3.1.5 \(\int \cos ^3(c+d x) (a+a \sin (c+d x)) \, dx\) [5]

Optimal. Leaf size=45 \[ \frac {2 (a+a \sin (c+d x))^3}{3 a^2 d}-\frac {(a+a \sin (c+d x))^4}{4 a^3 d} \]

[Out]

2/3*(a+a*sin(d*x+c))^3/a^2/d-1/4*(a+a*sin(d*x+c))^4/a^3/d

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Rubi [A]
time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2746, 45} \begin {gather*} \frac {2 (a \sin (c+d x)+a)^3}{3 a^2 d}-\frac {(a \sin (c+d x)+a)^4}{4 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.01, size = 44, normalized size = 0.98 \begin {gather*} -\frac {a \cos ^4(c+d x)}{4 d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^3(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(a*Cos[c + d*x]^4)/d + (a*Sin[c + d*x])/d - (a*Sin[c + d*x]^3)/(3*d)

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Maple [A]
time = 0.10, size = 36, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {-\frac {a \left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(36\)
default	\(\frac {-\frac {a \left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(36\)
risch	\(\frac {3 a \sin \left (d x +c \right )}{4 d}-\frac {a \cos \left (4 d x +4 c \right )}{32 d}+\frac {a \sin \left (3 d x +3 c \right )}{12 d}-\frac {a \cos \left (2 d x +2 c \right )}{8 d}\)	\(59\)
norman	\(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {10 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {10 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 48, normalized size = 1.07 \begin {gather*} -\frac {3 \, a \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 6 \, a \sin \left (d x + c\right )^{2} - 12 \, a \sin \left (d x + c\right )}{12 \, 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 39, normalized size = 0.87 \begin {gather*} -\frac {3 \, a \cos \left (d x + c\right )^{4} - 4 \, {\left (a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right )}{12 \, 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.20, size = 60, normalized size = 1.33 \begin {gather*} \begin {cases} \frac {2 a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {a \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \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)),x)

[Out]

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

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Giac [A]
time = 4.18, size = 48, normalized size = 1.07 \begin {gather*} -\frac {3 \, a \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 6 \, a \sin \left (d x + c\right )^{2} - 12 \, a \sin \left (d x + c\right )}{12 \, 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)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.06, size = 46, normalized size = 1.02 \begin {gather*} \frac {-\frac {a\,{\sin \left (c+d\,x\right )}^4}{4}-\frac {a\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {a\,{\sin \left (c+d\,x\right )}^2}{2}+a\,\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)),x)

[Out]

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

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