3.101 \(\int (b \cos (c+d x))^{5/2} \sec ^8(c+d x) \, dx\)

Optimal. Leaf size=128 \[ \frac{2 b^7 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 b^5 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 b^3 \sin (c+d x)}{15 d \sqrt{b \cos (c+d x)}}-\frac{14 b^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{15 d \sqrt{\cos (c+d x)}} \]

[Out]

(-14*b^2*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*Sqrt[Cos[c + d*x]]) + (2*b^7*Sin[c + d*x])/(9*d
*(b*Cos[c + d*x])^(9/2)) + (14*b^5*Sin[c + d*x])/(45*d*(b*Cos[c + d*x])^(5/2)) + (14*b^3*Sin[c + d*x])/(15*d*S
qrt[b*Cos[c + d*x]])

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Rubi [A]  time = 0.0948979, antiderivative size = 128, 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, 2636, 2640, 2639} \[ \frac{2 b^7 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 b^5 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 b^3 \sin (c+d x)}{15 d \sqrt{b \cos (c+d x)}}-\frac{14 b^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{15 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^8,x]

[Out]

(-14*b^2*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*Sqrt[Cos[c + d*x]]) + (2*b^7*Sin[c + d*x])/(9*d
*(b*Cos[c + d*x])^(9/2)) + (14*b^5*Sin[c + d*x])/(45*d*(b*Cos[c + d*x])^(5/2)) + (14*b^3*Sin[c + d*x])/(15*d*S
qrt[b*Cos[c + d*x]])

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 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[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} \sec ^8(c+d x) \, dx &=b^8 \int \frac{1}{(b \cos (c+d x))^{11/2}} \, dx\\ &=\frac{2 b^7 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{1}{9} \left (7 b^6\right ) \int \frac{1}{(b \cos (c+d x))^{7/2}} \, dx\\ &=\frac{2 b^7 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 b^5 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{1}{15} \left (7 b^4\right ) \int \frac{1}{(b \cos (c+d x))^{3/2}} \, dx\\ &=\frac{2 b^7 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 b^5 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 b^3 \sin (c+d x)}{15 d \sqrt{b \cos (c+d x)}}-\frac{1}{15} \left (7 b^2\right ) \int \sqrt{b \cos (c+d x)} \, dx\\ &=\frac{2 b^7 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 b^5 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 b^3 \sin (c+d x)}{15 d \sqrt{b \cos (c+d x)}}-\frac{\left (7 b^2 \sqrt{b \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{15 \sqrt{\cos (c+d x)}}\\ &=-\frac{14 b^2 \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^7 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 b^5 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 b^3 \sin (c+d x)}{15 d \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.25978, size = 79, normalized size = 0.62 \[ \frac{\sec ^7(c+d x) (b \cos (c+d x))^{5/2} \left (150 \sin (c+d x)+91 \sin (3 (c+d x))+21 \sin (5 (c+d x))-336 \cos ^{\frac{9}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{360 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^8,x]

[Out]

((b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^7*(-336*Cos[c + d*x]^(9/2)*EllipticE[(c + d*x)/2, 2] + 150*Sin[c + d*x] +
91*Sin[3*(c + d*x)] + 21*Sin[5*(c + d*x)]))/(360*d)

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Maple [B]  time = 3.356, size = 414, normalized size = 3.2 \begin{align*} -2\,{\frac{\sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{b}^{3}}{\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }d} \left ( -{\frac{\cos \left ( 1/2\,dx+c/2 \right ) \sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}{144\,b \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{5}}}-{\frac{7\,\cos \left ( 1/2\,dx+c/2 \right ) \sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}{180\,b \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{3}}}-{\frac{14\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) }{15\,\sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}+{\frac{7\,\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 ) }{15\,\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}}-{\frac{7\,\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} \left ({\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) }{15\,\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(5/2)*sec(d*x+c)^8,x)

[Out]

-2*(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*(-1/144*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/2*d
*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/
2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*
x+1/2*c)/(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1
/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c)
,2^(1/2))-7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(
1/2*d*x+1/2*c)^2))^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(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}} \sec \left (d x + c\right )^{8}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*sec(d*x+c)^8,x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c))^(5/2)*sec(d*x + c)^8, 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 )^{2} \sec \left (d x + c\right )^{8}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*sec(d*x+c)^8,x, algorithm="fricas")

[Out]

integral(sqrt(b*cos(d*x + c))*b^2*cos(d*x + c)^2*sec(d*x + c)^8, 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)*sec(d*x+c)**8,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}} \sec \left (d x + c\right )^{8}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*sec(d*x+c)^8,x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c))^(5/2)*sec(d*x + c)^8, x)