3.1.88 \(\int \cos ^5(c+d x) (b \sec (c+d x))^{3/2} \, dx\) [88]

Optimal. Leaf size=98 \[ \frac {10 b \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {2 b^4 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac {10 b^2 \sin (c+d x)}{21 d \sqrt {b \sec (c+d x)}} \]

[Out]

2/7*b^4*sin(d*x+c)/d/(b*sec(d*x+c))^(5/2)+10/21*b^2*sin(d*x+c)/d/(b*sec(d*x+c))^(1/2)+10/21*b*(cos(1/2*d*x+1/2
*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(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 = 98, 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, 2720} \begin {gather*} \frac {2 b^4 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac {10 b^2 \sin (c+d x)}{21 d \sqrt {b \sec (c+d x)}}+\frac {10 b \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(b*Sec[c + d*x])^(3/2),x]

[Out]

(10*b*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(21*d) + (2*b^4*Sin[c + d*x])/(7*d*(b
*Sec[c + d*x])^(5/2)) + (10*b^2*Sin[c + d*x])/(21*d*Sqrt[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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 ^5(c+d x) (b \sec (c+d x))^{3/2} \, dx &=b^5 \int \frac {1}{(b \sec (c+d x))^{7/2}} \, dx\\ &=\frac {2 b^4 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac {1}{7} \left (5 b^3\right ) \int \frac {1}{(b \sec (c+d x))^{3/2}} \, dx\\ &=\frac {2 b^4 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac {10 b^2 \sin (c+d x)}{21 d \sqrt {b \sec (c+d x)}}+\frac {1}{21} (5 b) \int \sqrt {b \sec (c+d x)} \, dx\\ &=\frac {2 b^4 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac {10 b^2 \sin (c+d x)}{21 d \sqrt {b \sec (c+d x)}}+\frac {1}{21} \left (5 b \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {10 b \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {2 b^4 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac {10 b^2 \sin (c+d x)}{21 d \sqrt {b \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 64, normalized size = 0.65 \begin {gather*} \frac {b \sqrt {b \sec (c+d x)} \left (40 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+26 \sin (2 (c+d x))+3 \sin (4 (c+d x))\right )}{84 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(b*Sec[c + d*x])^(3/2),x]

[Out]

(b*Sqrt[b*Sec[c + d*x]]*(40*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 26*Sin[2*(c + d*x)] + 3*Sin[4*(c +
d*x)]))/(84*d)

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Maple [C] Result contains complex when optimal does not.
time = 30.08, size = 151, normalized size = 1.54

method	result	size
default	\(\frac {2 \left (\cos \left (d x +c \right )+1\right )^{2} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos \left (d x +c \right )-1\right ) \cos \left (d x +c \right ) \left (-5 i \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )+3 \left (\cos ^{4}\left (d x +c \right )\right )-3 \left (\cos ^{3}\left (d x +c \right )\right )+5 \left (\cos ^{2}\left (d x +c \right )\right )-5 \cos \left (d x +c \right )\right )}{21 d \sin \left (d x +c \right )^{3}}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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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)^5*(b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c))^(3/2)*cos(d*x + c)^5, x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.62, size = 99, normalized size = 1.01 \begin {gather*} \frac {-5 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, b \cos \left (d x + c\right )^{3} + 5 \, b \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{21 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/21*(-5*I*sqrt(2)*b^(3/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*b^(3/2)*wei
erstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(3*b*cos(d*x + c)^3 + 5*b*cos(d*x + c))*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)**5*(b*sec(d*x+c))**(3/2),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)^5*(b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(3/2)*cos(d*x + c)^5, 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 )}^5\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(b/cos(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^5*(b/cos(c + d*x))^(3/2), x)

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