3.2.39 \(\int \cos ^7(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\) [139]

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

[Out]

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

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Rubi [A]
time = 0.06, 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 = {2746, 45} \begin {gather*} -\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} \end {gather*}

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 45

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 2746

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 \cos ^7(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=\frac {\text {Subst}\left (\int (a-x)^3 (a+x)^{13/2} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac {\text {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*}

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Mathematica [A]
time = 0.38, size = 64, normalized size = 0.66 \begin {gather*} -\frac {2 a^3 (1+\sin (c+d x))^7 \sqrt {a (1+\sin (c+d x))} \left (-3243+7365 \sin (c+d x)-5865 \sin ^2(c+d x)+1615 \sin ^3(c+d x)\right )}{33915 d} \end {gather*}

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 = 0.31, size = 57, normalized size = 0.59

method	result	size
default	\(\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {15}{2}} \left (1615 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-5865 \left (\cos ^{2}\left (d x +c \right )\right )-8980 \sin \left (d x +c \right )+9108\right )}{33915 a^{4} d}\)	\(57\)

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,method=_RETURNVERBOSE)

[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 = 0.32, size = 72, normalized size = 0.74 \begin {gather*} -\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 {gather*}

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 = 0.38, size = 154, normalized size = 1.59 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 [A]
time = 5.62, size = 140, normalized size = 1.44 \begin {gather*} -\frac {2048 \, \sqrt {2} {\left (1615 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{21} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 5355 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{19} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 5985 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{17} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 2261 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sqrt {a}}{33915 \, d} \end {gather*}

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]

-2048/33915*sqrt(2)*(1615*a^3*cos(-1/4*pi + 1/2*d*x + 1/2*c)^21*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 5355*a^3
*cos(-1/4*pi + 1/2*d*x + 1/2*c)^19*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) + 5985*a^3*cos(-1/4*pi + 1/2*d*x + 1/2*
c)^17*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 2261*a^3*cos(-1/4*pi + 1/2*d*x + 1/2*c)^15*sgn(cos(-1/4*pi + 1/2*d
*x + 1/2*c)))*sqrt(a)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^7\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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