3.96 \(\int \cos ^4(c+d x) (b \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=72 \[ \frac {2 b^3 \sin (c+d x)}{3 d \sqrt {b \sec (c+d x)}}+\frac {2 b^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d} \]

[Out]

2/3*b^3*sin(d*x+c)/d/(b*sec(d*x+c))^(1/2)+2/3*b^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(si
n(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(b*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.05, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 3769, 3771, 2641} \[ \frac {2 b^3 \sin (c+d x)}{3 d \sqrt {b \sec (c+d x)}}+\frac {2 b^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(b*Sec[c + d*x])^(5/2),x]

[Out]

(2*b^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*d) + (2*b^3*Sin[c + d*x])/(3*d*Sq
rt[b*Sec[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 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]

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

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Mathematica [A]  time = 0.05, size = 54, normalized size = 0.75 \[ \frac {b^2 \sqrt {b \sec (c+d x)} \left (\sin (2 (c+d x))+2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(b*Sec[c + d*x])^(5/2),x]

[Out]

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

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sec \left (d x + c\right )} b^{2} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(d*x + c))*b^2*cos(d*x + c)^4*sec(d*x + c)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}} \cos \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(5/2)*cos(d*x + c)^4, x)

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maple [C]  time = 0.82, size = 131, normalized size = 1.82 \[ \frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (-1+\cos \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (-i \sin \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\cos ^{2}\left (d x +c \right )-\cos \left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(b*sec(d*x+c))^(5/2),x)

[Out]

2/3/d*(1+cos(d*x+c))^2*(b/cos(d*x+c))^(5/2)*(-1+cos(d*x+c))*cos(d*x+c)^2*(-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)+cos(d*x+c)^2-cos(d*x+c))/sin(d*
x+c)^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}} \cos \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c))^(5/2)*cos(d*x + c)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^4\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(b/cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^4*(b/cos(c + d*x))^(5/2), 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)**4*(b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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