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

Optimal. Leaf size=51 \[ -\frac{f \sin (c+d x)}{a d^2}+\frac{(e+f x) \cos (c+d x)}{a d}+\frac{e x}{a}+\frac{f x^2}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.0641802, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {4523, 3296, 2637} \[ -\frac{f \sin (c+d x)}{a d^2}+\frac{(e+f x) \cos (c+d x)}{a d}+\frac{e x}{a}+\frac{f x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.505234, size = 53, normalized size = 1.04 \[ -\frac{(c+d x) (c f-2 d e-d f x)-2 d (e+f x) \cos (c+d x)+2 f \sin (c+d x)}{2 a d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.058, size = 78, normalized size = 1.5 \begin{align*} -{\frac{1}{a{d}^{2}} \left ( f \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) +cf\cos \left ( dx+c \right ) -de\cos \left ( dx+c \right ) -{\frac{f \left ( dx+c \right ) ^{2}}{2}}+cf \left ( dx+c \right ) -de \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d^2/a*(f*(sin(d*x+c)-(d*x+c)*cos(d*x+c))+c*f*cos(d*x+c)-d*e*cos(d*x+c)-1/2*f*(d*x+c)^2+c*f*(d*x+c)-d*e*(d*x
+c))

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Maxima [B]  time = 1.52251, size = 204, normalized size = 4. \begin{align*} -\frac{4 \, c f{\left (\frac{1}{a d + \frac{a d \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac{\arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a d}\right )} - 4 \, e{\left (\frac{\arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{1}{a + \frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}\right )} - \frac{{\left ({\left (d x + c\right )}^{2} + 2 \,{\left (d x + c\right )} \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right )\right )} f}{a d}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*c*f*(1/(a*d + a*d*sin(d*x + c)^2/(cos(d*x + c) + 1)^2) + arctan(sin(d*x + c)/(cos(d*x + c) + 1))/(a*d)
) - 4*e*(arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 1/(a + a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)) - ((d*x +
c)^2 + 2*(d*x + c)*cos(d*x + c) - 2*sin(d*x + c))*f/(a*d))/d

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Fricas [A]  time = 1.60205, size = 117, normalized size = 2.29 \begin{align*} \frac{d^{2} f x^{2} + 2 \, d^{2} e x + 2 \,{\left (d f x + d e\right )} \cos \left (d x + c\right ) - 2 \, f \sin \left (d x + c\right )}{2 \, a d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d^2*f*x^2 + 2*d^2*e*x + 2*(d*f*x + d*e)*cos(d*x + c) - 2*f*sin(d*x + c))/(a*d^2)

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Sympy [A]  time = 5.37991, size = 439, normalized size = 8.61 \begin{align*} \begin{cases} \frac{2 d^{2} e x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{2 d^{2} e x}{2 a d^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{d^{2} f x^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{d^{2} f x^{2}}{2 a d^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} - \frac{2 d e \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{2 d e}{2 a d^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} - \frac{2 d f x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{2 d f x}{2 a d^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{2 f \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} - \frac{4 f \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{2 f}{2 a d^{2} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} & \text{for}\: d \neq 0 \\\frac{\left (e x + \frac{f x^{2}}{2}\right ) \cos ^{2}{\left (c \right )}}{a \sin{\left (c \right )} + a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*d**2*e*x*tan(c/2 + d*x/2)**2/(2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2) + 2*d**2*e*x/(2*a*d**2*tan
(c/2 + d*x/2)**2 + 2*a*d**2) + d**2*f*x**2*tan(c/2 + d*x/2)**2/(2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2) + d**
2*f*x**2/(2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2) - 2*d*e*tan(c/2 + d*x/2)**2/(2*a*d**2*tan(c/2 + d*x/2)**2 +
2*a*d**2) + 2*d*e/(2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2) - 2*d*f*x*tan(c/2 + d*x/2)**2/(2*a*d**2*tan(c/2 +
d*x/2)**2 + 2*a*d**2) + 2*d*f*x/(2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2) + 2*f*tan(c/2 + d*x/2)**2/(2*a*d**2
*tan(c/2 + d*x/2)**2 + 2*a*d**2) - 4*f*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2) + 2*f/(2*a*d
**2*tan(c/2 + d*x/2)**2 + 2*a*d**2), Ne(d, 0)), ((e*x + f*x**2/2)*cos(c)**2/(a*sin(c) + a), True))

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Giac [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)*cos(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out