3.1.32 \(\int \cos (c+d x) (a+a \sin (c+d x))^3 \, dx\) [32]

Optimal. Leaf size=22 \[ \frac {(a+a \sin (c+d x))^4}{4 a d} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2746, 32} \begin {gather*} \frac {(a \sin (c+d x)+a)^4}{4 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a + a*Sin[c + d*x])^4/(4*a*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(47\) vs. \(2(22)=44\).
time = 0.27, size = 47, normalized size = 2.14 \begin {gather*} \frac {a^3 (-28 \cos (2 (c+d x))+\cos (4 (c+d x))+56 \sin (c+d x)-8 \sin (3 (c+d x)))}{32 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(-28*Cos[2*(c + d*x)] + Cos[4*(c + d*x)] + 56*Sin[c + d*x] - 8*Sin[3*(c + d*x)]))/(32*d)

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Maple [A]
time = 0.11, size = 21, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {\left (a +a \sin \left (d x +c \right )\right )^{4}}{4 d a}\)	\(21\)
default	\(\frac {\left (a +a \sin \left (d x +c \right )\right )^{4}}{4 d a}\)	\(21\)
risch	\(\frac {7 a^{3} \sin \left (d x +c \right )}{4 d}+\frac {a^{3} \cos \left (4 d x +4 c \right )}{32 d}-\frac {a^{3} \sin \left (3 d x +3 c \right )}{4 d}-\frac {7 a^{3} \cos \left (2 d x +2 c \right )}{8 d}\)	\(67\)
norman	\(\frac {\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {14 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {14 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(149\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 20, normalized size = 0.91 \begin {gather*} \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{4 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*sin(d*x + c) + a)^4/(a*d)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (20) = 40\).
time = 0.36, size = 57, normalized size = 2.59 \begin {gather*} \frac {a^{3} \cos \left (d x + c\right )^{4} - 8 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3}\right )} \sin \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^3*cos(d*x + c)^4 - 8*a^3*cos(d*x + c)^2 - 4*(a^3*cos(d*x + c)^2 - 2*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. 70 vs. \(2 (15) = 30\).
time = 0.17, size = 70, normalized size = 3.18 \begin {gather*} \begin {cases} \frac {a^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 a^{3} \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac {a^{3} \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{3} \cos {\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)*(a+a*sin(d*x+c))**3,x)

[Out]

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

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Giac [A]
time = 5.67, size = 20, normalized size = 0.91 \begin {gather*} \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{4 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*sin(d*x + c) + a)^4/(a*d)

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Mupad [B]
time = 0.06, size = 53, normalized size = 2.41 \begin {gather*} \frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^4}{4}+a^3\,{\sin \left (c+d\,x\right )}^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)*(a + a*sin(c + d*x))^3,x)

[Out]

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

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