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

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

[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)

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Rubi [A]  time = 0.302172, antiderivative size = 159, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.623684, size = 79, normalized size = 0.5 \[ -\frac{2 \left (3003 \sin ^4(c+d x)+15708 \sin ^3(c+d x)+33138 \sin ^2(c+d x)+34748 \sin (c+d x)+16363\right ) \cos ^7(c+d x) (a (\sin (c+d x)+1))^{3/2}}{45045 d (\sin (c+d x)+1)^5} \]

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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Maple [A]  time = 0.112, size = 87, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{4} \left ( 3003\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+15708\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+33138\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+34748\,\sin \left ( dx+c \right ) +16363 \right ) }{45045\,d\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{6}\,{d x} \end{align*}

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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Fricas [A]  time = 1.79285, size = 622, normalized size = 3.91 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{6}\,{d x} \end{align*}

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]

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