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

Optimal. Leaf size=280 \[ -\frac{2 f^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^3 \sqrt{a^2-b^2}}+\frac{2 f^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^3 \sqrt{a^2-b^2}}-\frac{2 i f (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^2 \sqrt{a^2-b^2}}+\frac{2 i f (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^2 \sqrt{a^2-b^2}}-\frac{(e+f x)^2}{b d (a+b \sin (c+d x))} \]

[Out]

((-2*I)*f*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^2) + ((2*I)*f*(
e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^2) - (2*f^2*PolyLog[2, (I*
b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^3) + (2*f^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(
a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^3) - (e + f*x)^2/(b*d*(a + b*Sin[c + d*x]))

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Rubi [A]  time = 0.527671, antiderivative size = 280, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {4422, 3323, 2264, 2190, 2279, 2391} \[ -\frac{2 f^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^3 \sqrt{a^2-b^2}}+\frac{2 f^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^3 \sqrt{a^2-b^2}}-\frac{2 i f (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b d^2 \sqrt{a^2-b^2}}+\frac{2 i f (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b d^2 \sqrt{a^2-b^2}}-\frac{(e+f x)^2}{b d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*f*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^2) + ((2*I)*f*(
e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^2) - (2*f^2*PolyLog[2, (I*
b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^3) + (2*f^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(
a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^3) - (e + f*x)^2/(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 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2190

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{(e+f x)^2 \cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=-\frac{(e+f x)^2}{b d (a+b \sin (c+d x))}+\frac{(2 f) \int \frac{e+f x}{a+b \sin (c+d x)} \, dx}{b d}\\ &=-\frac{(e+f x)^2}{b d (a+b \sin (c+d x))}+\frac{(4 f) \int \frac{e^{i (c+d x)} (e+f x)}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b d}\\ &=-\frac{(e+f x)^2}{b d (a+b \sin (c+d x))}-\frac{(4 i f) \int \frac{e^{i (c+d x)} (e+f x)}{2 a-2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\sqrt{a^2-b^2} d}+\frac{(4 i f) \int \frac{e^{i (c+d x)} (e+f x)}{2 a+2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\sqrt{a^2-b^2} d}\\ &=-\frac{2 i f (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}+\frac{2 i f (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}-\frac{(e+f x)^2}{b d (a+b \sin (c+d x))}+\frac{\left (2 i f^2\right ) \int \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{b \sqrt{a^2-b^2} d^2}-\frac{\left (2 i f^2\right ) \int \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{b \sqrt{a^2-b^2} d^2}\\ &=-\frac{2 i f (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}+\frac{2 i f (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}-\frac{(e+f x)^2}{b d (a+b \sin (c+d x))}+\frac{\left (2 f^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{2 i b x}{2 a-2 \sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b \sqrt{a^2-b^2} d^3}-\frac{\left (2 f^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{2 i b x}{2 a+2 \sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b \sqrt{a^2-b^2} d^3}\\ &=-\frac{2 i f (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}+\frac{2 i f (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^2}-\frac{2 f^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^3}+\frac{2 f^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2} d^3}-\frac{(e+f x)^2}{b d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.08874, size = 311, normalized size = 1.11 \[ -\frac{(e+f x)^2}{b d (a+b \sin (c+d x))}+\frac{2 i f \left (-f \sqrt{a^2-b^2} \text{PolyLog}\left (2,\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )+f \sqrt{a^2-b^2} \text{PolyLog}\left (2,-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}+i a}\right )-i d \left (2 e \sqrt{b^2-a^2} \tan ^{-1}\left (\frac{i a+b e^{i (c+d x)}}{\sqrt{a^2-b^2}}\right )+f x \sqrt{a^2-b^2} \left (\log \left (1-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )-\log \left (1+\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}+i a}\right )\right )\right )\right )}{b d^3 \sqrt{-\left (a^2-b^2\right )^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*I)*f*((-I)*d*(2*Sqrt[-a^2 + b^2]*e*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] + Sqrt[a^2 - b^2]*f*x
*(Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^
2])])) - Sqrt[a^2 - b^2]*f*PolyLog[2, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + Sqrt[a^2 - b^2]*f*Pol
yLog[2, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))]))/(b*Sqrt[-(a^2 - b^2)^2]*d^3) - (e + f*x)^2/(b*d*(a
+ b*Sin[c + d*x]))

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Maple [B]  time = 0.892, size = 606, normalized size = 2.2 \begin{align*}{\frac{-2\,i \left ({f}^{2}{x}^{2}+2\,fex+{e}^{2} \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{4\,ife}{b{d}^{2}}\arctan \left ({\frac{2\,ib{{\rm e}^{i \left ( dx+c \right ) }}-2\,a}{2}{\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}+2\,{\frac{{f}^{2}x}{b{d}^{2}\sqrt{-{a}^{2}+{b}^{2}}}\ln \left ({\frac{ia+b{{\rm e}^{i \left ( dx+c \right ) }}-\sqrt{-{a}^{2}+{b}^{2}}}{ia-\sqrt{-{a}^{2}+{b}^{2}}}} \right ) }+2\,{\frac{c{f}^{2}}{b{d}^{3}\sqrt{-{a}^{2}+{b}^{2}}}\ln \left ({\frac{ia+b{{\rm e}^{i \left ( dx+c \right ) }}-\sqrt{-{a}^{2}+{b}^{2}}}{ia-\sqrt{-{a}^{2}+{b}^{2}}}} \right ) }-2\,{\frac{{f}^{2}x}{b{d}^{2}\sqrt{-{a}^{2}+{b}^{2}}}\ln \left ({\frac{ia+b{{\rm e}^{i \left ( dx+c \right ) }}+\sqrt{-{a}^{2}+{b}^{2}}}{ia+\sqrt{-{a}^{2}+{b}^{2}}}} \right ) }-2\,{\frac{c{f}^{2}}{b{d}^{3}\sqrt{-{a}^{2}+{b}^{2}}}\ln \left ({\frac{ia+b{{\rm e}^{i \left ( dx+c \right ) }}+\sqrt{-{a}^{2}+{b}^{2}}}{ia+\sqrt{-{a}^{2}+{b}^{2}}}} \right ) }-{\frac{2\,i{f}^{2}}{b{d}^{3}}{\it dilog} \left ({ \left ( ia+b{{\rm e}^{i \left ( dx+c \right ) }}-\sqrt{-{a}^{2}+{b}^{2}} \right ) \left ( ia-\sqrt{-{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}+{\frac{2\,i{f}^{2}}{b{d}^{3}}{\it dilog} \left ({ \left ( ia+b{{\rm e}^{i \left ( dx+c \right ) }}+\sqrt{-{a}^{2}+{b}^{2}} \right ) \left ( ia+\sqrt{-{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}-{\frac{4\,i{f}^{2}c}{b{d}^{3}}\arctan \left ({\frac{2\,ib{{\rm e}^{i \left ( dx+c \right ) }}-2\,a}{2}{\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)^2*cos(d*x+c)/(a+b*sin(d*x+c))^2,x)

[Out]

-2*I*(f^2*x^2+2*e*f*x+e^2)*exp(I*(d*x+c))/b/d/(b*exp(2*I*(d*x+c))-b+2*I*a*exp(I*(d*x+c)))+4*I*f/b/d^2*e/(-a^2+
b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))+2*f^2/b/d^2/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(
I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x+2*f^2/b/d^3/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(I*(d*x+c))-(
-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*c-2*f^2/b/d^2/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1
/2))/(I*a+(-a^2+b^2)^(1/2)))*x-2*f^2/b/d^3/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-
a^2+b^2)^(1/2)))*c-2*I*f^2/b/d^3/(-a^2+b^2)^(1/2)*dilog((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2
)^(1/2)))+2*I*f^2/b/d^3/(-a^2+b^2)^(1/2)*dilog((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))
-4*I*f^2/b/d^3*c/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.8509, size = 3263, normalized size = 11.65 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2 - b^2)*d^2*f^2*x^2 + 2*(a^2 - b^2)*d^2*e*f*x + (a^2 - b^2)*d^2*e^2 + (-I*b^2*f^2*sin(d*x + c) - I*a*b*f
^2)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x
+ c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (I*b^2*f^2*sin(d*x + c) + I*a*b*f^2)*sqrt(-(a^2 - b^2)/b^2)*dilo
g(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) +
2*b)/b + 1) + (I*b^2*f^2*sin(d*x + c) + I*a*b*f^2)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2
*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (-I*b^2*f^2*sin
(d*x + c) - I*a*b*f^2)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*
x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - (a*b*d*e*f - a*b*c*f^2 + (b^2*d*e*f - b^2*c*
f^2)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^
2) + 2*I*a) - (a*b*d*e*f - a*b*c*f^2 + (b^2*d*e*f - b^2*c*f^2)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*log(2*b*co
s(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + (a*b*d*e*f - a*b*c*f^2 + (b^2*d*e*f -
b^2*c*f^2)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 -
b^2)/b^2) + 2*I*a) + (a*b*d*e*f - a*b*c*f^2 + (b^2*d*e*f - b^2*c*f^2)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*log
(-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) - (a*b*d*f^2*x + a*b*c*f^2 + (b^
2*d*f^2*x + b^2*c*f^2)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2
*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + (a*b*d*f^2*x + a*b*c*f^2 + (b^2*d*f^2*
x + b^2*c*f^2)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(
d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - (a*b*d*f^2*x + a*b*c*f^2 + (b^2*d*f^2*x + b^2*
c*f^2)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c
) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + (a*b*d*f^2*x + a*b*c*f^2 + (b^2*d*f^2*x + b^2*c*f^2)*
sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b
*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b))/((a^2*b^2 - b^4)*d^3*sin(d*x + c) + (a^3*b - a*b^3)*d^3)

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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)**2*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 )}^{2} \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)^2*cos(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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