3.363 \(\int (e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=34 \[ \frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m}}{d e m} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2671} \[ \frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m}}{d e m} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(-1 - m)*(a + a*Sin[c + d*x])^m,x]

[Out]

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

Rule 2671

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^m \, dx &=\frac {(e \cos (c+d x))^{-m} (a+a \sin (c+d x))^m}{d e m}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 34, normalized size = 1.00 \[ \frac {(a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-m}}{d e m} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(-1 - m)*(a + a*Sin[c + d*x])^m,x]

[Out]

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

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fricas [A]  time = 0.46, size = 39, normalized size = 1.15 \[ \frac {\left (e \cos \left (d x + c\right )\right )^{-m - 1} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )}{d m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{-m - 1} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(-m - 1)*(a*sin(d*x + c) + a)^m, x)

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{-1-m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^m,x)

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maxima [A]  time = 0.46, size = 65, normalized size = 1.91 \[ \frac {a^{m} e^{-m - 1} e^{\left (m \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) - m \log \left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )\right )}}{d m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

a^m*e^(-m - 1)*e^(m*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1) - m*log(-sin(d*x + c)/(cos(d*x + c) + 1) + 1))/(d
*m)

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mupad [B]  time = 0.29, size = 34, normalized size = 1.00 \[ \frac {{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m}{d\,e\,m\,{\left (e\,\cos \left (c+d\,x\right )\right )}^m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(m + 1),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(-1-m)*(a+a*sin(d*x+c))**m,x)

[Out]

Timed out

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