3.142 \(\int \cos ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=191 \[ -\frac {16384 a^6 \cos ^5(c+d x)}{45045 d (a \sin (c+d x)+a)^{5/2}}-\frac {4096 a^5 \cos ^5(c+d x)}{9009 d (a \sin (c+d x)+a)^{3/2}}-\frac {512 a^4 \cos ^5(c+d x)}{1287 d \sqrt {a \sin (c+d x)+a}}-\frac {128 a^3 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{429 d}-\frac {8 a^2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{39 d}-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d} \]

[Out]

-16384/45045*a^6*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-4096/9009*a^5*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)-8/3
9*a^2*cos(d*x+c)^5*(a+a*sin(d*x+c))^(3/2)/d-2/15*a*cos(d*x+c)^5*(a+a*sin(d*x+c))^(5/2)/d-512/1287*a^4*cos(d*x+
c)^5/d/(a+a*sin(d*x+c))^(1/2)-128/429*a^3*cos(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.37, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2674, 2673} \[ -\frac {16384 a^6 \cos ^5(c+d x)}{45045 d (a \sin (c+d x)+a)^{5/2}}-\frac {4096 a^5 \cos ^5(c+d x)}{9009 d (a \sin (c+d x)+a)^{3/2}}-\frac {512 a^4 \cos ^5(c+d x)}{1287 d \sqrt {a \sin (c+d x)+a}}-\frac {128 a^3 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{429 d}-\frac {8 a^2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{39 d}-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(-16384*a^6*Cos[c + d*x]^5)/(45045*d*(a + a*Sin[c + d*x])^(5/2)) - (4096*a^5*Cos[c + d*x]^5)/(9009*d*(a + a*Si
n[c + d*x])^(3/2)) - (512*a^4*Cos[c + d*x]^5)/(1287*d*Sqrt[a + a*Sin[c + d*x]]) - (128*a^3*Cos[c + d*x]^5*Sqrt
[a + a*Sin[c + d*x]])/(429*d) - (8*a^2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(39*d) - (2*a*Cos[c + d*x]^5
*(a + a*Sin[c + d*x])^(5/2))/(15*d)

Rule 2673

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 - 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 2674

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 ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}+\frac {1}{3} (4 a) \int \cos ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\\ &=-\frac {8 a^2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}+\frac {1}{39} \left (64 a^2\right ) \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac {128 a^3 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{429 d}-\frac {8 a^2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}+\frac {1}{143} \left (256 a^3\right ) \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {512 a^4 \cos ^5(c+d x)}{1287 d \sqrt {a+a \sin (c+d x)}}-\frac {128 a^3 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{429 d}-\frac {8 a^2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}+\frac {\left (2048 a^4\right ) \int \frac {\cos ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{1287}\\ &=-\frac {4096 a^5 \cos ^5(c+d x)}{9009 d (a+a \sin (c+d x))^{3/2}}-\frac {512 a^4 \cos ^5(c+d x)}{1287 d \sqrt {a+a \sin (c+d x)}}-\frac {128 a^3 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{429 d}-\frac {8 a^2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}+\frac {\left (8192 a^5\right ) \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{9009}\\ &=-\frac {16384 a^6 \cos ^5(c+d x)}{45045 d (a+a \sin (c+d x))^{5/2}}-\frac {4096 a^5 \cos ^5(c+d x)}{9009 d (a+a \sin (c+d x))^{3/2}}-\frac {512 a^4 \cos ^5(c+d x)}{1287 d \sqrt {a+a \sin (c+d x)}}-\frac {128 a^3 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{429 d}-\frac {8 a^2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 92, normalized size = 0.48 \[ -\frac {2 a^3 \left (3003 \sin ^5(c+d x)+19635 \sin ^4(c+d x)+55230 \sin ^3(c+d x)+86870 \sin ^2(c+d x)+81815 \sin (c+d x)+41735\right ) \cos ^5(c+d x) \sqrt {a (\sin (c+d x)+1)}}{45045 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(-2*a^3*Cos[c + d*x]^5*Sqrt[a*(1 + Sin[c + d*x])]*(41735 + 81815*Sin[c + d*x] + 86870*Sin[c + d*x]^2 + 55230*S
in[c + d*x]^3 + 19635*Sin[c + d*x]^4 + 3003*Sin[c + d*x]^5))/(45045*d*(1 + Sin[c + d*x])^3)

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fricas [A]  time = 0.70, size = 244, normalized size = 1.28 \[ \frac {2 \, {\left (3003 \, a^{3} \cos \left (d x + c\right )^{8} + 13629 \, a^{3} \cos \left (d x + c\right )^{7} - 17346 \, a^{3} \cos \left (d x + c\right )^{6} - 36932 \, a^{3} \cos \left (d x + c\right )^{5} + 1280 \, a^{3} \cos \left (d x + c\right )^{4} - 2048 \, a^{3} \cos \left (d x + c\right )^{3} + 4096 \, a^{3} \cos \left (d x + c\right )^{2} - 16384 \, a^{3} \cos \left (d x + c\right ) - 32768 \, a^{3} + {\left (3003 \, a^{3} \cos \left (d x + c\right )^{7} - 10626 \, a^{3} \cos \left (d x + c\right )^{6} - 27972 \, a^{3} \cos \left (d x + c\right )^{5} + 8960 \, a^{3} \cos \left (d x + c\right )^{4} + 10240 \, a^{3} \cos \left (d x + c\right )^{3} + 12288 \, a^{3} \cos \left (d x + c\right )^{2} + 16384 \, a^{3} \cos \left (d x + c\right ) + 32768 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{45045 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

2/45045*(3003*a^3*cos(d*x + c)^8 + 13629*a^3*cos(d*x + c)^7 - 17346*a^3*cos(d*x + c)^6 - 36932*a^3*cos(d*x + c
)^5 + 1280*a^3*cos(d*x + c)^4 - 2048*a^3*cos(d*x + c)^3 + 4096*a^3*cos(d*x + c)^2 - 16384*a^3*cos(d*x + c) - 3
2768*a^3 + (3003*a^3*cos(d*x + c)^7 - 10626*a^3*cos(d*x + c)^6 - 27972*a^3*cos(d*x + c)^5 + 8960*a^3*cos(d*x +
c)^4 + 10240*a^3*cos(d*x + c)^3 + 12288*a^3*cos(d*x + c)^2 + 16384*a^3*cos(d*x + c) + 32768*a^3)*sin(d*x + c)
)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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giac [B]  time = 2.28, size = 504, normalized size = 2.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

1/2882880*sqrt(2)*(3465*a^3*cos(1/4*pi + 13/2*d*x + 13/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 55055*a^3*
cos(1/4*pi + 9/2*d*x + 9/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 351351*a^3*cos(1/4*pi + 5/2*d*x + 5/2*c)
*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 1216215*a^3*cos(1/4*pi + 1/2*d*x + 1/2*c)*sgn(cos(-1/4*pi + 1/2*d*x +
1/2*c))/d + 3003*a^3*cos(-1/4*pi + 15/2*d*x + 15/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 45045*a^3*cos(-
1/4*pi + 11/2*d*x + 11/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 250965*a^3*cos(-1/4*pi + 7/2*d*x + 7/2*c)*
sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 405405*a^3*cos(-1/4*pi + 3/2*d*x + 3/2*c)*sgn(cos(-1/4*pi + 1/2*d*x +
1/2*c))/d - 24570*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 11/2*d*x + 11/2*c)/d - 25740*a^3*sgn(co
s(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 7/2*d*x + 7/2*c)/d + 570570*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))
*sin(1/4*pi + 3/2*d*x + 3/2*c)/d - 20790*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 13/2*d*x + 13/2
*c)/d - 20020*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 9/2*d*x + 9/2*c)/d + 342342*a^3*sgn(cos(-1
/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 5/2*d*x + 5/2*c)/d + 3243240*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*s
in(-1/4*pi + 1/2*d*x + 1/2*c)/d)*sqrt(a)

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maple [A]  time = 0.20, size = 97, normalized size = 0.51 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{4} \left (\sin \left (d x +c \right )-1\right )^{3} \left (3003 \left (\sin ^{5}\left (d x +c \right )\right )+19635 \left (\sin ^{4}\left (d x +c \right )\right )+55230 \left (\sin ^{3}\left (d x +c \right )\right )+86870 \left (\sin ^{2}\left (d x +c \right )\right )+81815 \sin \left (d x +c \right )+41735\right )}{45045 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*sin(d*x+c))^(7/2),x)

[Out]

2/45045*(1+sin(d*x+c))*a^4*(sin(d*x+c)-1)^3*(3003*sin(d*x+c)^5+19635*sin(d*x+c)^4+55230*sin(d*x+c)^3+86870*sin
(d*x+c)^2+81815*sin(d*x+c)+41735)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \cos \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(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)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + a*sin(c + d*x))^(7/2),x)

[Out]

int(cos(c + d*x)^4*(a + a*sin(c + d*x))^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+a*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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