3.171 \(\int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=73 \[ \frac {2 (a \sin (c+d x)+a)^{7/2}}{7 a^5 d}-\frac {8 (a \sin (c+d x)+a)^{5/2}}{5 a^4 d}+\frac {8 (a \sin (c+d x)+a)^{3/2}}{3 a^3 d} \]

[Out]

8/3*(a+a*sin(d*x+c))^(3/2)/a^3/d-8/5*(a+a*sin(d*x+c))^(5/2)/a^4/d+2/7*(a+a*sin(d*x+c))^(7/2)/a^5/d

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Rubi [A]  time = 0.07, antiderivative size = 73, 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} \[ \frac {2 (a \sin (c+d x)+a)^{7/2}}{7 a^5 d}-\frac {8 (a \sin (c+d x)+a)^{5/2}}{5 a^4 d}+\frac {8 (a \sin (c+d x)+a)^{3/2}}{3 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(8*(a + a*Sin[c + d*x])^(3/2))/(3*a^3*d) - (8*(a + a*Sin[c + d*x])^(5/2))/(5*a^4*d) + (2*(a + a*Sin[c + d*x])^
(7/2))/(7*a^5*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 ^5(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int (a-x)^2 \sqrt {a+x} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (4 a^2 \sqrt {a+x}-4 a (a+x)^{3/2}+(a+x)^{5/2}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {8 (a+a \sin (c+d x))^{3/2}}{3 a^3 d}-\frac {8 (a+a \sin (c+d x))^{5/2}}{5 a^4 d}+\frac {2 (a+a \sin (c+d x))^{7/2}}{7 a^5 d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 44, normalized size = 0.60 \[ \frac {2 \left (15 \sin ^2(c+d x)-54 \sin (c+d x)+71\right ) (a (\sin (c+d x)+1))^{3/2}}{105 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(2*(a*(1 + Sin[c + d*x]))^(3/2)*(71 - 54*Sin[c + d*x] + 15*Sin[c + d*x]^2))/(105*a^3*d)

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fricas [A]  time = 0.73, size = 52, normalized size = 0.71 \[ \frac {2 \, {\left (39 \, \cos \left (d x + c\right )^{2} - {\left (15 \, \cos \left (d x + c\right )^{2} - 32\right )} \sin \left (d x + c\right ) + 32\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{105 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/105*(39*cos(d*x + c)^2 - (15*cos(d*x + c)^2 - 32)*sin(d*x + c) + 32)*sqrt(a*sin(d*x + c) + a)/(a^2*d)

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giac [B]  time = 2.21, size = 250, normalized size = 3.42 \[ \frac {2 \, {\left ({\left ({\left ({\left ({\left ({\left ({\left (\frac {71 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {105 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {91 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {245 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {245 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {91 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {105 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {71 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )}}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

2/105*(((((((71*a^2*tan(1/2*d*x + 1/2*c)/sgn(tan(1/2*d*x + 1/2*c) + 1) + 105*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1)
)*tan(1/2*d*x + 1/2*c) + 91*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 245*a^2/sgn(tan(1/2*d*x
+ 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 245*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 91*a^2/sgn
(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 105*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c)
+ 71*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))/((a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2)*d)

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maple [A]  time = 0.16, size = 41, normalized size = 0.56 \[ -\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \left (15 \left (\cos ^{2}\left (d x +c \right )\right )+54 \sin \left (d x +c \right )-86\right )}{105 a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(a+a*sin(d*x+c))^(3/2),x)

[Out]

-2/105/a^3*(a+a*sin(d*x+c))^(3/2)*(15*cos(d*x+c)^2+54*sin(d*x+c)-86)/d

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maxima [A]  time = 0.41, size = 55, normalized size = 0.75 \[ \frac {2 \, {\left (15 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 84 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 140 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2}\right )}}{105 \, a^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

2/105*(15*(a*sin(d*x + c) + a)^(7/2) - 84*(a*sin(d*x + c) + a)^(5/2)*a + 140*(a*sin(d*x + c) + a)^(3/2)*a^2)/(
a^5*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^5}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(a + a*sin(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^5/(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)**5/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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