3.2.44 \(\int \cos ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\) [144]

Optimal. Leaf size=159 \[ -\frac {4096 a^5 \cos ^3(c+d x)}{3465 d (a+a \sin (c+d x))^{3/2}}-\frac {1024 a^4 \cos ^3(c+d x)}{1155 d \sqrt {a+a \sin (c+d x)}}-\frac {128 a^3 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{231 d}-\frac {32 a^2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{99 d}-\frac {2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d} \]

[Out]

-4096/3465*a^5*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-32/99*a^2*cos(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/d-2/11*a*co
s(d*x+c)^3*(a+a*sin(d*x+c))^(5/2)/d-1024/1155*a^4*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-128/231*a^3*cos(d*x+c)
^3*(a+a*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]
time = 0.20, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {4096 a^5 \cos ^3(c+d x)}{3465 d (a \sin (c+d x)+a)^{3/2}}-\frac {1024 a^4 \cos ^3(c+d x)}{1155 d \sqrt {a \sin (c+d x)+a}}-\frac {128 a^3 \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{231 d}-\frac {32 a^2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{99 d}-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4096*a^5*Cos[c + d*x]^3)/(3465*d*(a + a*Sin[c + d*x])^(3/2)) - (1024*a^4*Cos[c + d*x]^3)/(1155*d*Sqrt[a + a*
Sin[c + d*x]]) - (128*a^3*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(231*d) - (32*a^2*Cos[c + d*x]^3*(a + a*Sin
[c + d*x])^(3/2))/(99*d) - (2*a*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2))/(11*d)

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 ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=-\frac {2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d}+\frac {1}{11} (16 a) \int \cos ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\\ &=-\frac {32 a^2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{99 d}-\frac {2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d}+\frac {1}{33} \left (64 a^2\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac {128 a^3 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{231 d}-\frac {32 a^2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{99 d}-\frac {2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d}+\frac {1}{231} \left (512 a^3\right ) \int \cos ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {1024 a^4 \cos ^3(c+d x)}{1155 d \sqrt {a+a \sin (c+d x)}}-\frac {128 a^3 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{231 d}-\frac {32 a^2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{99 d}-\frac {2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d}+\frac {\left (2048 a^4\right ) \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{1155}\\ &=-\frac {4096 a^5 \cos ^3(c+d x)}{3465 d (a+a \sin (c+d x))^{3/2}}-\frac {1024 a^4 \cos ^3(c+d x)}{1155 d \sqrt {a+a \sin (c+d x)}}-\frac {128 a^3 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{231 d}-\frac {32 a^2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{99 d}-\frac {2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.14, size = 82, normalized size = 0.52 \begin {gather*} -\frac {2 a^3 \cos ^3(c+d x) \sqrt {a (1+\sin (c+d x))} \left (5419+6396 \sin (c+d x)+4530 \sin ^2(c+d x)+1820 \sin ^3(c+d x)+315 \sin ^4(c+d x)\right )}{3465 d (1+\sin (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*Cos[c + d*x]^3*Sqrt[a*(1 + Sin[c + d*x])]*(5419 + 6396*Sin[c + d*x] + 4530*Sin[c + d*x]^2 + 1820*Sin[c
+ d*x]^3 + 315*Sin[c + d*x]^4))/(3465*d*(1 + Sin[c + d*x])^2)

________________________________________________________________________________________

Maple [A]
time = 0.42, size = 87, normalized size = 0.55

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{4} \left (\sin \left (d x +c \right )-1\right )^{2} \left (315 \left (\sin ^{4}\left (d x +c \right )\right )+1820 \left (\sin ^{3}\left (d x +c \right )\right )+4530 \left (\sin ^{2}\left (d x +c \right )\right )+6396 \sin \left (d x +c \right )+5419\right )}{3465 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*(1+sin(d*x+c))*a^4*(sin(d*x+c)-1)^2*(315*sin(d*x+c)^4+1820*sin(d*x+c)^3+4530*sin(d*x+c)^2+6396*sin(d*x
+c)+5419)/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)^2*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 192, normalized size = 1.21 \begin {gather*} \frac {2 \, {\left (315 \, a^{3} \cos \left (d x + c\right )^{6} + 1505 \, a^{3} \cos \left (d x + c\right )^{5} - 2150 \, a^{3} \cos \left (d x + c\right )^{4} - 4876 \, a^{3} \cos \left (d x + c\right )^{3} + 512 \, a^{3} \cos \left (d x + c\right )^{2} - 2048 \, a^{3} \cos \left (d x + c\right ) - 4096 \, a^{3} + {\left (315 \, a^{3} \cos \left (d x + c\right )^{5} - 1190 \, a^{3} \cos \left (d x + c\right )^{4} - 3340 \, a^{3} \cos \left (d x + c\right )^{3} + 1536 \, a^{3} \cos \left (d x + c\right )^{2} + 2048 \, a^{3} \cos \left (d x + c\right ) + 4096 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3465 \, {\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)^2*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

2/3465*(315*a^3*cos(d*x + c)^6 + 1505*a^3*cos(d*x + c)^5 - 2150*a^3*cos(d*x + c)^4 - 4876*a^3*cos(d*x + c)^3 +
512*a^3*cos(d*x + c)^2 - 2048*a^3*cos(d*x + c) - 4096*a^3 + (315*a^3*cos(d*x + c)^5 - 1190*a^3*cos(d*x + c)^4
- 3340*a^3*cos(d*x + c)^3 + 1536*a^3*cos(d*x + c)^2 + 2048*a^3*cos(d*x + c) + 4096*a^3)*sin(d*x + c))*sqrt(a*
sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4845 deep

________________________________________________________________________________________

Giac [A]
time = 6.48, size = 172, normalized size = 1.08 \begin {gather*} \frac {64 \, \sqrt {2} {\left (315 \, a^{3} \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 )^{11} - 1540 \, a^{3} \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 )^{9} + 2970 \, a^{3} \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 )^{7} - 2772 \, a^{3} \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 )^{5} + 1155 \, a^{3} \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 )^{3}\right )} \sqrt {a}}{3465 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

64/3465*sqrt(2)*(315*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^11 - 1540*a^3*sgn(
cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 + 2970*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)
)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 2772*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*
c)^5 + 1155*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3)*sqrt(a)/d

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________