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]

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

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Rubi [A]  time = 0.08, antiderivative size = 97, normalized size of antiderivative = 1.00, 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 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])

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])

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*}

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Mathematica [A]  time = 0.19, 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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fricas [A]  time = 0.61, size = 62, normalized size = 0.64 \[ -\frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{4} - 226 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (65 \, \cos \left (d x + c\right )^{2} - 64\right )} \sin \left (d x + c\right ) - 128\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, a^{3} d} \]

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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giac [B]  time = 10.72, size = 310, normalized size = 3.20 \[ \frac {2 \, {\left ({\left ({\left ({\left ({\left ({\left ({\left ({\left ({\left (\frac {319 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {315 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {648 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {1680 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {1134 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {1134 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {1680 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {648 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {315 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {319 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )}}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {9}{2}} d} \]

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*(((((((((319*a^2*tan(1/2*d*x + 1/2*c)/sgn(tan(1/2*d*x + 1/2*c) + 1) + 315*a^2/sgn(tan(1/2*d*x + 1/2*c) +
1))*tan(1/2*d*x + 1/2*c) + 648*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 1680*a^2/sgn(tan(1/2
*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 1134*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 1134
*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 1680*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x
+ 1/2*c) + 648*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 315*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1
))*tan(1/2*d*x + 1/2*c) + 319*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))/((a*tan(1/2*d*x + 1/2*c)^2 + a)^(9/2)*d)

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maple [A]  time = 0.17, size = 57, normalized size = 0.59 \[ \frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \left (35 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-165 \left (\cos ^{2}\left (d x +c \right )\right )-356 \sin \left (d x +c \right )+484\right )}{315 a^{4} d} \]

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 = 1.31, size = 72, normalized size = 0.74 \[ -\frac {2 \, {\left (35 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 270 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 756 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 840 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3}\right )}}{315 \, a^{7} d} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^7}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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