3.185 \(\int \frac{\cos ^7(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=97 \[ -\frac{2 (a \sin (c+d x)+a)^{9/2}}{9 a^7 d}+\frac{12 (a \sin (c+d x)+a)^{7/2}}{7 a^6 d}-\frac{24 (a \sin (c+d x)+a)^{5/2}}{5 a^5 d}+\frac{16 (a \sin (c+d x)+a)^{3/2}}{3 a^4 d} \]

[Out]

(16*(a + a*Sin[c + d*x])^(3/2))/(3*a^4*d) - (24*(a + a*Sin[c + d*x])^(5/2))/(5*a^5*d) + (12*(a + a*Sin[c + d*x
])^(7/2))/(7*a^6*d) - (2*(a + a*Sin[c + d*x])^(9/2))/(9*a^7*d)

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Rubi [A]  time = 0.0831496, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2667, 43} \[ -\frac{2 (a \sin (c+d x)+a)^{9/2}}{9 a^7 d}+\frac{12 (a \sin (c+d x)+a)^{7/2}}{7 a^6 d}-\frac{24 (a \sin (c+d x)+a)^{5/2}}{5 a^5 d}+\frac{16 (a \sin (c+d x)+a)^{3/2}}{3 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(16*(a + a*Sin[c + d*x])^(3/2))/(3*a^4*d) - (24*(a + a*Sin[c + d*x])^(5/2))/(5*a^5*d) + (12*(a + a*Sin[c + d*x
])^(7/2))/(7*a^6*d) - (2*(a + a*Sin[c + d*x])^(9/2))/(9*a^7*d)

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^7(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int (a-x)^3 \sqrt{a+x} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (8 a^3 \sqrt{a+x}-12 a^2 (a+x)^{3/2}+6 a (a+x)^{5/2}-(a+x)^{7/2}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{16 (a+a \sin (c+d x))^{3/2}}{3 a^4 d}-\frac{24 (a+a \sin (c+d x))^{5/2}}{5 a^5 d}+\frac{12 (a+a \sin (c+d x))^{7/2}}{7 a^6 d}-\frac{2 (a+a \sin (c+d x))^{9/2}}{9 a^7 d}\\ \end{align*}

Mathematica [A]  time = 0.174714, size = 54, normalized size = 0.56 \[ -\frac{2 \left (35 \sin ^3(c+d x)-165 \sin ^2(c+d x)+321 \sin (c+d x)-319\right ) (a (\sin (c+d x)+1))^{3/2}}{315 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-2*(a*(1 + Sin[c + d*x]))^(3/2)*(-319 + 321*Sin[c + d*x] - 165*Sin[c + d*x]^2 + 35*Sin[c + d*x]^3))/(315*a^4*
d)

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Maple [A]  time = 0.161, size = 57, normalized size = 0.6 \begin{align*}{\frac{70\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -330\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-712\,\sin \left ( dx+c \right ) +968}{315\,{a}^{4}d} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7/(a+a*sin(d*x+c))^(5/2),x)

[Out]

2/315/a^4*(a+a*sin(d*x+c))^(3/2)*(35*cos(d*x+c)^2*sin(d*x+c)-165*cos(d*x+c)^2-356*sin(d*x+c)+484)/d

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Maxima [A]  time = 0.936901, size = 97, normalized size = 1. \begin{align*} -\frac{2 \,{\left (35 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{9}{2}} - 270 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a + 756 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{2} - 840 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{3}\right )}}{315 \, a^{7} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/315*(35*(a*sin(d*x + c) + a)^(9/2) - 270*(a*sin(d*x + c) + a)^(7/2)*a + 756*(a*sin(d*x + c) + a)^(5/2)*a^2
- 840*(a*sin(d*x + c) + a)^(3/2)*a^3)/(a^7*d)

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Fricas [A]  time = 2.2687, size = 176, normalized size = 1.81 \begin{align*} -\frac{2 \,{\left (35 \, \cos \left (d x + c\right )^{4} - 226 \, \cos \left (d x + c\right )^{2} + 2 \,{\left (65 \, \cos \left (d x + c\right )^{2} - 64\right )} \sin \left (d x + c\right ) - 128\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{315 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/315*(35*cos(d*x + c)^4 - 226*cos(d*x + c)^2 + 2*(65*cos(d*x + c)^2 - 64)*sin(d*x + c) - 128)*sqrt(a*sin(d*x
+ c) + a)/(a^3*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)**7/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.24723, size = 97, normalized size = 1. \begin{align*} -\frac{2 \,{\left (35 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{9}{2}} - 270 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a + 756 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{2} - 840 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{3}\right )}}{315 \, a^{7} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-2/315*(35*(a*sin(d*x + c) + a)^(9/2) - 270*(a*sin(d*x + c) + a)^(7/2)*a + 756*(a*sin(d*x + c) + a)^(5/2)*a^2
- 840*(a*sin(d*x + c) + a)^(3/2)*a^3)/(a^7*d)