3.1.99 \(\int \cos ^7(c+d x) (b \sec (c+d x))^{5/2} \, dx\) [99]

Optimal. Leaf size=100 \[ \frac {14 b^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac {14 b^4 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}} \]

[Out]

2/9*b^6*sin(d*x+c)/d/(b*sec(d*x+c))^(7/2)+14/45*b^4*sin(d*x+c)/d/(b*sec(d*x+c))^(3/2)+14/15*b^3*(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))/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2
)

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Rubi [A]
time = 0.05, antiderivative size = 100, 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, 3854, 3856, 2719} \begin {gather*} \frac {2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac {14 b^4 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}}+\frac {14 b^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(b*Sec[c + d*x])^(5/2),x]

[Out]

(14*b^3*EllipticE[(c + d*x)/2, 2])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b^6*Sin[c + d*x])/(9*d*
(b*Sec[c + d*x])^(7/2)) + (14*b^4*Sin[c + d*x])/(45*d*(b*Sec[c + d*x])^(3/2))

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 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 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \cos ^7(c+d x) (b \sec (c+d x))^{5/2} \, dx &=b^7 \int \frac {1}{(b \sec (c+d x))^{9/2}} \, dx\\ &=\frac {2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac {1}{9} \left (7 b^5\right ) \int \frac {1}{(b \sec (c+d x))^{5/2}} \, dx\\ &=\frac {2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac {14 b^4 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}}+\frac {1}{15} \left (7 b^3\right ) \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx\\ &=\frac {2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac {14 b^4 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}}+\frac {\left (7 b^3\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\\ &=\frac {14 b^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac {14 b^4 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 74, normalized size = 0.74 \begin {gather*} \frac {b^2 \sqrt {b \sec (c+d x)} \left (84 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\cos ^2(c+d x) (33 \sin (c+d x)+5 \sin (3 (c+d x)))\right )}{90 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(b*Sec[c + d*x])^(5/2),x]

[Out]

(b^2*Sqrt[b*Sec[c + d*x]]*(84*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + Cos[c + d*x]^2*(33*Sin[c + d*x] +
5*Sin[3*(c + d*x)])))/(90*d)

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Maple [C] Result contains complex when optimal does not.
time = 30.57, size = 333, normalized size = 3.33

method	result	size
default	\(-\frac {2 \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \left (5 \left (\cos ^{6}\left (d x +c \right )\right )+21 i \cos \left (d x +c \right ) \EllipticE \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )-21 i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right )+21 i \EllipticE \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-21 i \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )+2 \left (\cos ^{4}\left (d x +c \right )\right )+14 \left (\cos ^{2}\left (d x +c \right )\right )-21 \cos \left (d x +c \right )\right )}{45 d \sin \left (d x +c \right )}\)	\(333\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/45/d*(b/cos(d*x+c))^(5/2)*cos(d*x+c)^2*(5*cos(d*x+c)^6+21*I*cos(d*x+c)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c
),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-21*I*cos(d*x+c)*EllipticF(I*(cos(d*
x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+21*I*EllipticE(I*(
cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-21*I*Ellipti
cF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*cos(
d*x+c)^4+14*cos(d*x+c)^2-21*cos(d*x+c))/sin(d*x+c)

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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)^7*(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c))^(5/2)*cos(d*x + c)^7, x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.86, size = 111, normalized size = 1.11 \begin {gather*} \frac {21 i \, \sqrt {2} b^{\frac {5}{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 {5}{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^{2} \cos \left (d x + c\right )^{4} + 7 \, b^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {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)^7*(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*(b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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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)^7*(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(5/2)*cos(d*x + c)^7, 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 )}^7\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7*(b/cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^7*(b/cos(c + d*x))^(5/2), x)

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