3.120 \(\int \frac {\cos ^3(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=97 \[ \frac {2 b^2 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac {14 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}}+\frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 97, 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, 3769, 3771, 2639} \[ \frac {2 b^2 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac {14 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}}+\frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(14*EllipticE[(c + d*x)/2, 2])/(15*b*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b^2*Sin[c + d*x])/(9*d*(b
*Sec[c + d*x])^(7/2)) + (14*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 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 3769

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 3771

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

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Mathematica [A]  time = 0.08, size = 73, normalized size = 0.75 \[ \frac {84 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(33 \sin (c+d x)+5 \sin (3 (c+d x))) \cos ^{\frac {3}{2}}(c+d x)}{90 d \cos ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (d x + c\right )} \cos \left (d x + c\right )^{3}}{b^{2} \sec \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{3}}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.98, size = 333, normalized size = 3.43 \[ \frac {\frac {14 i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}}{15}-\frac {14 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{15}-\frac {2 \left (\cos ^{6}\left (d x +c \right )\right )}{9}+\frac {14 i \sin \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{15}-\frac {14 i \sin \left (d x +c \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{15}-\frac {4 \left (\cos ^{4}\left (d x +c \right )\right )}{45}-\frac {28 \left (\cos ^{2}\left (d x +c \right )\right )}{45}+\frac {14 \cos \left (d x +c \right )}{15}}{d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(b*sec(d*x+c))^(3/2),x)

[Out]

2/45/d*(21*I*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)
*(1/(1+cos(d*x+c)))^(1/2)-21*I*cos(d*x+c)*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c
)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-5*cos(d*x+c)^6+21*I*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x
+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-21*I*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/
sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-2*cos(d*x+c)^4-14*cos(d*x+c)^2+21*cos
(d*x+c))/sin(d*x+c)/cos(d*x+c)^2/(b/cos(d*x+c))^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{3}}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^3}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(c + d*x)^3/(b/cos(c + d*x))^(3/2), 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(cos(d*x+c)**3/(b*sec(d*x+c))**(3/2),x)

[Out]

Timed out

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