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

Optimal. Leaf size=296 \[ -\frac {a f \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac {a f \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}-\frac {f \log (a+b \cos (c+d x))}{d^2 \left (a^2-b^2\right )}-\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {b (e+f x) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.52, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3324, 3321, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac {a f \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac {a f \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}-\frac {f \log (a+b \cos (c+d x))}{d^2 \left (a^2-b^2\right )}-\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {b (e+f x) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*a*(e + f*x)*Log[1 + (b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) + (I*a*(e + f*x)*L
og[1 + (b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) - (f*Log[a + b*Cos[c + d*x]])/((a^2 -
b^2)*d^2) - (a*f*PolyLog[2, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2]))])/((a^2 - b^2)^(3/2)*d^2) + (a*f*Pol
yLog[2, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))])/((a^2 - b^2)^(3/2)*d^2) - (b*(e + f*x)*Sin[c + d*x])/((
a^2 - b^2)*d*(a + b*Cos[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 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
+ d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && 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]

Rubi steps

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

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Mathematica [B]  time = 9.75, size = 933, normalized size = 3.15 \[ \frac {\left (\frac {2 a (d e-c f) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+f \log \left (\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )-f \log \left ((a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )-\frac {i a f \left (\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {\sqrt {a+b}-\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )}{i \sqrt {b-a}+\sqrt {a+b}}\right )+\text {Li}_2\left (\frac {\sqrt {b-a} \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {b-a}-i \sqrt {a+b}}\right )\right )}{\sqrt {b-a} \sqrt {a+b}}+\frac {i a f \left (\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {i \left (\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt {a+b}\right )}{\sqrt {b-a}+i \sqrt {a+b}}\right )+\text {Li}_2\left (\frac {\sqrt {b-a} \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {b-a}+i \sqrt {a+b}}\right )\right )}{\sqrt {b-a} \sqrt {a+b}}-\frac {i a f \left (\log \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right ) \log \left (\frac {\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt {a+b}}{i \sqrt {b-a}+\sqrt {a+b}}\right )+\text {Li}_2\left (\frac {\sqrt {b-a} \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right )}{\sqrt {b-a}-i \sqrt {a+b}}\right )\right )}{\sqrt {b-a} \sqrt {a+b}}+\frac {i a f \left (\log \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right ) \log \left (\frac {i \left (\sqrt {a+b}-\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {b-a}+i \sqrt {a+b}}\right )+\text {Li}_2\left (\frac {\sqrt {b-a} \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right )}{\sqrt {b-a}+i \sqrt {a+b}}\right )\right )}{\sqrt {b-a} \sqrt {a+b}}\right ) (a d e+a d f x+b f \sin (c+d x)) \left (\sqrt {a+b}-\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt {a+b}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )}{\left (a^2-b^2\right ) d^2 (a+b \cos (c+d x)) \left (a \left (d e-c f+i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )-i f \log \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right )\right )+b f \sin (c+d x)\right )}+\frac {-b d e \sin (c+d x)+b c f \sin (c+d x)-b f (c+d x) \sin (c+d x)}{(a-b) (a+b) d^2 (a+b \cos (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(b*d*e*Sin[c + d*x]) + b*c*f*Sin[c + d*x] - b*f*(c + d*x)*Sin[c + d*x])/((a - b)*(a + b)*d^2*(a + b*Cos[c +
d*x])) + (Cos[(c + d*x)/2]^2*((2*a*(d*e - c*f)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b
]*Sqrt[a + b]) + f*Log[Sec[(c + d*x)/2]^2] - f*Log[(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2] - (I*a*f*(Log[1 -
I*Tan[(c + d*x)/2]]*Log[(Sqrt[a + b] - Sqrt[-a + b]*Tan[(c + d*x)/2])/(I*Sqrt[-a + b] + Sqrt[a + b])] + PolyLo
g[2, (Sqrt[-a + b]*(1 - I*Tan[(c + d*x)/2]))/(Sqrt[-a + b] - I*Sqrt[a + b])]))/(Sqrt[-a + b]*Sqrt[a + b]) + (I
*a*f*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(I*(Sqrt[a + b] + Sqrt[-a + b]*Tan[(c + d*x)/2]))/(Sqrt[-a + b] + I*Sqrt
[a + b])] + PolyLog[2, (Sqrt[-a + b]*(1 - I*Tan[(c + d*x)/2]))/(Sqrt[-a + b] + I*Sqrt[a + b])]))/(Sqrt[-a + b]
*Sqrt[a + b]) - (I*a*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(Sqrt[a + b] + Sqrt[-a + b]*Tan[(c + d*x)/2])/(I*Sqrt[
-a + b] + Sqrt[a + b])] + PolyLog[2, (Sqrt[-a + b]*(1 + I*Tan[(c + d*x)/2]))/(Sqrt[-a + b] - I*Sqrt[a + b])]))
/(Sqrt[-a + b]*Sqrt[a + b]) + (I*a*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(I*(Sqrt[a + b] - Sqrt[-a + b]*Tan[(c +
d*x)/2]))/(Sqrt[-a + b] + I*Sqrt[a + b])] + PolyLog[2, (Sqrt[-a + b]*(1 + I*Tan[(c + d*x)/2]))/(Sqrt[-a + b] +
I*Sqrt[a + b])]))/(Sqrt[-a + b]*Sqrt[a + b]))*(a*d*e + a*d*f*x + b*f*Sin[c + d*x])*(Sqrt[a + b] - Sqrt[-a + b
]*Tan[(c + d*x)/2])*(Sqrt[a + b] + Sqrt[-a + b]*Tan[(c + d*x)/2]))/((a^2 - b^2)*d^2*(a + b*Cos[c + d*x])*(a*(d
*e - c*f + I*f*Log[1 - I*Tan[(c + d*x)/2]] - I*f*Log[1 + I*Tan[(c + d*x)/2]]) + b*f*Sin[c + d*x]))

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fricas [B]  time = 1.59, size = 1482, normalized size = 5.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a*b^2*f*cos(d*x + c) + a^2*b*f)*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*x + c) + I*a*sin(d*x + c) + (b*co
s(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) - (a*b^2*f*cos(d*x + c) + a^2*b*f)*sqrt((a^2
- b^2)/b^2)*dilog(-(a*cos(d*x + c) + I*a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b
^2) + b)/b + 1) + (a*b^2*f*cos(d*x + c) + a^2*b*f)*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*x + c) - I*a*sin(d*x
+ c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) - (a*b^2*f*cos(d*x + c) + a^2*b*f
)*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*x + c) - I*a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((
a^2 - b^2)/b^2) + b)/b + 1) - (-I*a^2*b*d*f*x - I*a^2*b*c*f + (-I*a*b^2*d*f*x - I*a*b^2*c*f)*cos(d*x + c))*sqr
t((a^2 - b^2)/b^2)*log((a*cos(d*x + c) + I*a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^
2)/b^2) + b)/b) - (I*a^2*b*d*f*x + I*a^2*b*c*f + (I*a*b^2*d*f*x + I*a*b^2*c*f)*cos(d*x + c))*sqrt((a^2 - b^2)/
b^2)*log((a*cos(d*x + c) + I*a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b
) - (I*a^2*b*d*f*x + I*a^2*b*c*f + (I*a*b^2*d*f*x + I*a*b^2*c*f)*cos(d*x + c))*sqrt((a^2 - b^2)/b^2)*log((a*co
s(d*x + c) - I*a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b) - (-I*a^2*b*
d*f*x - I*a^2*b*c*f + (-I*a*b^2*d*f*x - I*a*b^2*c*f)*cos(d*x + c))*sqrt((a^2 - b^2)/b^2)*log((a*cos(d*x + c) -
I*a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b) + ((a^2*b - b^3)*f*cos(d
*x + c) + (a^3 - a*b^2)*f - (I*a^2*b*d*e - I*a^2*b*c*f + (I*a*b^2*d*e - I*a*b^2*c*f)*cos(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*a) + ((a^2*b - b^3)*f*co
s(d*x + c) + (a^3 - a*b^2)*f - (-I*a^2*b*d*e + I*a^2*b*c*f + (-I*a*b^2*d*e + I*a*b^2*c*f)*cos(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*a) + ((a^2*b - b^3)
*f*cos(d*x + c) + (a^3 - a*b^2)*f - (I*a^2*b*d*e - I*a^2*b*c*f + (I*a*b^2*d*e - I*a*b^2*c*f)*cos(d*x + c))*sqr
t((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*a) + ((a^2*b -
b^3)*f*cos(d*x + c) + (a^3 - a*b^2)*f - (-I*a^2*b*d*e + I*a^2*b*c*f + (-I*a*b^2*d*e + I*a*b^2*c*f)*cos(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*a) + 2*((
a^2*b - b^3)*d*f*x + (a^2*b - b^3)*d*e)*sin(d*x + c))/((a^4*b - 2*a^2*b^3 + b^5)*d^2*cos(d*x + c) + (a^5 - 2*a
^3*b^2 + a*b^4)*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.70, size = 674, normalized size = 2.28 \[ \frac {2 i \left (f x +e \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {2 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} \left (-a^{2}+b^{2}\right )}+\frac {f \ln \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d^{2} \left (-a^{2}+b^{2}\right )}+\frac {2 i a e \arctan \left (\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d \left (-a^{2}+b^{2}\right )^{\frac {3}{2}}}+\frac {i a f \ln \left (\frac {-b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}+\frac {i a f \ln \left (\frac {-b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}-\frac {i a f \ln \left (\frac {b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}-\frac {i a f \ln \left (\frac {b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}+\frac {a f \dilog \left (\frac {-b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}-\frac {a f \dilog \left (\frac {b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}-\frac {2 i a f c \arctan \left (\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} \left (-a^{2}+b^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(a+b*cos(d*x+c))^2,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?

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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((e + f*x)/(a + b*cos(c + d*x))^2,x)

[Out]

\text{Hanged}

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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)/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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