\(\int \cos (c+d x) (b \cos (c+d x))^{5/2} \, dx\) [92]

Optimal result

Integrand size = 19, antiderivative size = 97 \[ \int \cos (c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {10 b^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {10 b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d} \]

[Out]

2/7*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/d+10/21*b^3*(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)/d/(b*cos(d*x+c))^(1/2)+10/21*b^2*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {16, 2715, 2721, 2720} \[ \int \cos (c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {10 b^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {10 b^2 \sin (c+d x) \sqrt {b \cos (c+d x)}}{21 d}+\frac {2 \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d} \]

[In]

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

[Out]

(10*b^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (10*b^2*Sqrt[b*Cos[c + d*x
]]*Sin[c + d*x])/(21*d) + (2*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d)

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 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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 2721

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int (b \cos (c+d x))^{7/2} \, dx}{b} \\ & = \frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{7} (5 b) \int (b \cos (c+d x))^{3/2} \, dx \\ & = \frac {10 b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{21} \left (5 b^3\right ) \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx \\ & = \frac {10 b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {\left (5 b^3 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 \sqrt {b \cos (c+d x)}} \\ & = \frac {10 b^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {10 b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {b^2 \sqrt {b \cos (c+d x)} \left (20 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (23 \sin (c+d x)+3 \sin (3 (c+d x)))\right )}{42 d \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

(b^2*Sqrt[b*Cos[c + d*x]]*(20*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(23*Sin[c + d*x] + 3*Sin[3*(c + d
*x)])))/(42*d*Sqrt[Cos[c + d*x]])

Maple [A] (verified)

Time = 3.61 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.16

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \left (48 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+128 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(210\)

[In]

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

[Out]

-2/21*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*(48*cos(1/2*d*x+1/2*c)^9-120*cos(1/2*d*x+1
/2*c)^7+128*cos(1/2*d*x+1/2*c)^5-72*cos(1/2*d*x+1/2*c)^3+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)
^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+16*cos(1/2*d*x+1/2*c))/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d
*x+1/2*c)^2))^(1/2)/sin(1/2*d*x+1/2*c)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \cos (c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {-5 i \, \sqrt {2} b^{\frac {5}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} b^{\frac {5}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, b^{2} \cos \left (d x + c\right )^{2} + 5 \, b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{21 \, d} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \cos (c+d x) (b \cos (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \cos (c+d x) (b \cos (c+d x))^{5/2} \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \cos (c+d x) (b \cos (c+d x))^{5/2} \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \cos (c+d x) (b \cos (c+d x))^{5/2} \, dx=\int \cos \left (c+d\,x\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]

[In]

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

[Out]

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