3.2.14 \(\int \cos ^7(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [114]

Optimal. Leaf size=97 \[ \frac {16 (a+a \sin (c+d x))^{11/2}}{11 a^4 d}-\frac {24 (a+a \sin (c+d x))^{13/2}}{13 a^5 d}+\frac {4 (a+a \sin (c+d x))^{15/2}}{5 a^6 d}-\frac {2 (a+a \sin (c+d x))^{17/2}}{17 a^7 d} \]

[Out]

16/11*(a+a*sin(d*x+c))^(11/2)/a^4/d-24/13*(a+a*sin(d*x+c))^(13/2)/a^5/d+4/5*(a+a*sin(d*x+c))^(15/2)/a^6/d-2/17
*(a+a*sin(d*x+c))^(17/2)/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)^{17/2}}{17 a^7 d}+\frac {4 (a \sin (c+d x)+a)^{15/2}}{5 a^6 d}-\frac {24 (a \sin (c+d x)+a)^{13/2}}{13 a^5 d}+\frac {16 (a \sin (c+d x)+a)^{11/2}}{11 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(16*(a + a*Sin[c + d*x])^(11/2))/(11*a^4*d) - (24*(a + a*Sin[c + d*x])^(13/2))/(13*a^5*d) + (4*(a + a*Sin[c +
d*x])^(15/2))/(5*a^6*d) - (2*(a + a*Sin[c + d*x])^(17/2))/(17*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))^{3/2} \, dx &=\frac {\text {Subst}\left (\int (a-x)^3 (a+x)^{9/2} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac {\text {Subst}\left (\int \left (8 a^3 (a+x)^{9/2}-12 a^2 (a+x)^{11/2}+6 a (a+x)^{13/2}-(a+x)^{15/2}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac {16 (a+a \sin (c+d x))^{11/2}}{11 a^4 d}-\frac {24 (a+a \sin (c+d x))^{13/2}}{13 a^5 d}+\frac {4 (a+a \sin (c+d x))^{15/2}}{5 a^6 d}-\frac {2 (a+a \sin (c+d x))^{17/2}}{17 a^7 d}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 61, normalized size = 0.63 \begin {gather*} -\frac {2 (1+\sin (c+d x))^4 (a (1+\sin (c+d x)))^{3/2} \left (-1767+3641 \sin (c+d x)-2717 \sin ^2(c+d x)+715 \sin ^3(c+d x)\right )}{12155 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-2*(1 + Sin[c + d*x])^4*(a*(1 + Sin[c + d*x]))^(3/2)*(-1767 + 3641*Sin[c + d*x] - 2717*Sin[c + d*x]^2 + 715*S
in[c + d*x]^3))/(12155*d)

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Maple [A]
time = 0.26, size = 57, normalized size = 0.59

method	result	size
default	\(\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {11}{2}} \left (715 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-2717 \left (\cos ^{2}\left (d x +c \right )\right )-4356 \sin \left (d x +c \right )+4484\right )}{12155 a^{4} d}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/12155/a^4*(a+a*sin(d*x+c))^(11/2)*(715*cos(d*x+c)^2*sin(d*x+c)-2717*cos(d*x+c)^2-4356*sin(d*x+c)+4484)/d

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Maxima [A]
time = 0.29, size = 72, normalized size = 0.74 \begin {gather*} -\frac {2 \, {\left (715 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {17}{2}} - 4862 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {15}{2}} a + 11220 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a^{2} - 8840 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{3}\right )}}{12155 \, 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))^(3/2),x, algorithm="maxima")

[Out]

-2/12155*(715*(a*sin(d*x + c) + a)^(17/2) - 4862*(a*sin(d*x + c) + a)^(15/2)*a + 11220*(a*sin(d*x + c) + a)^(1
3/2)*a^2 - 8840*(a*sin(d*x + c) + a)^(11/2)*a^3)/(a^7*d)

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Fricas [A]
time = 0.38, size = 110, normalized size = 1.13 \begin {gather*} -\frac {2 \, {\left (715 \, a \cos \left (d x + c\right )^{8} - 66 \, a \cos \left (d x + c\right )^{6} - 112 \, a \cos \left (d x + c\right )^{4} - 256 \, a \cos \left (d x + c\right )^{2} - 2 \, {\left (429 \, a \cos \left (d x + c\right )^{6} + 504 \, a \cos \left (d x + c\right )^{4} + 640 \, a \cos \left (d x + c\right )^{2} + 1024 \, a\right )} \sin \left (d x + c\right ) - 2048 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{12155 \, 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))^(3/2),x, algorithm="fricas")

[Out]

-2/12155*(715*a*cos(d*x + c)^8 - 66*a*cos(d*x + c)^6 - 112*a*cos(d*x + c)^4 - 256*a*cos(d*x + c)^2 - 2*(429*a*
cos(d*x + c)^6 + 504*a*cos(d*x + c)^4 + 640*a*cos(d*x + c)^2 + 1024*a)*sin(d*x + c) - 2048*a)*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))**(3/2),x)

[Out]

Timed out

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Giac [A]
time = 7.12, size = 132, normalized size = 1.36 \begin {gather*} -\frac {512 \, \sqrt {2} {\left (715 \, a \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 ) - 2431 \, a \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 ) + 2805 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 1105 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sqrt {a}}{12155 \, 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))^(3/2),x, algorithm="giac")

[Out]

-512/12155*sqrt(2)*(715*a*cos(-1/4*pi + 1/2*d*x + 1/2*c)^17*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 2431*a*cos(-
1/4*pi + 1/2*d*x + 1/2*c)^15*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) + 2805*a*cos(-1/4*pi + 1/2*d*x + 1/2*c)^13*sg
n(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 1105*a*cos(-1/4*pi + 1/2*d*x + 1/2*c)^11*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 )}^{3/2} \,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))^(3/2),x)

[Out]

int(cos(c + d*x)^7*(a + a*sin(c + d*x))^(3/2), x)

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