3.101 \(\int \cos ^7(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=97 \[ -\frac{2 (a \sin (c+d x)+a)^{15/2}}{15 a^7 d}+\frac{12 (a \sin (c+d x)+a)^{13/2}}{13 a^6 d}-\frac{24 (a \sin (c+d x)+a)^{11/2}}{11 a^5 d}+\frac{16 (a \sin (c+d x)+a)^{9/2}}{9 a^4 d} \]

[Out]

(16*(a + a*Sin[c + d*x])^(9/2))/(9*a^4*d) - (24*(a + a*Sin[c + d*x])^(11/2))/(11*a^5*d) + (12*(a + a*Sin[c + d
*x])^(13/2))/(13*a^6*d) - (2*(a + a*Sin[c + d*x])^(15/2))/(15*a^7*d)

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Rubi [A]  time = 0.0829884, 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} \[ -\frac{2 (a \sin (c+d x)+a)^{15/2}}{15 a^7 d}+\frac{12 (a \sin (c+d x)+a)^{13/2}}{13 a^6 d}-\frac{24 (a \sin (c+d x)+a)^{11/2}}{11 a^5 d}+\frac{16 (a \sin (c+d x)+a)^{9/2}}{9 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(16*(a + a*Sin[c + d*x])^(9/2))/(9*a^4*d) - (24*(a + a*Sin[c + d*x])^(11/2))/(11*a^5*d) + (12*(a + a*Sin[c + d
*x])^(13/2))/(13*a^6*d) - (2*(a + a*Sin[c + d*x])^(15/2))/(15*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 \cos ^7(c+d x) \sqrt{a+a \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int (a-x)^3 (a+x)^{7/2} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (8 a^3 (a+x)^{7/2}-12 a^2 (a+x)^{9/2}+6 a (a+x)^{11/2}-(a+x)^{13/2}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{16 (a+a \sin (c+d x))^{9/2}}{9 a^4 d}-\frac{24 (a+a \sin (c+d x))^{11/2}}{11 a^5 d}+\frac{12 (a+a \sin (c+d x))^{13/2}}{13 a^6 d}-\frac{2 (a+a \sin (c+d x))^{15/2}}{15 a^7 d}\\ \end{align*}

Mathematica [A]  time = 4.2383, size = 74, normalized size = 0.76 \[ \frac{\sqrt{a (\sin (c+d x)+1)} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8 (-10755 \sin (c+d x)+429 \sin (3 (c+d x))-3366 \cos (2 (c+d x))+8330)}{12870 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8*Sqrt[a*(1 + Sin[c + d*x])]*(8330 - 3366*Cos[2*(c + d*x)] - 10755*Sin[
c + d*x] + 429*Sin[3*(c + d*x)]))/(12870*d)

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Maple [A]  time = 0.091, size = 57, normalized size = 0.6 \begin{align*}{\frac{858\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -3366\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-5592\,\sin \left ( dx+c \right ) +5848}{6435\,{a}^{4}d} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{{\frac{9}{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))^(1/2),x)

[Out]

2/6435/a^4*(a+a*sin(d*x+c))^(9/2)*(429*cos(d*x+c)^2*sin(d*x+c)-1683*cos(d*x+c)^2-2796*sin(d*x+c)+2924)/d

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Maxima [A]  time = 0.948102, size = 97, normalized size = 1. \begin{align*} -\frac{2 \,{\left (429 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{15}{2}} - 2970 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{13}{2}} a + 7020 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{11}{2}} a^{2} - 5720 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{9}{2}} a^{3}\right )}}{6435 \, 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))^(1/2),x, algorithm="maxima")

[Out]

-2/6435*(429*(a*sin(d*x + c) + a)^(15/2) - 2970*(a*sin(d*x + c) + a)^(13/2)*a + 7020*(a*sin(d*x + c) + a)^(11/
2)*a^2 - 5720*(a*sin(d*x + c) + a)^(9/2)*a^3)/(a^7*d)

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Fricas [A]  time = 1.74841, size = 254, normalized size = 2.62 \begin{align*} \frac{2 \,{\left (33 \, \cos \left (d x + c\right )^{6} + 56 \, \cos \left (d x + c\right )^{4} + 128 \, \cos \left (d x + c\right )^{2} +{\left (429 \, \cos \left (d x + c\right )^{6} + 504 \, \cos \left (d x + c\right )^{4} + 640 \, \cos \left (d x + c\right )^{2} + 1024\right )} \sin \left (d x + c\right ) + 1024\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{6435 \, 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))^(1/2),x, algorithm="fricas")

[Out]

2/6435*(33*cos(d*x + c)^6 + 56*cos(d*x + c)^4 + 128*cos(d*x + c)^2 + (429*cos(d*x + c)^6 + 504*cos(d*x + c)^4
+ 640*cos(d*x + c)^2 + 1024)*sin(d*x + c) + 1024)*sqrt(a*sin(d*x + c) + a)/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))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 2.16911, size = 104, normalized size = 1.07 \begin{align*} -\frac{2 \,{\left (\frac{429 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{15}{2}}}{a^{6}} - \frac{2970 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{13}{2}}}{a^{5}} + \frac{7020 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{11}{2}}}{a^{4}} - \frac{5720 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{9}{2}}}{a^{3}}\right )}}{6435 \, a 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))^(1/2),x, algorithm="giac")

[Out]

-2/6435*(429*(a*sin(d*x + c) + a)^(15/2)/a^6 - 2970*(a*sin(d*x + c) + a)^(13/2)/a^5 + 7020*(a*sin(d*x + c) + a
)^(11/2)/a^4 - 5720*(a*sin(d*x + c) + a)^(9/2)/a^3)/(a*d)