3.7.33 \(\int \cos (c+d x) (a+b \sin (c+d x))^m \, dx\) [633]

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

[Out]

(a+b*sin(d*x+c))^(1+m)/b/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 = {2747, 32} \begin {gather*} \frac {(a+b \sin (c+d x))^{m+1}}{b d (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*Sin[c + d*x])^(1 + m)/(b*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 2747

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*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 26, normalized size = 1.00 \begin {gather*} \frac {(a+b \sin (c+d x))^{1+m}}{b d (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 27, normalized size = 1.04

method result size
derivativedivides \(\frac {\left (a +b \sin \left (d x +c \right )\right )^{1+m}}{b d \left (1+m \right )}\) \(27\)
default \(\frac {\left (a +b \sin \left (d x +c \right )\right )^{1+m}}{b d \left (1+m \right )}\) \(27\)
norman \(\frac {\frac {a \,{\mathrm e}^{m \ln \left (a +\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\right )}}{b d \left (1+m \right )}+\frac {a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) {\mathrm e}^{m \ln \left (a +\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\right )}}{b d \left (1+m \right )}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) {\mathrm e}^{m \ln \left (a +\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\right )}}{d \left (1+m \right )}}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\) \(173\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 33, normalized size = 1.27 \begin {gather*} \frac {{\left (b \sin \left (d x + c\right ) + a\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{b d m + b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (19) = 38\).
time = 0.53, size = 99, normalized size = 3.81 \begin {gather*} \begin {cases} \frac {x \cos {\left (c \right )}}{a} & \text {for}\: b = 0 \wedge d = 0 \wedge m = -1 \\\frac {a^{m} \sin {\left (c + d x \right )}}{d} & \text {for}\: b = 0 \\x \left (a + b \sin {\left (c \right )}\right )^{m} \cos {\left (c \right )} & \text {for}\: d = 0 \\\frac {\log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{b d} & \text {for}\: m = -1 \\\frac {a \left (a + b \sin {\left (c + d x \right )}\right )^{m}}{b d m + b d} + \frac {b \left (a + b \sin {\left (c + d x \right )}\right )^{m} \sin {\left (c + d x \right )}}{b d m + b d} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.94, size = 26, normalized size = 1.00 \begin {gather*} \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{m + 1}}{b d {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 6.32, size = 26, normalized size = 1.00 \begin {gather*} \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{m+1}}{b\,d\,\left (m+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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