3.115 \(\int \cos ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=159 \[ -\frac {4096 a^5 \cos ^7(c+d x)}{45045 d (a \sin (c+d x)+a)^{7/2}}-\frac {1024 a^4 \cos ^7(c+d x)}{6435 d (a \sin (c+d x)+a)^{5/2}}-\frac {128 a^3 \cos ^7(c+d x)}{715 d (a \sin (c+d x)+a)^{3/2}}-\frac {32 a^2 \cos ^7(c+d x)}{195 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a \cos ^7(c+d x) \sqrt {a \sin (c+d x)+a}}{15 d} \]

[Out]

-4096/45045*a^5*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(7/2)-1024/6435*a^4*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(5/2)-128/
715*a^3*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(3/2)-32/195*a^2*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(1/2)-2/15*a*cos(d*x+
c)^7*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.30, 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 = {2674, 2673} \[ -\frac {32 a^2 \cos ^7(c+d x)}{195 d \sqrt {a \sin (c+d x)+a}}-\frac {128 a^3 \cos ^7(c+d x)}{715 d (a \sin (c+d x)+a)^{3/2}}-\frac {1024 a^4 \cos ^7(c+d x)}{6435 d (a \sin (c+d x)+a)^{5/2}}-\frac {4096 a^5 \cos ^7(c+d x)}{45045 d (a \sin (c+d x)+a)^{7/2}}-\frac {2 a \cos ^7(c+d x) \sqrt {a \sin (c+d x)+a}}{15 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-4096*a^5*Cos[c + d*x]^7)/(45045*d*(a + a*Sin[c + d*x])^(7/2)) - (1024*a^4*Cos[c + d*x]^7)/(6435*d*(a + a*Sin
[c + d*x])^(5/2)) - (128*a^3*Cos[c + d*x]^7)/(715*d*(a + a*Sin[c + d*x])^(3/2)) - (32*a^2*Cos[c + d*x]^7)/(195
*d*Sqrt[a + a*Sin[c + d*x]]) - (2*a*Cos[c + d*x]^7*Sqrt[a + a*Sin[c + d*x]])/(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 ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac {2 a \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}+\frac {1}{15} (16 a) \int \cos ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {32 a^2 \cos ^7(c+d x)}{195 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}+\frac {1}{65} \left (64 a^2\right ) \int \frac {\cos ^6(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {128 a^3 \cos ^7(c+d x)}{715 d (a+a \sin (c+d x))^{3/2}}-\frac {32 a^2 \cos ^7(c+d x)}{195 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}+\frac {1}{715} \left (512 a^3\right ) \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {1024 a^4 \cos ^7(c+d x)}{6435 d (a+a \sin (c+d x))^{5/2}}-\frac {128 a^3 \cos ^7(c+d x)}{715 d (a+a \sin (c+d x))^{3/2}}-\frac {32 a^2 \cos ^7(c+d x)}{195 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}+\frac {\left (2048 a^4\right ) \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx}{6435}\\ &=-\frac {4096 a^5 \cos ^7(c+d x)}{45045 d (a+a \sin (c+d x))^{7/2}}-\frac {1024 a^4 \cos ^7(c+d x)}{6435 d (a+a \sin (c+d x))^{5/2}}-\frac {128 a^3 \cos ^7(c+d x)}{715 d (a+a \sin (c+d x))^{3/2}}-\frac {32 a^2 \cos ^7(c+d x)}{195 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 79, normalized size = 0.50 \[ -\frac {2 \left (3003 \sin ^4(c+d x)+15708 \sin ^3(c+d x)+33138 \sin ^2(c+d x)+34748 \sin (c+d x)+16363\right ) \cos ^7(c+d x) (a (\sin (c+d x)+1))^{3/2}}{45045 d (\sin (c+d x)+1)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-2*Cos[c + d*x]^7*(a*(1 + Sin[c + d*x]))^(3/2)*(16363 + 34748*Sin[c + d*x] + 33138*Sin[c + d*x]^2 + 15708*Sin
[c + d*x]^3 + 3003*Sin[c + d*x]^4))/(45045*d*(1 + Sin[c + d*x])^5)

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fricas [A]  time = 0.57, size = 210, normalized size = 1.32 \[ -\frac {2 \, {\left (3003 \, a \cos \left (d x + c\right )^{8} + 6699 \, a \cos \left (d x + c\right )^{7} - 336 \, a \cos \left (d x + c\right )^{6} + 448 \, a \cos \left (d x + c\right )^{5} - 640 \, a \cos \left (d x + c\right )^{4} + 1024 \, a \cos \left (d x + c\right )^{3} - 2048 \, a \cos \left (d x + c\right )^{2} + 8192 \, a \cos \left (d x + c\right ) + {\left (3003 \, a \cos \left (d x + c\right )^{7} - 3696 \, a \cos \left (d x + c\right )^{6} - 4032 \, a \cos \left (d x + c\right )^{5} - 4480 \, a \cos \left (d x + c\right )^{4} - 5120 \, a \cos \left (d x + c\right )^{3} - 6144 \, a \cos \left (d x + c\right )^{2} - 8192 \, a \cos \left (d x + c\right ) - 16384 \, a\right )} \sin \left (d x + c\right ) + 16384 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{45045 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-2/45045*(3003*a*cos(d*x + c)^8 + 6699*a*cos(d*x + c)^7 - 336*a*cos(d*x + c)^6 + 448*a*cos(d*x + c)^5 - 640*a*
cos(d*x + c)^4 + 1024*a*cos(d*x + c)^3 - 2048*a*cos(d*x + c)^2 + 8192*a*cos(d*x + c) + (3003*a*cos(d*x + c)^7
- 3696*a*cos(d*x + c)^6 - 4032*a*cos(d*x + c)^5 - 4480*a*cos(d*x + c)^4 - 5120*a*cos(d*x + c)^3 - 6144*a*cos(d
*x + c)^2 - 8192*a*cos(d*x + c) - 16384*a)*sin(d*x + c) + 16384*a)*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 = 1.34, size = 474, normalized size = 2.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/2882880*sqrt(2)*(3465*a*cos(1/4*pi + 13/2*d*x + 13/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 25025*a*cos
(1/4*pi + 9/2*d*x + 9/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 81081*a*cos(1/4*pi + 5/2*d*x + 5/2*c)*sgn(c
os(-1/4*pi + 1/2*d*x + 1/2*c))/d + 225225*a*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*cos(-1/4*pi + 15/2*d*x + 15/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 20475*a*cos(-1/4*pi + 11/2
*d*x + 11/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 57915*a*cos(-1/4*pi + 7/2*d*x + 7/2*c)*sgn(cos(-1/4*pi
+ 1/2*d*x + 1/2*c))/d + 75075*a*cos(-1/4*pi + 3/2*d*x + 3/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 8190*a*
sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 11/2*d*x + 11/2*c)/d - 77220*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/
2*c))*sin(1/4*pi + 7/2*d*x + 7/2*c)/d - 450450*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 3/2*d*x + 3/
2*c)/d - 6930*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 13/2*d*x + 13/2*c)/d - 60060*a*sgn(cos(-1/4*
pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 9/2*d*x + 9/2*c)/d - 270270*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/
4*pi + 5/2*d*x + 5/2*c)/d - 1801800*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)/d)*sq
rt(a)

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maple [A]  time = 0.21, size = 87, normalized size = 0.55 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{4} \left (3003 \left (\sin ^{4}\left (d x +c \right )\right )+15708 \left (\sin ^{3}\left (d x +c \right )\right )+33138 \left (\sin ^{2}\left (d x +c \right )\right )+34748 \sin \left (d x +c \right )+16363\right )}{45045 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+a*sin(d*x+c))^(3/2),x)

[Out]

-2/45045*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^4*(3003*sin(d*x+c)^4+15708*sin(d*x+c)^3+33138*sin(d*x+c)^2+34748*si
n(d*x+c)+16363)/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 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^6, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^6\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*(a + a*sin(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^6*(a + a*sin(c + d*x))^(3/2), 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)**6*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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