3.4.46 \(\int \cos (c+d x) (a+a \sin (c+d x))^m \, dx\) [346]

Optimal. Leaf size=26 \[ \frac {(a+a \sin (c+d x))^{1+m}}{a d (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 26, 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)^{m+1}}{a d (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a + a*Sin[c + d*x])^(1 + m)/(a*d*(1 + m))

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))^m \, dx &=\frac {\text {Subst}\left (\int (a+x)^m \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {(a+a \sin (c+d x))^{1+m}}{a d (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 26, normalized size = 1.00 \begin {gather*} \frac {(a (1+\sin (c+d x)))^{1+m}}{a d (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(1 + Sin[c + d*x]))^(1 + m)/(a*d*(1 + m))

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Maple [A]
time = 3.10, size = 27, normalized size = 1.04

method	result	size
derivativedivides	\(\frac {\left (a +a \sin \left (d x +c \right )\right )^{1+m}}{a d \left (1+m \right )}\)	\(27\)
default	\(\frac {\left (a +a \sin \left (d x +c \right )\right )^{1+m}}{a d \left (1+m \right )}\)	\(27\)
norman	\(\frac {\frac {{\mathrm e}^{m \ln \left (a +\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\right )}}{d \left (1+m \right )}+\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) {\mathrm e}^{m \ln \left (a +\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\right )}}{d \left (1+m \right )}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) {\mathrm e}^{m \ln \left (a +\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\right )}}{d \left (1+m \right )}}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\)	\(165\)
risch	\(\text {Expression too large to display}\)	\(1742\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 26, normalized size = 1.00 \begin {gather*} \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m + 1}}{a d {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 28, normalized size = 1.08 \begin {gather*} \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m} {\left (\sin \left (d x + c\right ) + 1\right )}}{d m + 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))^m,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (19) = 38\).
time = 0.46, size = 80, normalized size = 3.08 \begin {gather*} \begin {cases} \frac {x \cos {\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {for}\: d = 0 \wedge m = -1 \\x \left (a \sin {\left (c \right )} + a\right )^{m} \cos {\left (c \right )} & \text {for}\: d = 0 \\\frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a d} & \text {for}\: m = -1 \\\frac {\left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin {\left (c + d x \right )}}{d m + d} + \frac {\left (a \sin {\left (c + d x \right )} + a\right )^{m}}{d m + d} & \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))**m,x)

[Out]

Piecewise((x*cos(c)/(a*sin(c) + a), Eq(d, 0) & Eq(m, -1)), (x*(a*sin(c) + a)**m*cos(c), Eq(d, 0)), (log(sin(c
+ d*x) + 1)/(a*d), Eq(m, -1)), ((a*sin(c + d*x) + a)**m*sin(c + d*x)/(d*m + d) + (a*sin(c + d*x) + a)**m/(d*m
+ d), True))

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Giac [A]
time = 5.62, size = 26, normalized size = 1.00 \begin {gather*} \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m + 1}}{a d {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.22, size = 29, normalized size = 1.12 \begin {gather*} \frac {{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (\sin \left (c+d\,x\right )+1\right )}{d\,\left (m+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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