\(\int \frac {\arctan (a+b x)}{c+d x^2} \, dx\) [53]

Optimal result

Integrand size = 16, antiderivative size = 319 \[ \int \frac {\arctan (a+b x)}{c+d x^2} \, dx=\frac {\arctan (a+b x) \log \left (\frac {2 b \left (\sqrt {-c}-\sqrt {d} x\right )}{\left (b \sqrt {-c}-(i-a) \sqrt {d}\right ) (1-i (a+b x))}\right )}{2 \sqrt {-c} \sqrt {d}}-\frac {\arctan (a+b x) \log \left (\frac {2 b \left (\sqrt {-c}+\sqrt {d} x\right )}{\left (b \sqrt {-c}+(i-a) \sqrt {d}\right ) (1-i (a+b x))}\right )}{2 \sqrt {-c} \sqrt {d}}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b \left (\sqrt {-c}-\sqrt {d} x\right )}{\left (b \sqrt {-c}-(i-a) \sqrt {d}\right ) (1-i (a+b x))}\right )}{4 \sqrt {-c} \sqrt {d}}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b \left (\sqrt {-c}+\sqrt {d} x\right )}{\left (b \sqrt {-c}+(i-a) \sqrt {d}\right ) (1-i (a+b x))}\right )}{4 \sqrt {-c} \sqrt {d}} \] Output:

1/2*arctan(b*x+a)*ln(2*b*((-c)^(1/2)-d^(1/2)*x)/(b*(-c)^(1/2)-(I-a)*d^(1/2 
))/(1-I*(b*x+a)))/(-c)^(1/2)/d^(1/2)-1/2*arctan(b*x+a)*ln(2*b*((-c)^(1/2)+ 
d^(1/2)*x)/(b*(-c)^(1/2)+(I-a)*d^(1/2))/(1-I*(b*x+a)))/(-c)^(1/2)/d^(1/2)- 
1/4*I*polylog(2,1-2*b*((-c)^(1/2)-d^(1/2)*x)/(b*(-c)^(1/2)-(I-a)*d^(1/2))/ 
(1-I*(b*x+a)))/(-c)^(1/2)/d^(1/2)+1/4*I*polylog(2,1-2*b*((-c)^(1/2)+d^(1/2 
)*x)/(b*(-c)^(1/2)+(I-a)*d^(1/2))/(1-I*(b*x+a)))/(-c)^(1/2)/d^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.23 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.28 \[ \int \frac {\arctan (a+b x)}{c+d x^2} \, dx=-\frac {i \left (\log (1+i a+i b x) \log \left (\frac {b \left (\sqrt {-c}-\sqrt {d} x\right )}{b \sqrt {-c}+(-i+a) \sqrt {d}}\right )-\log (-i (i+a+b x)) \log \left (\frac {b \left (\sqrt {-c}-\sqrt {d} x\right )}{b \sqrt {-c}+(i+a) \sqrt {d}}\right )-\log (1+i a+i b x) \log \left (\frac {b \left (\sqrt {-c}+\sqrt {d} x\right )}{b \sqrt {-c}-(-i+a) \sqrt {d}}\right )+\log (-i (i+a+b x)) \log \left (\frac {b \left (\sqrt {-c}+\sqrt {d} x\right )}{b \sqrt {-c}-(i+a) \sqrt {d}}\right )-\operatorname {PolyLog}\left (2,\frac {\sqrt {d} (-i+a+b x)}{-b \sqrt {-c}+(-i+a) \sqrt {d}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {d} (-i+a+b x)}{b \sqrt {-c}+(-i+a) \sqrt {d}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {d} (i+a+b x)}{-b \sqrt {-c}+(i+a) \sqrt {d}}\right )-\operatorname {PolyLog}\left (2,\frac {\sqrt {d} (i+a+b x)}{b \sqrt {-c}+(i+a) \sqrt {d}}\right )\right )}{4 \sqrt {-c} \sqrt {d}} \] Input:

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

Output:

((-1/4*I)*(Log[1 + I*a + I*b*x]*Log[(b*(Sqrt[-c] - Sqrt[d]*x))/(b*Sqrt[-c] 
 + (-I + a)*Sqrt[d])] - Log[(-I)*(I + a + b*x)]*Log[(b*(Sqrt[-c] - Sqrt[d] 
*x))/(b*Sqrt[-c] + (I + a)*Sqrt[d])] - Log[1 + I*a + I*b*x]*Log[(b*(Sqrt[- 
c] + Sqrt[d]*x))/(b*Sqrt[-c] - (-I + a)*Sqrt[d])] + Log[(-I)*(I + a + b*x) 
]*Log[(b*(Sqrt[-c] + Sqrt[d]*x))/(b*Sqrt[-c] - (I + a)*Sqrt[d])] - PolyLog 
[2, (Sqrt[d]*(-I + a + b*x))/(-(b*Sqrt[-c]) + (-I + a)*Sqrt[d])] + PolyLog 
[2, (Sqrt[d]*(-I + a + b*x))/(b*Sqrt[-c] + (-I + a)*Sqrt[d])] + PolyLog[2, 
 (Sqrt[d]*(I + a + b*x))/(-(b*Sqrt[-c]) + (I + a)*Sqrt[d])] - PolyLog[2, ( 
Sqrt[d]*(I + a + b*x))/(b*Sqrt[-c] + (I + a)*Sqrt[d])]))/(Sqrt[-c]*Sqrt[d] 
)
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 541, normalized size of antiderivative = 1.70, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5574, 2856, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\arctan (a+b x)}{c+d x^2} \, dx\)

\(\Big \downarrow \) 5574

\(\displaystyle \frac {1}{2} i \int \frac {\log (-i a-i b x+1)}{d x^2+c}dx-\frac {1}{2} i \int \frac {\log (i a+i b x+1)}{d x^2+c}dx\)

\(\Big \downarrow \) 2856

\(\displaystyle \frac {1}{2} i \int \left (\frac {\sqrt {-c} \log (-i a-i b x+1)}{2 c \left (\sqrt {-c}-\sqrt {d} x\right )}+\frac {\sqrt {-c} \log (-i a-i b x+1)}{2 c \left (\sqrt {d} x+\sqrt {-c}\right )}\right )dx-\frac {1}{2} i \int \left (\frac {\sqrt {-c} \log (i a+i b x+1)}{2 c \left (\sqrt {-c}-\sqrt {d} x\right )}+\frac {\sqrt {-c} \log (i a+i b x+1)}{2 c \left (\sqrt {d} x+\sqrt {-c}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} i \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {\sqrt {d} (a+b x+i)}{b \sqrt {-c}-(a+i) \sqrt {d}}\right )}{2 \sqrt {-c} \sqrt {d}}+\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt {d} (a+b x+i)}{\sqrt {d} (a+i)+b \sqrt {-c}}\right )}{2 \sqrt {-c} \sqrt {d}}+\frac {\log (-i a-i b x+1) \log \left (\frac {b \left (\sqrt {-c}-\sqrt {d} x\right )}{b \sqrt {-c}+(a+i) \sqrt {d}}\right )}{2 \sqrt {-c} \sqrt {d}}-\frac {\log (-i a-i b x+1) \log \left (\frac {b \left (\sqrt {-c}+\sqrt {d} x\right )}{b \sqrt {-c}-(a+i) \sqrt {d}}\right )}{2 \sqrt {-c} \sqrt {d}}\right )-\frac {1}{2} i \left (\frac {\operatorname {PolyLog}\left (2,-\frac {\sqrt {d} (-a-b x+i)}{b \sqrt {-c}-(i-a) \sqrt {d}}\right )}{2 \sqrt {-c} \sqrt {d}}-\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt {d} (-a-b x+i)}{\sqrt {d} (i-a)+b \sqrt {-c}}\right )}{2 \sqrt {-c} \sqrt {d}}+\frac {\log (i a+i b x+1) \log \left (\frac {b \left (\sqrt {-c}-\sqrt {d} x\right )}{b \sqrt {-c}-(-a+i) \sqrt {d}}\right )}{2 \sqrt {-c} \sqrt {d}}-\frac {\log (i a+i b x+1) \log \left (\frac {b \left (\sqrt {-c}+\sqrt {d} x\right )}{b \sqrt {-c}+(-a+i) \sqrt {d}}\right )}{2 \sqrt {-c} \sqrt {d}}\right )\)

Input:

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

Output:

(-1/2*I)*((Log[1 + I*a + I*b*x]*Log[(b*(Sqrt[-c] - Sqrt[d]*x))/(b*Sqrt[-c] 
 - (I - a)*Sqrt[d])])/(2*Sqrt[-c]*Sqrt[d]) - (Log[1 + I*a + I*b*x]*Log[(b* 
(Sqrt[-c] + Sqrt[d]*x))/(b*Sqrt[-c] + (I - a)*Sqrt[d])])/(2*Sqrt[-c]*Sqrt[ 
d]) + PolyLog[2, -((Sqrt[d]*(I - a - b*x))/(b*Sqrt[-c] - (I - a)*Sqrt[d])) 
]/(2*Sqrt[-c]*Sqrt[d]) - PolyLog[2, (Sqrt[d]*(I - a - b*x))/(b*Sqrt[-c] + 
(I - a)*Sqrt[d])]/(2*Sqrt[-c]*Sqrt[d])) + (I/2)*((Log[1 - I*a - I*b*x]*Log 
[(b*(Sqrt[-c] - Sqrt[d]*x))/(b*Sqrt[-c] + (I + a)*Sqrt[d])])/(2*Sqrt[-c]*S 
qrt[d]) - (Log[1 - I*a - I*b*x]*Log[(b*(Sqrt[-c] + Sqrt[d]*x))/(b*Sqrt[-c] 
 - (I + a)*Sqrt[d])])/(2*Sqrt[-c]*Sqrt[d]) - PolyLog[2, -((Sqrt[d]*(I + a 
+ b*x))/(b*Sqrt[-c] - (I + a)*Sqrt[d]))]/(2*Sqrt[-c]*Sqrt[d]) + PolyLog[2, 
 (Sqrt[d]*(I + a + b*x))/(b*Sqrt[-c] + (I + a)*Sqrt[d])]/(2*Sqrt[-c]*Sqrt[ 
d]))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2856

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. 
)*(x_)^(r_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x) 
^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x] && I 
GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
 

rule 5574

Int[ArcTan[(a_) + (b_.)*(x_)]/((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[ 
I/2   Int[Log[1 - I*a - I*b*x]/(c + d*x^n), x], x] - Simp[I/2   Int[Log[1 + 
 I*a + I*b*x]/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[n]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 541 vs. \(2 (253 ) = 506\).

Time = 1.14 (sec) , antiderivative size = 542, normalized size of antiderivative = 1.70

method	result	size
risch	\(\frac {\ln \left (-b x i-a i+1\right ) \ln \left (\frac {i a d -b \sqrt {c d}+\left (-b x i-a i+1\right ) d -d}{i a d -b \sqrt {c d}-d}\right ) \sqrt {c d}}{4 c d}-\frac {\ln \left (-b x i-a i+1\right ) \ln \left (\frac {i a d +b \sqrt {c d}+\left (-b x i-a i+1\right ) d -d}{i a d +b \sqrt {c d}-d}\right ) \sqrt {c d}}{4 c d}-\frac {\operatorname {dilog}\left (\frac {i a d +b \sqrt {c d}+\left (-b x i-a i+1\right ) d -d}{i a d +b \sqrt {c d}-d}\right ) \sqrt {c d}}{4 c d}+\frac {\operatorname {dilog}\left (\frac {i a d -b \sqrt {c d}+\left (-b x i-a i+1\right ) d -d}{i a d -b \sqrt {c d}-d}\right ) \sqrt {c d}}{4 c d}+\frac {\ln \left (b x i+a i+1\right ) \ln \left (\frac {i a d +b \sqrt {c d}-\left (b x i+a i+1\right ) d +d}{i a d +b \sqrt {c d}+d}\right ) \sqrt {c d}}{4 c d}-\frac {\ln \left (b x i+a i+1\right ) \ln \left (\frac {i a d -b \sqrt {c d}-\left (b x i+a i+1\right ) d +d}{i a d -b \sqrt {c d}+d}\right ) \sqrt {c d}}{4 c d}+\frac {\operatorname {dilog}\left (\frac {i a d +b \sqrt {c d}-\left (b x i+a i+1\right ) d +d}{i a d +b \sqrt {c d}+d}\right ) \sqrt {c d}}{4 c d}-\frac {\operatorname {dilog}\left (\frac {i a d -b \sqrt {c d}-\left (b x i+a i+1\right ) d +d}{i a d -b \sqrt {c d}+d}\right ) \sqrt {c d}}{4 c d}\)	\(542\)
derivativedivides	\(\text {Expression too large to display}\)	\(2184\)
default	\(\text {Expression too large to display}\)	\(2184\)

Input:

int(arctan(b*x+a)/(d*x^2+c),x,method=_RETURNVERBOSE)
 

Output:

1/4*ln(1-I*a-I*b*x)/c/d*ln((I*a*d-b*(c*d)^(1/2)+(1-I*a-I*b*x)*d-d)/(I*a*d- 
b*(c*d)^(1/2)-d))*(c*d)^(1/2)-1/4*ln(1-I*a-I*b*x)/c/d*ln((I*a*d+b*(c*d)^(1 
/2)+(1-I*a-I*b*x)*d-d)/(I*a*d+b*(c*d)^(1/2)-d))*(c*d)^(1/2)-1/4/c/d*dilog( 
(I*a*d+b*(c*d)^(1/2)+(1-I*a-I*b*x)*d-d)/(I*a*d+b*(c*d)^(1/2)-d))*(c*d)^(1/ 
2)+1/4/c/d*dilog((I*a*d-b*(c*d)^(1/2)+(1-I*a-I*b*x)*d-d)/(I*a*d-b*(c*d)^(1 
/2)-d))*(c*d)^(1/2)+1/4*ln(1+I*a+I*b*x)/c/d*ln((I*a*d+b*(c*d)^(1/2)-(1+I*a 
+I*b*x)*d+d)/(I*a*d+b*(c*d)^(1/2)+d))*(c*d)^(1/2)-1/4*ln(1+I*a+I*b*x)/c/d* 
ln((I*a*d-b*(c*d)^(1/2)-(1+I*a+I*b*x)*d+d)/(I*a*d-b*(c*d)^(1/2)+d))*(c*d)^ 
(1/2)+1/4/c/d*dilog((I*a*d+b*(c*d)^(1/2)-(1+I*a+I*b*x)*d+d)/(I*a*d+b*(c*d) 
^(1/2)+d))*(c*d)^(1/2)-1/4/c/d*dilog((I*a*d-b*(c*d)^(1/2)-(1+I*a+I*b*x)*d+ 
d)/(I*a*d-b*(c*d)^(1/2)+d))*(c*d)^(1/2)
 

Fricas [F]

\[ \int \frac {\arctan (a+b x)}{c+d x^2} \, dx=\int { \frac {\arctan \left (b x + a\right )}{d x^{2} + c} \,d x } \] Input:

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

Output:

integral(arctan(b*x + a)/(d*x^2 + c), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\arctan (a+b x)}{c+d x^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 8520 vs. \(2 (235) = 470\).

Time = 4.49 (sec) , antiderivative size = 8520, normalized size of antiderivative = 26.71 \[ \int \frac {\arctan (a+b x)}{c+d x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*b*(8*arctan(d*x/sqrt(c*d))*arctan((b^2*x + a*b)/b)/b - (4*arctan(sqrt( 
d)*x/sqrt(c))*arctan2((2*a*b^2*c*d + (a*b^3*c + (a^3 + a)*b*d + (b^4*c + ( 
a^2 + 3)*b^2*d)*x)*sqrt(c)*sqrt(d) + (3*b^3*c*d + (a^2 + 1)*b*d^2)*x)/(b^4 
*c^2 + 2*(a^2 + 3)*b^2*c*d + (a^4 + 2*a^2 + 1)*d^2 + 4*(b^3*c + (a^2 + 1)* 
b*d)*sqrt(c)*sqrt(d)), ((a^2 + 3)*b^2*c*d + (a^4 + 2*a^2 + 1)*d^2 + (2*a*b 
^2*d*x + b^3*c + 3*(a^2 + 1)*b*d)*sqrt(c)*sqrt(d) + (a*b^3*c*d + (a^3 + a) 
*b*d^2)*x)/(b^4*c^2 + 2*(a^2 + 3)*b^2*c*d + (a^4 + 2*a^2 + 1)*d^2 + 4*(b^3 
*c + (a^2 + 1)*b*d)*sqrt(c)*sqrt(d))) + 4*arctan(sqrt(d)*x/sqrt(c))*arctan 
2((2*a*b^2*c*d - (a*b^3*c + (a^3 + a)*b*d + (b^4*c + (a^2 + 3)*b^2*d)*x)*s 
qrt(c)*sqrt(d) + (3*b^3*c*d + (a^2 + 1)*b*d^2)*x)/(b^4*c^2 + 2*(a^2 + 3)*b 
^2*c*d + (a^4 + 2*a^2 + 1)*d^2 - 4*(b^3*c + (a^2 + 1)*b*d)*sqrt(c)*sqrt(d) 
), ((a^2 + 3)*b^2*c*d + (a^4 + 2*a^2 + 1)*d^2 - (2*a*b^2*d*x + b^3*c + 3*( 
a^2 + 1)*b*d)*sqrt(c)*sqrt(d) + (a*b^3*c*d + (a^3 + a)*b*d^2)*x)/(b^4*c^2 
+ 2*(a^2 + 3)*b^2*c*d + (a^4 + 2*a^2 + 1)*d^2 - 4*(b^3*c + (a^2 + 1)*b*d)* 
sqrt(c)*sqrt(d))) + log(d*x^2 + c)*log(((a^2 + 1)*b^22*c^11*d + 11*(a^4 + 
22*a^2 + 21)*b^20*c^10*d^2 + 55*(a^6 + 39*a^4 + 171*a^2 + 133)*b^18*c^9*d^ 
3 + 33*(5*a^8 + 260*a^6 + 1870*a^4 + 3876*a^2 + 2261)*b^16*c^8*d^4 + 330*( 
a^10 + 61*a^8 + 570*a^6 + 1802*a^4 + 2261*a^2 + 969)*b^14*c^7*d^5 + 22*(21 
*a^12 + 1386*a^10 + 15015*a^8 + 60060*a^6 + 109395*a^4 + 92378*a^2 + 29393 
)*b^12*c^6*d^6 + 22*(21*a^14 + 1407*a^12 + 16401*a^10 + 75075*a^8 + 169...
 

Giac [F]

\[ \int \frac {\arctan (a+b x)}{c+d x^2} \, dx=\int { \frac {\arctan \left (b x + a\right )}{d x^{2} + c} \,d x } \] Input:

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

Output:

integrate(arctan(b*x + a)/(d*x^2 + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (a+b x)}{c+d x^2} \, dx=\int \frac {\mathrm {atan}\left (a+b\,x\right )}{d\,x^2+c} \,d x \] Input:

int(atan(a + b*x)/(c + d*x^2),x)
 

Output:

int(atan(a + b*x)/(c + d*x^2), x)
 

Reduce [F]

\[ \int \frac {\arctan (a+b x)}{c+d x^2} \, dx=\int \frac {\mathit {atan} \left (b x +a \right )}{d \,x^{2}+c}d x \] Input:

int(atan(b*x+a)/(d*x^2+c),x)
 

Output:

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