3.323 \(\int \frac {(e+f x)^2 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=357 \[ -\frac {a f^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^3 \left (a^2-b^2\right )^{3/2}}+\frac {a f^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^3 \left (a^2-b^2\right )^{3/2}}-\frac {f^2 \log (a+b \sin (c+d x))}{b d^3 \left (a^2-b^2\right )}-\frac {i a f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {i a f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {f (e+f x) \cos (c+d x)}{d^2 \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {(e+f x)^2}{2 b d (a+b \sin (c+d x))^2} \]

[Out]

-f^2*ln(a+b*sin(d*x+c))/b/(a^2-b^2)/d^3-I*a*f*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)
^(3/2)/d^2+I*a*f*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2-a*f^2*polylog(2,I*
b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^3+a*f^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1
/2)))/b/(a^2-b^2)^(3/2)/d^3-1/2*(f*x+e)^2/b/d/(a+b*sin(d*x+c))^2+f*(f*x+e)*cos(d*x+c)/(a^2-b^2)/d^2/(a+b*sin(d
*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.61, antiderivative size = 357, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {4422, 3324, 3323, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac {a f^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^3 \left (a^2-b^2\right )^{3/2}}+\frac {a f^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^3 \left (a^2-b^2\right )^{3/2}}-\frac {i a f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {i a f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {f (e+f x) \cos (c+d x)}{d^2 \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {f^2 \log (a+b \sin (c+d x))}{b d^3 \left (a^2-b^2\right )}-\frac {(e+f x)^2}{2 b d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^2*Cos[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 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 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]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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 3324

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

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 15.34, size = 1104, normalized size = 3.09 \[ \frac {x \cot (c) f^2}{b \left (b^2-a^2\right ) d^2}-\frac {x \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) f^2}{2 b (b-a) (a+b) d^2}-\frac {i e^{i c} \left (4 e^{i c} f x-\frac {2 i a e^{2 i c} f \log \left (\frac {e^{i (2 c+d x)} b}{i a e^{i c}-\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}+1\right ) x}{\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}+\frac {2 i a f \log \left (\frac {e^{i (2 c+d x)} b}{i a e^{i c}-\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}+1\right ) x}{\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}+\frac {2 i a e^{2 i c} f \log \left (\frac {e^{i (2 c+d x)} b}{i e^{i c} a+\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}+1\right ) x}{\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}-\frac {2 i a f \log \left (\frac {e^{i (2 c+d x)} b}{i e^{i c} a+\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}+1\right ) x}{\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}+\frac {4 i a e e^{-i c} \tan ^{-1}\left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {4 i a e e^{i c} \tan ^{-1}\left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {2 e^{-i c} f \tan ^{-1}\left (\frac {2 a e^{i (c+d x)}}{b \left (-1+e^{2 i (c+d x)}\right )}\right )}{d}-\frac {2 e^{i c} f \tan ^{-1}\left (\frac {2 a e^{i (c+d x)}}{b \left (-1+e^{2 i (c+d x)}\right )}\right )}{d}-\frac {i e^{-i c} f \log \left (4 e^{2 i (c+d x)} a^2+b^2 \left (-1+e^{2 i (c+d x)}\right )^2\right )}{d}+\frac {i e^{i c} f \log \left (4 e^{2 i (c+d x)} a^2+b^2 \left (-1+e^{2 i (c+d x)}\right )^2\right )}{d}-\frac {2 a \left (-1+e^{2 i c}\right ) f \text {Li}_2\left (\frac {i b e^{i (2 c+d x)}}{e^{i c} a+i \sqrt {\left (b^2-a^2\right ) e^{2 i c}}}\right )}{d \sqrt {\left (b^2-a^2\right ) e^{2 i c}}}+\frac {2 a \left (-1+e^{2 i c}\right ) f \text {Li}_2\left (-\frac {b e^{i (2 c+d x)}}{i e^{i c} a+\sqrt {\left (b^2-a^2\right ) e^{2 i c}}}\right )}{d \sqrt {\left (b^2-a^2\right ) e^{2 i c}}}\right ) f}{2 b \left (b^2-a^2\right ) d^2 \left (-1+e^{2 i c}\right )}+\frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (-a x \cos (c) f^2-b x \sin (d x) f^2-a e \cos (c) f-b e \sin (d x) f\right )}{2 (a-b) b (a+b) d^2 (a+b \sin (c+d x))}-\frac {(e+f x)^2}{2 b d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^2*Cos[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

(f^2*x*Cot[c])/(b*(-a^2 + b^2)*d^2) - ((I/2)*E^(I*c)*f*(4*E^(I*c)*f*x + ((4*I)*a*e*ArcTan[(I*a + b*E^(I*(c + d
*x)))/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*E^(I*c)) - ((4*I)*a*e*E^(I*c)*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a
^2 - b^2]])/Sqrt[a^2 - b^2] + (2*f*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))])/(d*E^(I*c)) -
(2*E^(I*c)*f*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))])/d - (I*f*Log[4*a^2*E^((2*I)*(c + d
*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2])/(d*E^(I*c)) + (I*E^(I*c)*f*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 +
E^((2*I)*(c + d*x)))^2])/d + ((2*I)*a*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((
2*I)*c)])])/Sqrt[(-a^2 + b^2)*E^((2*I)*c)] - ((2*I)*a*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*
c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])])/Sqrt[(-a^2 + b^2)*E^((2*I)*c)] - ((2*I)*a*f*x*Log[1 + (b*E^(I*(2*c + d*
x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])])/Sqrt[(-a^2 + b^2)*E^((2*I)*c)] + ((2*I)*a*E^((2*I)*c)*f*
x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])])/Sqrt[(-a^2 + b^2)*E^((2*I)*c)
] - (2*a*(-1 + E^((2*I)*c))*f*PolyLog[2, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]
)])/(d*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]) + (2*a*(-1 + E^((2*I)*c))*f*PolyLog[2, -((b*E^(I*(2*c + d*x)))/(I*a*E^(
I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))])/(d*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])))/(b*(-a^2 + b^2)*d^2*(-1 + E^((2
*I)*c))) - (f^2*x*Csc[c/2]*Sec[c/2]*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2]))/(2*b*(-a + b)*(a + b)*d^2) -
(e + f*x)^2/(2*b*d*(a + b*Sin[c + d*x])^2) + (Csc[c/2]*Sec[c/2]*(-(a*e*f*Cos[c]) - a*f^2*x*Cos[c] - b*e*f*Sin[
d*x] - b*f^2*x*Sin[d*x]))/(2*(a - b)*b*(a + b)*d^2*(a + b*Sin[c + d*x]))

________________________________________________________________________________________

fricas [B]  time = 1.80, size = 2383, normalized size = 6.68 \[ \text {result too large to display} \]

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))^3,x, algorithm="fricas")

[Out]

1/2*((a^4 - 2*a^2*b^2 + b^4)*d^2*f^2*x^2 + 2*(a^4 - 2*a^2*b^2 + b^4)*d^2*e*f*x + (a^4 - 2*a^2*b^2 + b^4)*d^2*e
^2 - 2*((a^2*b^2 - b^4)*d*f^2*x + (a^2*b^2 - b^4)*d*e*f)*cos(d*x + c)*sin(d*x + c) - (-I*a*b^3*f^2*cos(d*x + c
)^2 + 2*I*a^2*b^2*f^2*sin(d*x + c) + I*(a^3*b + a*b^3)*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*a*b^3
*f^2*cos(d*x + c)^2 - 2*I*a^2*b^2*f^2*sin(d*x + c) - I*(a^3*b + a*b^3)*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*a*b^3*f^2*cos(d*x + c)^2 - 2*I*a^2*b^2*f^2*sin(d*x + c) - I*(a^3*b + a*b^3)*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*a*b^3*f^2*cos(d*x + c)^2 + 2*I*a^2*b^2*f^2*sin(d*x + c) + I*(a^3*b + a*b^3)*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^3*b + a*b^3)*d*f^2*x + (a^3*b + a*b^3)*c*f^2 - (a*b^3*d*f^2*x +
a*b^3*c*f^2)*cos(d*x + c)^2 + 2*(a^2*b^2*d*f^2*x + a^2*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^3*b + a*b^3)*d*f^2*x + (a^3*b + a*b^3)*c*f^2 - (a*b^3*d*f^2*x + a*b^3*c*f^2)*cos(d*x + c)^2 + 2*(a^2
*b^2*d*f^2*x + a^2*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^3*b + a*b^3)*d*f^2*x + (a^3
*b + a*b^3)*c*f^2 - (a*b^3*d*f^2*x + a*b^3*c*f^2)*cos(d*x + c)^2 + 2*(a^2*b^2*d*f^2*x + a^2*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^3*b + a*b^3)*d*f^2*x + (a^3*b + a*b^3)*c*f^2 - (a*b^3*d*f^2*x +
a*b^3*c*f^2)*cos(d*x + c)^2 + 2*(a^2*b^2*d*f^2*x + a^2*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) - 2*((a^3*b - a*b^3)*d*f^2*x + (a^3*b - a*b^3)*d*e*f)*cos(d*x + c) - ((a^2*b^2 - b^4)*f^2*cos(d*x + c)^2
- 2*(a^3*b - a*b^3)*f^2*sin(d*x + c) - (a^4 - b^4)*f^2 + ((a^3*b + a*b^3)*d*e*f - (a^3*b + a*b^3)*c*f^2 - (a*b
^3*d*e*f - a*b^3*c*f^2)*cos(d*x + c)^2 + 2*(a^2*b^2*d*e*f - a^2*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^2*b^2 - b^4)*f^2*cos(
d*x + c)^2 - 2*(a^3*b - a*b^3)*f^2*sin(d*x + c) - (a^4 - b^4)*f^2 + ((a^3*b + a*b^3)*d*e*f - (a^3*b + a*b^3)*c
*f^2 - (a*b^3*d*e*f - a*b^3*c*f^2)*cos(d*x + c)^2 + 2*(a^2*b^2*d*e*f - a^2*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^2*b^2 - b^
4)*f^2*cos(d*x + c)^2 - 2*(a^3*b - a*b^3)*f^2*sin(d*x + c) - (a^4 - b^4)*f^2 - ((a^3*b + a*b^3)*d*e*f - (a^3*b
+ a*b^3)*c*f^2 - (a*b^3*d*e*f - a*b^3*c*f^2)*cos(d*x + c)^2 + 2*(a^2*b^2*d*e*f - a^2*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^2*b^2 - b^4)*f^2*cos(d*x + c)^2 - 2*(a^3*b - a*b^3)*f^2*sin(d*x + c) - (a^4 - b^4)*f^2 - ((a^3*b + a*b^3)*d*
e*f - (a^3*b + a*b^3)*c*f^2 - (a*b^3*d*e*f - a*b^3*c*f^2)*cos(d*x + c)^2 + 2*(a^2*b^2*d*e*f - a^2*b^2*c*f^2)*s
in(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^4*b^3 - 2*a^2*b^5 + b^7)*d^3*cos(d*x + c)^2 - 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d^3*sin(d*x + c) -
(a^6*b - a^4*b^3 - a^2*b^5 + b^7)*d^3)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{2} \cos \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

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))^3,x, algorithm="giac")

[Out]

integrate((f*x + e)^2*cos(d*x + c)/(b*sin(d*x + c) + a)^3, x)

________________________________________________________________________________________

maple [B]  time = 2.94, size = 946, normalized size = 2.65 \[ \frac {2 a^{2} d \,f^{2} x^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 b^{2} d \,f^{2} x^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 i a^{2} e f \,{\mathrm e}^{2 i \left (d x +c \right )}+4 i a^{2} f^{2} x \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a^{2} d e f x \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b a \,f^{2} x \,{\mathrm e}^{3 i \left (d x +c \right )}-4 b^{2} d e f x \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i b^{2} e f +2 i b^{2} f^{2} x \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a^{2} d \,e^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 b a e f \,{\mathrm e}^{3 i \left (d x +c \right )}-2 b^{2} d \,e^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 i b^{2} f^{2} x -6 a b \,f^{2} x \,{\mathrm e}^{i \left (d x +c \right )}+2 i b^{2} e f \,{\mathrm e}^{2 i \left (d x +c \right )}-6 a b e f \,{\mathrm e}^{i \left (d x +c \right )}}{\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d^{2} \left (a^{2}-b^{2}\right ) b}+\frac {f^{2} \ln \left (i b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}-i b \right )}{b \left (-a^{2}+b^{2}\right ) d^{3}}-\frac {2 f^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{b \left (-a^{2}+b^{2}\right ) d^{3}}-\frac {f^{2} a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{2}}-\frac {f^{2} a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) c}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{3}}+\frac {2 i f^{2} a c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{3}}-\frac {2 i f a e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{2}}-\frac {i f^{2} a \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{3}}+\frac {f^{2} a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{2}}+\frac {f^{2} a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{3}}+\frac {i f^{2} a \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{3}} \]

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))^3,x)

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

________________________________________________________________________________________

mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(e + f*x)^2)/(a + b*sin(c + d*x))^3,x)

[Out]

\text{Hanged}

________________________________________________________________________________________

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((f*x+e)**2*cos(d*x+c)/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________