3.2.2 \(\int \cos ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [102]

Optimal. Leaf size=127 \[ -\frac {256 a^4 \cos ^7(c+d x)}{3003 d (a+a \sin (c+d x))^{7/2}}-\frac {64 a^3 \cos ^7(c+d x)}{429 d (a+a \sin (c+d x))^{5/2}}-\frac {24 a^2 \cos ^7(c+d x)}{143 d (a+a \sin (c+d x))^{3/2}}-\frac {2 a \cos ^7(c+d x)}{13 d \sqrt {a+a \sin (c+d x)}} \]

[Out]

-256/3003*a^4*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(7/2)-64/429*a^3*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(5/2)-24/143*a^
2*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(3/2)-2/13*a*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.17, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2753, 2752} \begin {gather*} -\frac {256 a^4 \cos ^7(c+d x)}{3003 d (a \sin (c+d x)+a)^{7/2}}-\frac {64 a^3 \cos ^7(c+d x)}{429 d (a \sin (c+d x)+a)^{5/2}}-\frac {24 a^2 \cos ^7(c+d x)}{143 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 a \cos ^7(c+d x)}{13 d \sqrt {a \sin (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-256*a^4*Cos[c + d*x]^7)/(3003*d*(a + a*Sin[c + d*x])^(7/2)) - (64*a^3*Cos[c + d*x]^7)/(429*d*(a + a*Sin[c +
d*x])^(5/2)) - (24*a^2*Cos[c + d*x]^7)/(143*d*(a + a*Sin[c + d*x])^(3/2)) - (2*a*Cos[c + d*x]^7)/(13*d*Sqrt[a
+ a*Sin[c + d*x]])

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2753

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-b)*(
g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Dist[a*((2*m + p - 1)/(m + p)), Int
[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2,
0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rubi steps

\begin {align*} \int \cos ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=-\frac {2 a \cos ^7(c+d x)}{13 d \sqrt {a+a \sin (c+d x)}}+\frac {1}{13} (12 a) \int \frac {\cos ^6(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {24 a^2 \cos ^7(c+d x)}{143 d (a+a \sin (c+d x))^{3/2}}-\frac {2 a \cos ^7(c+d x)}{13 d \sqrt {a+a \sin (c+d x)}}+\frac {1}{143} \left (96 a^2\right ) \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {64 a^3 \cos ^7(c+d x)}{429 d (a+a \sin (c+d x))^{5/2}}-\frac {24 a^2 \cos ^7(c+d x)}{143 d (a+a \sin (c+d x))^{3/2}}-\frac {2 a \cos ^7(c+d x)}{13 d \sqrt {a+a \sin (c+d x)}}+\frac {1}{429} \left (128 a^3\right ) \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\\ &=-\frac {256 a^4 \cos ^7(c+d x)}{3003 d (a+a \sin (c+d x))^{7/2}}-\frac {64 a^3 \cos ^7(c+d x)}{429 d (a+a \sin (c+d x))^{5/2}}-\frac {24 a^2 \cos ^7(c+d x)}{143 d (a+a \sin (c+d x))^{3/2}}-\frac {2 a \cos ^7(c+d x)}{13 d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 2.37, size = 99, normalized size = 0.78 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^7 \sqrt {a (1+\sin (c+d x))} (-5230+1890 \cos (2 (c+d x))-6377 \sin (c+d x)+231 \sin (3 (c+d x)))}{6006 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^7*Sqrt[a*(1 + Sin[c + d*x])]*(-5230 + 1890*Cos[2*(c + d*x)] - 6377*Sin[
c + d*x] + 231*Sin[3*(c + d*x)]))/(6006*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

________________________________________________________________________________________

Maple [A]
time = 0.38, size = 75, normalized size = 0.59

method result size
default \(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{4} \left (231 \left (\sin ^{3}\left (d x +c \right )\right )+945 \left (\sin ^{2}\left (d x +c \right )\right )+1421 \sin \left (d x +c \right )+835\right )}{3003 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) \(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/3003*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^4*(231*sin(d*x+c)^3+945*sin(d*x+c)^2+1421*sin(d*x+c)+835)/cos(d*x+c)/(
a+a*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

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)^6*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*cos(d*x + c)^6, x)

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 172, normalized size = 1.35 \begin {gather*} -\frac {2 \, {\left (231 \, \cos \left (d x + c\right )^{7} - 21 \, \cos \left (d x + c\right )^{6} + 28 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{4} + 64 \, \cos \left (d x + c\right )^{3} - 128 \, \cos \left (d x + c\right )^{2} - {\left (231 \, \cos \left (d x + c\right )^{6} + 252 \, \cos \left (d x + c\right )^{5} + 280 \, \cos \left (d x + c\right )^{4} + 320 \, \cos \left (d x + c\right )^{3} + 384 \, \cos \left (d x + c\right )^{2} + 512 \, \cos \left (d x + c\right ) + 1024\right )} \sin \left (d x + c\right ) + 512 \, \cos \left (d x + c\right ) + 1024\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3003 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-2/3003*(231*cos(d*x + c)^7 - 21*cos(d*x + c)^6 + 28*cos(d*x + c)^5 - 40*cos(d*x + c)^4 + 64*cos(d*x + c)^3 -
128*cos(d*x + c)^2 - (231*cos(d*x + c)^6 + 252*cos(d*x + c)^5 + 280*cos(d*x + c)^4 + 320*cos(d*x + c)^3 + 384*
cos(d*x + c)^2 + 512*cos(d*x + c) + 1024)*sin(d*x + c) + 512*cos(d*x + c) + 1024)*sqrt(a*sin(d*x + c) + a)/(d*
cos(d*x + c) + d*sin(d*x + c) + d)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \cos ^{6}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*cos(c + d*x)**6, x)

________________________________________________________________________________________

Giac [A]
time = 6.71, size = 201, normalized size = 1.58 \begin {gather*} \frac {\sqrt {2} {\left (60060 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15015 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 9009 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 2574 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 2002 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right ) - 273 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {11}{4} \, \pi + \frac {11}{2} \, d x + \frac {11}{2} \, c\right ) - 231 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {13}{4} \, \pi + \frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )} \sqrt {a}}{96096 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/96096*sqrt(2)*(60060*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) - 15015*sgn(cos(-1/4
*pi + 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) - 9009*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi
+ 5/2*d*x + 5/2*c) + 2574*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-7/4*pi + 7/2*d*x + 7/2*c) + 2002*sgn(cos(-1
/4*pi + 1/2*d*x + 1/2*c))*sin(-9/4*pi + 9/2*d*x + 9/2*c) - 273*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-11/4*p
i + 11/2*d*x + 11/2*c) - 231*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-13/4*pi + 13/2*d*x + 13/2*c))*sqrt(a)/d

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^6\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*(a + a*sin(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^6*(a + a*sin(c + d*x))^(1/2), x)

________________________________________________________________________________________