3.1.30 \(\int \cos ^3(c+d x) (a+a \sin (c+d x))^3 \, dx\) [30]

Optimal. Leaf size=45 \[ \frac {2 (a+a \sin (c+d x))^5}{5 a^2 d}-\frac {(a+a \sin (c+d x))^6}{6 a^3 d} \]

[Out]

2/5*(a+a*sin(d*x+c))^5/a^2/d-1/6*(a+a*sin(d*x+c))^6/a^3/d

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Rubi [A]
time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2746, 45} \begin {gather*} \frac {2 (a \sin (c+d x)+a)^5}{5 a^2 d}-\frac {(a \sin (c+d x)+a)^6}{6 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + a*Sin[c + d*x])^5)/(5*a^2*d) - (a + a*Sin[c + d*x])^6/(6*a^3*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 ^3(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac {\text {Subst}\left (\int (a-x) (a+x)^4 \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\text {Subst}\left (\int \left (2 a (a+x)^4-(a+x)^5\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {2 (a+a \sin (c+d x))^5}{5 a^2 d}-\frac {(a+a \sin (c+d x))^6}{6 a^3 d}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 43, normalized size = 0.96 \begin {gather*} -\frac {a^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{10} (-7+5 \sin (c+d x))}{30 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*(a^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^10*(-7 + 5*Sin[c + d*x]))/d

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(112\) vs. \(2(41)=82\).
time = 0.21, size = 113, normalized size = 2.51

method	result	size
risch	\(\frac {9 a^{3} \sin \left (d x +c \right )}{8 d}+\frac {a^{3} \cos \left (6 d x +6 c \right )}{192 d}-\frac {3 a^{3} \sin \left (5 d x +5 c \right )}{80 d}-\frac {3 a^{3} \cos \left (4 d x +4 c \right )}{32 d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{48 d}-\frac {27 a^{3} \cos \left (2 d x +2 c \right )}{64 d}\)	\(101\)
derivativedivides	\(\frac {a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {3 \left (\cos ^{4}\left (d x +c \right )\right ) a^{3}}{4}+\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(113\)
default	\(\frac {a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {3 \left (\cos ^{4}\left (d x +c \right )\right ) a^{3}}{4}+\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(113\)
norman	\(\frac {\frac {16 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {46 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {84 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {84 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {46 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(225\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)+3*a^3*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*
x+c)^2)*sin(d*x+c))-3/4*cos(d*x+c)^4*a^3+1/3*a^3*(2+cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]
time = 0.32, size = 82, normalized size = 1.82 \begin {gather*} -\frac {5 \, a^{3} \sin \left (d x + c\right )^{6} + 18 \, a^{3} \sin \left (d x + c\right )^{5} + 15 \, a^{3} \sin \left (d x + c\right )^{4} - 20 \, a^{3} \sin \left (d x + c\right )^{3} - 45 \, a^{3} \sin \left (d x + c\right )^{2} - 30 \, a^{3} \sin \left (d x + c\right )}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(5*a^3*sin(d*x + c)^6 + 18*a^3*sin(d*x + c)^5 + 15*a^3*sin(d*x + c)^4 - 20*a^3*sin(d*x + c)^3 - 45*a^3*s
in(d*x + c)^2 - 30*a^3*sin(d*x + c))/d

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Fricas [A]
time = 0.36, size = 72, normalized size = 1.60 \begin {gather*} \frac {5 \, a^{3} \cos \left (d x + c\right )^{6} - 30 \, a^{3} \cos \left (d x + c\right )^{4} - 2 \, {\left (9 \, a^{3} \cos \left (d x + c\right )^{4} - 8 \, a^{3} \cos \left (d x + c\right )^{2} - 16 \, a^{3}\right )} \sin \left (d x + c\right )}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(5*a^3*cos(d*x + c)^6 - 30*a^3*cos(d*x + c)^4 - 2*(9*a^3*cos(d*x + c)^4 - 8*a^3*cos(d*x + c)^2 - 16*a^3)*
sin(d*x + c))/d

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (37) = 74\).
time = 0.43, size = 146, normalized size = 3.24 \begin {gather*} \begin {cases} \frac {2 a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {2 a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {a^{3} \cos ^{6}{\left (c + d x \right )}}{12 d} - \frac {3 a^{3} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{3} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((2*a**3*sin(c + d*x)**5/(5*d) + a**3*sin(c + d*x)**3*cos(c + d*x)**2/d + 2*a**3*sin(c + d*x)**3/(3*d
) - a**3*sin(c + d*x)**2*cos(c + d*x)**4/(4*d) + a**3*sin(c + d*x)*cos(c + d*x)**2/d - a**3*cos(c + d*x)**6/(1
2*d) - 3*a**3*cos(c + d*x)**4/(4*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*cos(c)**3, True))

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Giac [A]
time = 3.72, size = 82, normalized size = 1.82 \begin {gather*} -\frac {5 \, a^{3} \sin \left (d x + c\right )^{6} + 18 \, a^{3} \sin \left (d x + c\right )^{5} + 15 \, a^{3} \sin \left (d x + c\right )^{4} - 20 \, a^{3} \sin \left (d x + c\right )^{3} - 45 \, a^{3} \sin \left (d x + c\right )^{2} - 30 \, a^{3} \sin \left (d x + c\right )}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(5*a^3*sin(d*x + c)^6 + 18*a^3*sin(d*x + c)^5 + 15*a^3*sin(d*x + c)^4 - 20*a^3*sin(d*x + c)^3 - 45*a^3*s
in(d*x + c)^2 - 30*a^3*sin(d*x + c))/d

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Mupad [B]
time = 4.47, size = 80, normalized size = 1.78 \begin {gather*} \frac {-\frac {a^3\,{\sin \left (c+d\,x\right )}^6}{6}-\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {a^3\,{\sin \left (c+d\,x\right )}^4}{2}+\frac {2\,a^3\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^2}{2}+a^3\,\sin \left (c+d\,x\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*sin(c + d*x) + (3*a^3*sin(c + d*x)^2)/2 + (2*a^3*sin(c + d*x)^3)/3 - (a^3*sin(c + d*x)^4)/2 - (3*a^3*sin(
c + d*x)^5)/5 - (a^3*sin(c + d*x)^6)/6)/d

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