3.289 \(\int \frac{(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=154 \[ \frac{e^{i \left (c-\frac{d e}{f}\right )} (e+f x)^m \left (-\frac{i d (e+f x)}{f}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i d (e+f x)}{f}\right )}{2 a d}+\frac{e^{-i \left (c-\frac{d e}{f}\right )} (e+f x)^m \left (\frac{i d (e+f x)}{f}\right )^{-m} \text{Gamma}\left (m+1,\frac{i d (e+f x)}{f}\right )}{2 a d}+\frac{(e+f x)^{m+1}}{a f (m+1)} \]

[Out]

(e + f*x)^(1 + m)/(a*f*(1 + m)) + (E^(I*(c - (d*e)/f))*(e + f*x)^m*Gamma[1 + m, ((-I)*d*(e + f*x))/f])/(2*a*d*
(((-I)*d*(e + f*x))/f)^m) + ((e + f*x)^m*Gamma[1 + m, (I*d*(e + f*x))/f])/(2*a*d*E^(I*(c - (d*e)/f))*((I*d*(e
+ f*x))/f)^m)

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Rubi [A]  time = 0.176591, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4523, 32, 3308, 2181} \[ \frac{e^{i \left (c-\frac{d e}{f}\right )} (e+f x)^m \left (-\frac{i d (e+f x)}{f}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i d (e+f x)}{f}\right )}{2 a d}+\frac{e^{-i \left (c-\frac{d e}{f}\right )} (e+f x)^m \left (\frac{i d (e+f x)}{f}\right )^{-m} \text{Gamma}\left (m+1,\frac{i d (e+f x)}{f}\right )}{2 a d}+\frac{(e+f x)^{m+1}}{a f (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^m*Cos[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

(e + f*x)^(1 + m)/(a*f*(1 + m)) + (E^(I*(c - (d*e)/f))*(e + f*x)^m*Gamma[1 + m, ((-I)*d*(e + f*x))/f])/(2*a*d*
(((-I)*d*(e + f*x))/f)^m) + ((e + f*x)^m*Gamma[1 + m, (I*d*(e + f*x))/f])/(2*a*d*E^(I*(c - (d*e)/f))*((I*d*(e
+ f*x))/f)^m)

Rule 4523

Int[(Cos[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol
] :> Dist[1/a, Int[(e + f*x)^m*Cos[c + d*x]^(n - 2), x], x] - Dist[1/b, Int[(e + f*x)^m*Cos[c + d*x]^(n - 2)*S
in[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 1] && EqQ[a^2 - b^2, 0]

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 3308

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

\begin{align*} \int \frac{(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int (e+f x)^m \, dx}{a}-\frac{\int (e+f x)^m \sin (c+d x) \, dx}{a}\\ &=\frac{(e+f x)^{1+m}}{a f (1+m)}-\frac{i \int e^{-i (c+d x)} (e+f x)^m \, dx}{2 a}+\frac{i \int e^{i (c+d x)} (e+f x)^m \, dx}{2 a}\\ &=\frac{(e+f x)^{1+m}}{a f (1+m)}+\frac{e^{i \left (c-\frac{d e}{f}\right )} (e+f x)^m \left (-\frac{i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,-\frac{i d (e+f x)}{f}\right )}{2 a d}+\frac{e^{-i \left (c-\frac{d e}{f}\right )} (e+f x)^m \left (\frac{i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,\frac{i d (e+f x)}{f}\right )}{2 a d}\\ \end{align*}

Mathematica [A]  time = 0.983611, size = 220, normalized size = 1.43 \[ \frac{e^{i \left (c-\frac{d e}{f}\right )} (e+f x)^m \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (\frac{d^2 (e+f x)^2}{f^2}\right )^{-m} \left (f (m+1) e^{-2 i \left (c-\frac{d e}{f}\right )} \left (-\frac{i d (e+f x)}{f}\right )^m \text{Gamma}\left (m+1,\frac{i d (e+f x)}{f}\right )+f (m+1) \left (\frac{i d (e+f x)}{f}\right )^m \text{Gamma}\left (m+1,-\frac{i d (e+f x)}{f}\right )+2 d (e+f x) e^{-i \left (c-\frac{d e}{f}\right )} \left (\frac{d^2 (e+f x)^2}{f^2}\right )^m\right )}{2 a d f (m+1) (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^m*Cos[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

(E^(I*(c - (d*e)/f))*(e + f*x)^m*((2*d*(e + f*x)*((d^2*(e + f*x)^2)/f^2)^m)/E^(I*(c - (d*e)/f)) + f*(1 + m)*((
I*d*(e + f*x))/f)^m*Gamma[1 + m, ((-I)*d*(e + f*x))/f] + (f*(1 + m)*(((-I)*d*(e + f*x))/f)^m*Gamma[1 + m, (I*d
*(e + f*x))/f])/E^((2*I)*(c - (d*e)/f)))*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(2*a*d*f*(1 + m)*((d^2*(e +
f*x)^2)/f^2)^m*(1 + Sin[c + d*x]))

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Maple [F]  time = 0.116, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{m} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a+a\sin \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^m*cos(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

int((f*x+e)^m*cos(d*x+c)^2/(a+a*sin(d*x+c)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{m} \cos \left (d x + c\right )^{2}}{a \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^m*cos(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

integrate((f*x + e)^m*cos(d*x + c)^2/(a*sin(d*x + c) + a), x)

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Fricas [A]  time = 1.8099, size = 306, normalized size = 1.99 \begin{align*} \frac{{\left (f m + f\right )} e^{\left (-\frac{f m \log \left (\frac{i \, d}{f}\right ) - i \, d e + i \, c f}{f}\right )} \Gamma \left (m + 1, \frac{i \, d f x + i \, d e}{f}\right ) +{\left (f m + f\right )} e^{\left (-\frac{f m \log \left (-\frac{i \, d}{f}\right ) + i \, d e - i \, c f}{f}\right )} \Gamma \left (m + 1, \frac{-i \, d f x - i \, d e}{f}\right ) + 2 \,{\left (d f x + d e\right )}{\left (f x + e\right )}^{m}}{2 \,{\left (a d f m + a d f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^m*cos(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2*((f*m + f)*e^(-(f*m*log(I*d/f) - I*d*e + I*c*f)/f)*gamma(m + 1, (I*d*f*x + I*d*e)/f) + (f*m + f)*e^(-(f*m*
log(-I*d/f) + I*d*e - I*c*f)/f)*gamma(m + 1, (-I*d*f*x - I*d*e)/f) + 2*(d*f*x + d*e)*(f*x + e)^m)/(a*d*f*m + a
*d*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**m*cos(d*x+c)**2/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{m} \cos \left (d x + c\right )^{2}}{a \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^m*cos(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^m*cos(d*x + c)^2/(a*sin(d*x + c) + a), x)