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

Optimal. Leaf size=77 \[ \frac{2 f \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b d^2 \sqrt{a^2-b^2}}-\frac{e+f x}{b d (a+b \sin (c+d x))} \]

[Out]

(2*f*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b*Sqrt[a^2 - b^2]*d^2) - (e + f*x)/(b*d*(a + b*Sin[c +
d*x]))

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Rubi [A]  time = 0.0717487, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4422, 2660, 618, 204} \[ \frac{2 f \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b d^2 \sqrt{a^2-b^2}}-\frac{e+f x}{b d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*f*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b*Sqrt[a^2 - b^2]*d^2) - (e + f*x)/(b*d*(a + b*Sin[c +
d*x]))

Rule 4422

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(e+f x) \cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=-\frac{e+f x}{b d (a+b \sin (c+d x))}+\frac{f \int \frac{1}{a+b \sin (c+d x)} \, dx}{b d}\\ &=-\frac{e+f x}{b d (a+b \sin (c+d x))}+\frac{(2 f) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b d^2}\\ &=-\frac{e+f x}{b d (a+b \sin (c+d x))}-\frac{(4 f) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{b d^2}\\ &=\frac{2 f \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}-\frac{e+f x}{b d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.445768, size = 73, normalized size = 0.95 \[ \frac{\frac{2 f \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{d (e+f x)}{a+b \sin (c+d x)}}{b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*f*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - (d*(e + f*x))/(a + b*Sin[c + d*x]))/
(b*d^2)

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Maple [C]  time = 0.889, size = 194, normalized size = 2.5 \begin{align*}{\frac{-2\,i \left ( fx+e \right ){{\rm e}^{i \left ( dx+c \right ) }}}{bd \left ( b{{\rm e}^{2\,i \left ( dx+c \right ) }}-b+2\,ia{{\rm e}^{i \left ( dx+c \right ) }} \right ) }}-{\frac{f}{b{d}^{2}}\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+{\frac{1}{b} \left ( ia\sqrt{-{a}^{2}+{b}^{2}}-{a}^{2}+{b}^{2} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}+{\frac{f}{b{d}^{2}}\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+{\frac{1}{b} \left ( ia\sqrt{-{a}^{2}+{b}^{2}}+{a}^{2}-{b}^{2} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*(f*x+e)*exp(I*(d*x+c))/b/d/(b*exp(2*I*(d*x+c))-b+2*I*a*exp(I*(d*x+c)))-1/(-a^2+b^2)^(1/2)*f/d^2/b*ln(exp(
I*(d*x+c))+(I*a*(-a^2+b^2)^(1/2)-a^2+b^2)/(-a^2+b^2)^(1/2)/b)+1/(-a^2+b^2)^(1/2)*f/d^2/b*ln(exp(I*(d*x+c))+(I*
a*(-a^2+b^2)^(1/2)+a^2-b^2)/(-a^2+b^2)^(1/2)/b)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.07617, size = 736, normalized size = 9.56 \begin{align*} \left [-\frac{2 \,{\left (a^{2} - b^{2}\right )} d f x + 2 \,{\left (a^{2} - b^{2}\right )} d e +{\left (b f \sin \left (d x + c\right ) + a f\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right )}{2 \,{\left ({\left (a^{2} b^{2} - b^{4}\right )} d^{2} \sin \left (d x + c\right ) +{\left (a^{3} b - a b^{3}\right )} d^{2}\right )}}, -\frac{{\left (a^{2} - b^{2}\right )} d f x +{\left (a^{2} - b^{2}\right )} d e +{\left (b f \sin \left (d x + c\right ) + a f\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right )}{{\left (a^{2} b^{2} - b^{4}\right )} d^{2} \sin \left (d x + c\right ) +{\left (a^{3} b - a b^{3}\right )} d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(a^2 - b^2)*d*f*x + 2*(a^2 - b^2)*d*e + (b*f*sin(d*x + c) + a*f)*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*
cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 +
b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)))/((a^2*b^2 - b^4)*d^2*sin(d*x + c) + (a^3*b - a*
b^3)*d^2), -((a^2 - b^2)*d*f*x + (a^2 - b^2)*d*e + (b*f*sin(d*x + c) + a*f)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x
+ c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))))/((a^2*b^2 - b^4)*d^2*sin(d*x + c) + (a^3*b - a*b^3)*d^2)]

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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)*cos(d*x+c)/(a+b*sin(d*x+c))**2,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 )} \cos \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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