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

Optimal. Leaf size=97 \[ -\frac{2 (a \sin (c+d x)+a)^{21/2}}{21 a^7 d}+\frac{12 (a \sin (c+d x)+a)^{19/2}}{19 a^6 d}-\frac{24 (a \sin (c+d x)+a)^{17/2}}{17 a^5 d}+\frac{16 (a \sin (c+d x)+a)^{15/2}}{15 a^4 d} \]

[Out]

(16*(a + a*Sin[c + d*x])^(15/2))/(15*a^4*d) - (24*(a + a*Sin[c + d*x])^(17/2))/(17*a^5*d) + (12*(a + a*Sin[c +
d*x])^(19/2))/(19*a^6*d) - (2*(a + a*Sin[c + d*x])^(21/2))/(21*a^7*d)

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Rubi [A]  time = 0.079213, 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} \[ -\frac{2 (a \sin (c+d x)+a)^{21/2}}{21 a^7 d}+\frac{12 (a \sin (c+d x)+a)^{19/2}}{19 a^6 d}-\frac{24 (a \sin (c+d x)+a)^{17/2}}{17 a^5 d}+\frac{16 (a \sin (c+d x)+a)^{15/2}}{15 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*(a + a*Sin[c + d*x])^(15/2))/(15*a^4*d) - (24*(a + a*Sin[c + d*x])^(17/2))/(17*a^5*d) + (12*(a + a*Sin[c +
d*x])^(19/2))/(19*a^6*d) - (2*(a + a*Sin[c + d*x])^(21/2))/(21*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) (a+a \sin (c+d x))^{7/2} \, dx &=\frac{\operatorname{Subst}\left (\int (a-x)^3 (a+x)^{13/2} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (8 a^3 (a+x)^{13/2}-12 a^2 (a+x)^{15/2}+6 a (a+x)^{17/2}-(a+x)^{19/2}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{16 (a+a \sin (c+d x))^{15/2}}{15 a^4 d}-\frac{24 (a+a \sin (c+d x))^{17/2}}{17 a^5 d}+\frac{12 (a+a \sin (c+d x))^{19/2}}{19 a^6 d}-\frac{2 (a+a \sin (c+d x))^{21/2}}{21 a^7 d}\\ \end{align*}

Mathematica [A]  time = 0.601816, size = 64, normalized size = 0.66 \[ -\frac{2 a^3 (\sin (c+d x)+1)^7 \left (1615 \sin ^3(c+d x)-5865 \sin ^2(c+d x)+7365 \sin (c+d x)-3243\right ) \sqrt{a (\sin (c+d x)+1)}}{33915 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*(1 + Sin[c + d*x])^7*Sqrt[a*(1 + Sin[c + d*x])]*(-3243 + 7365*Sin[c + d*x] - 5865*Sin[c + d*x]^2 + 161
5*Sin[c + d*x]^3))/(33915*d)

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Maple [A]  time = 0.131, size = 57, normalized size = 0.6 \begin{align*}{\frac{3230\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -11730\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-17960\,\sin \left ( dx+c \right ) +18216}{33915\,{a}^{4}d} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{{\frac{15}{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))^(7/2),x)

[Out]

2/33915/a^4*(a+a*sin(d*x+c))^(15/2)*(1615*cos(d*x+c)^2*sin(d*x+c)-5865*cos(d*x+c)^2-8980*sin(d*x+c)+9108)/d

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Maxima [A]  time = 0.944939, size = 97, normalized size = 1. \begin{align*} -\frac{2 \,{\left (1615 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{21}{2}} - 10710 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{19}{2}} a + 23940 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{17}{2}} a^{2} - 18088 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{15}{2}} a^{3}\right )}}{33915 \, 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))^(7/2),x, algorithm="maxima")

[Out]

-2/33915*(1615*(a*sin(d*x + c) + a)^(21/2) - 10710*(a*sin(d*x + c) + a)^(19/2)*a + 23940*(a*sin(d*x + c) + a)^
(17/2)*a^2 - 18088*(a*sin(d*x + c) + a)^(15/2)*a^3)/(a^7*d)

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Fricas [A]  time = 1.9035, size = 410, normalized size = 4.23 \begin{align*} \frac{2 \,{\left (1615 \, a^{3} \cos \left (d x + c\right )^{10} - 8300 \, a^{3} \cos \left (d x + c\right )^{8} + 264 \, a^{3} \cos \left (d x + c\right )^{6} + 448 \, a^{3} \cos \left (d x + c\right )^{4} + 1024 \, a^{3} \cos \left (d x + c\right )^{2} + 8192 \, a^{3} - 8 \,{\left (680 \, a^{3} \cos \left (d x + c\right )^{8} - 429 \, a^{3} \cos \left (d x + c\right )^{6} - 504 \, a^{3} \cos \left (d x + c\right )^{4} - 640 \, a^{3} \cos \left (d x + c\right )^{2} - 1024 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{33915 \, 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))^(7/2),x, algorithm="fricas")

[Out]

2/33915*(1615*a^3*cos(d*x + c)^10 - 8300*a^3*cos(d*x + c)^8 + 264*a^3*cos(d*x + c)^6 + 448*a^3*cos(d*x + c)^4
+ 1024*a^3*cos(d*x + c)^2 + 8192*a^3 - 8*(680*a^3*cos(d*x + c)^8 - 429*a^3*cos(d*x + c)^6 - 504*a^3*cos(d*x +
c)^4 - 640*a^3*cos(d*x + c)^2 - 1024*a^3)*sin(d*x + c))*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))**(7/2),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{7}{2}} \cos \left (d x + c\right )^{7}\,{d x} \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))^(7/2),x, algorithm="giac")

[Out]

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