\(\int \cos ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\) [141]

Optimal result

Integrand size = 23, antiderivative size = 73 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {8 (a+a \sin (c+d x))^{13/2}}{13 a^3 d}-\frac {8 (a+a \sin (c+d x))^{15/2}}{15 a^4 d}+\frac {2 (a+a \sin (c+d x))^{17/2}}{17 a^5 d} \]

[Out]

8/13*(a+a*sin(d*x+c))^(13/2)/a^3/d-8/15*(a+a*sin(d*x+c))^(15/2)/a^4/d+2/17*(a+a*sin(d*x+c))^(17/2)/a^5/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2746, 45} \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 (a \sin (c+d x)+a)^{17/2}}{17 a^5 d}-\frac {8 (a \sin (c+d x)+a)^{15/2}}{15 a^4 d}+\frac {8 (a \sin (c+d x)+a)^{13/2}}{13 a^3 d} \]

[In]

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

[Out]

(8*(a + a*Sin[c + d*x])^(13/2))/(13*a^3*d) - (8*(a + a*Sin[c + d*x])^(15/2))/(15*a^4*d) + (2*(a + a*Sin[c + d*
x])^(17/2))/(17*a^5*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a-x)^2 (a+x)^{11/2} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \left (4 a^2 (a+x)^{11/2}-4 a (a+x)^{13/2}+(a+x)^{15/2}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {8 (a+a \sin (c+d x))^{13/2}}{13 a^3 d}-\frac {8 (a+a \sin (c+d x))^{15/2}}{15 a^4 d}+\frac {2 (a+a \sin (c+d x))^{17/2}}{17 a^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 a^3 (1+\sin (c+d x))^6 \sqrt {a (1+\sin (c+d x))} \left (331-494 \sin (c+d x)+195 \sin ^2(c+d x)\right )}{3315 d} \]

[In]

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

[Out]

(2*a^3*(1 + Sin[c + d*x])^6*Sqrt[a*(1 + Sin[c + d*x])]*(331 - 494*Sin[c + d*x] + 195*Sin[c + d*x]^2))/(3315*d)

Maple [A] (verified)

Time = 58.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {17}{2}}}{17}-\frac {8 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {15}{2}}}{15}+\frac {8 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {13}{2}}}{13}}{d \,a^{5}}\)	\(56\)
default	\(\frac {\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {17}{2}}}{17}-\frac {8 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {15}{2}}}{15}+\frac {8 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {13}{2}}}{13}}{d \,a^{5}}\)	\(56\)

[In]

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

[Out]

2/d/a^5*(1/17*(a+a*sin(d*x+c))^(17/2)-4/15*a*(a+a*sin(d*x+c))^(15/2)+4/13*a^2*(a+a*sin(d*x+c))^(13/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (61) = 122\).

Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.75 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 \, {\left (195 \, a^{3} \cos \left (d x + c\right )^{8} - 1072 \, a^{3} \cos \left (d x + c\right )^{6} + 56 \, a^{3} \cos \left (d x + c\right )^{4} + 128 \, a^{3} \cos \left (d x + c\right )^{2} + 1024 \, a^{3} - 4 \, {\left (169 \, a^{3} \cos \left (d x + c\right )^{6} - 126 \, a^{3} \cos \left (d x + c\right )^{4} - 160 \, a^{3} \cos \left (d x + c\right )^{2} - 256 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3315 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

2/3315*(195*a^3*cos(d*x + c)^8 - 1072*a^3*cos(d*x + c)^6 + 56*a^3*cos(d*x + c)^4 + 128*a^3*cos(d*x + c)^2 + 10
24*a^3 - 4*(169*a^3*cos(d*x + c)^6 - 126*a^3*cos(d*x + c)^4 - 160*a^3*cos(d*x + c)^2 - 256*a^3)*sin(d*x + c))*
sqrt(a*sin(d*x + c) + a)/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*(a+a*sin(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 \, {\left (195 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {17}{2}} - 884 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {15}{2}} a + 1020 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a^{2}\right )}}{3315 \, a^{5} d} \]

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2/3315*(195*(a*sin(d*x + c) + a)^(17/2) - 884*(a*sin(d*x + c) + a)^(15/2)*a + 1020*(a*sin(d*x + c) + a)^(13/2)
*a^2)/(a^5*d)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.48 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {512 \, \sqrt {2} {\left (195 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{17} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 442 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 255 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sqrt {a}}{3315 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

512/3315*sqrt(2)*(195*a^3*cos(-1/4*pi + 1/2*d*x + 1/2*c)^17*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 442*a^3*cos(
-1/4*pi + 1/2*d*x + 1/2*c)^15*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) + 255*a^3*cos(-1/4*pi + 1/2*d*x + 1/2*c)^13*
sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int {\cos \left (c+d\,x\right )}^5\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \]

[In]

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

[Out]

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