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]

2/9*b^7*sin(d*x+c)/d/(b*cos(d*x+c))^(9/2)+14/45*b^5*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+14/15*b^3*sin(d*x+c)/d/(
b*cos(d*x+c))^(1/2)-14/15*b^2*(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 = 0.09, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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]

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]

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*}

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Mathematica [A]  time = 0.24, 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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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \cos \left (d x + c\right )} b^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{8}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{8}\,{d x} \]

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)

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maple [B]  time = 0.23, size = 414, normalized size = 3.23 \[ -\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 (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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 )}}{144 b \left (-\frac {1}{2}+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {7 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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 )}}{180 b \left (-\frac {1}{2}+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {14 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 \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 )}}+\frac {7 \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}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{15 \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 )}}-\frac {7 \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}\, \left (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \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 )}}\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} \]

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)/(-1/2+cos(1/2*d*x+1/2*c)^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)/(-1/2+cos(1/2*d*x+1/2*c)^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-si
n(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{8}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^8} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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