2*d*x+1/2*c)^5-72*cos(1/2*d*x+1/2*c)^3+5*(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))+16*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{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c))^(7/2), 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^{3} \cos \left (d x + c\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*cos(d*x + c))*b^3*cos(d*x + c)^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((b*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [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))^(7/2),x, algorithm="giac")

[Out]

Timed out