### 3.6 Test ﬁle Number [151] 5-Inverse-trig-functions/5.3-Inverse-tangent/5.3.5-u-a+b-arctan-c+d-x-^p

#### 3.6.1 Mathematica

Integral number [65] $\int \frac {\tan ^{-1}(a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx$

[B]   time = 0.421783 (sec), size = 163 ,normalized size = 7.09 $\frac {\frac {5 \sqrt [3]{2} \sqrt {\pi } \Gamma \left (\frac {5}{3}\right ) \, _3F_2\left (1,\frac {4}{3},\frac {4}{3};\frac {11}{6},\frac {7}{3};\frac {1}{(a+b x)^2+1}\right )}{(a+b x)^2+1}+6 \Gamma \left (\frac {11}{6}\right ) \Gamma \left (\frac {7}{3}\right ) \left (\frac {4 (a+b x) \, _2F_1\left (1,\frac {4}{3};\frac {11}{6};\frac {1}{(a+b x)^2+1}\right ) \tan ^{-1}(a+b x)}{(a+b x)^2+1}+10 (a+b x) \tan ^{-1}(a+b x)+15\right )}{20 b \Gamma \left (\frac {11}{6}\right ) \Gamma \left (\frac {7}{3}\right ) \sqrt [3]{a^2+2 a b x+b^2 x^2+1}}$

[In]

Integrate[ArcTan[a + b*x]/(1 + a^2 + 2*a*b*x + b^2*x^2)^(1/3),x]

[Out]

(6*Gamma[11/6]*Gamma[7/3]*(15 + 10*(a + b*x)*ArcTan[a + b*x] + (4*(a + b*x)*ArcTan[a + b*x]*Hypergeometric2F1[
1, 4/3, 11/6, (1 + (a + b*x)^2)^(-1)])/(1 + (a + b*x)^2)) + (5*2^(1/3)*Sqrt[Pi]*Gamma[5/3]*HypergeometricPFQ[{
1, 4/3, 4/3}, {11/6, 7/3}, (1 + (a + b*x)^2)^(-1)])/(1 + (a + b*x)^2))/(20*b*(1 + a^2 + 2*a*b*x + b^2*x^2)^(1/
3)*Gamma[11/6]*Gamma[7/3])

Integral number [66] $\int \frac {\tan ^{-1}(a+b x)}{\sqrt [3]{\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx$

[B]   time = 0.0997431 (sec), size = 165 ,normalized size = 6.6 $\frac {\frac {5 \sqrt [3]{2} \sqrt {\pi } \Gamma \left (\frac {5}{3}\right ) \, _3F_2\left (1,\frac {4}{3},\frac {4}{3};\frac {11}{6},\frac {7}{3};\frac {1}{(a+b x)^2+1}\right )}{(a+b x)^2+1}+6 \Gamma \left (\frac {11}{6}\right ) \Gamma \left (\frac {7}{3}\right ) \left (\frac {4 (a+b x) \, _2F_1\left (1,\frac {4}{3};\frac {11}{6};\frac {1}{(a+b x)^2+1}\right ) \tan ^{-1}(a+b x)}{(a+b x)^2+1}+10 (a+b x) \tan ^{-1}(a+b x)+15\right )}{20 b \Gamma \left (\frac {11}{6}\right ) \Gamma \left (\frac {7}{3}\right ) \sqrt [3]{c \left (a^2+2 a b x+b^2 x^2+1\right )}}$

[In]

Integrate[ArcTan[a + b*x]/((1 + a^2)*c + 2*a*b*c*x + b^2*c*x^2)^(1/3),x]

[Out]

(6*Gamma[11/6]*Gamma[7/3]*(15 + 10*(a + b*x)*ArcTan[a + b*x] + (4*(a + b*x)*ArcTan[a + b*x]*Hypergeometric2F1[
1, 4/3, 11/6, (1 + (a + b*x)^2)^(-1)])/(1 + (a + b*x)^2)) + (5*2^(1/3)*Sqrt[Pi]*Gamma[5/3]*HypergeometricPFQ[{
1, 4/3, 4/3}, {11/6, 7/3}, (1 + (a + b*x)^2)^(-1)])/(1 + (a + b*x)^2))/(20*b*(c*(1 + a^2 + 2*a*b*x + b^2*x^2))
^(1/3)*Gamma[11/6]*Gamma[7/3])

Integral number [69] $\int \frac {(a+b x)^2 \tan ^{-1}(a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx$

[B]   time = 1.57668 (sec), size = 181 ,normalized size = 6.03 $-\frac {3 \left ((a+b x)^2+1\right )^{2/3} \left (\frac {5 \sqrt [3]{2} \sqrt {\pi } \Gamma \left (\frac {5}{3}\right ) \, _3F_2\left (1,\frac {4}{3},\frac {4}{3};\frac {11}{6},\frac {7}{3};\frac {1}{(a+b x)^2+1}\right )}{\left ((a+b x)^2+1\right )^2}+\Gamma \left (\frac {11}{6}\right ) \Gamma \left (\frac {7}{3}\right ) \left (\frac {24 (a+b x) \, _2F_1\left (1,\frac {4}{3};\frac {11}{6};\frac {1}{(a+b x)^2+1}\right ) \tan ^{-1}(a+b x)}{\left ((a+b x)^2+1\right )^2}+\frac {90}{(a+b x)^2+1}+5 \tan ^{-1}(a+b x) \left (6 \sin \left (2 \tan ^{-1}(a+b x)\right )-4 (a+b x)\right )+15\right )\right )}{140 b \Gamma \left (\frac {11}{6}\right ) \Gamma \left (\frac {7}{3}\right )}$

[In]

Integrate[((a + b*x)^2*ArcTan[a + b*x])/(1 + a^2 + 2*a*b*x + b^2*x^2)^(1/3),x]

[Out]

(-3*(1 + (a + b*x)^2)^(2/3)*((5*2^(1/3)*Sqrt[Pi]*Gamma[5/3]*HypergeometricPFQ[{1, 4/3, 4/3}, {11/6, 7/3}, (1 +
(a + b*x)^2)^(-1)])/(1 + (a + b*x)^2)^2 + Gamma[11/6]*Gamma[7/3]*(15 + 90/(1 + (a + b*x)^2) + (24*(a + b*x)*A
rcTan[a + b*x]*Hypergeometric2F1[1, 4/3, 11/6, (1 + (a + b*x)^2)^(-1)])/(1 + (a + b*x)^2)^2 + 5*ArcTan[a + b*x
]*(-4*(a + b*x) + 6*Sin[2*ArcTan[a + b*x]]))))/(140*b*Gamma[11/6]*Gamma[7/3])

Integral number [70] $\int \frac {(a+b x)^2 \tan ^{-1}(a+b x)}{\sqrt [3]{\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx$

[B]   time = 0.800008 (sec), size = 225 ,normalized size = 7.03 $-\frac {3 \sqrt [3]{a^2+2 a b x+b^2 x^2+1} \left ((a+b x)^2+1\right )^{2/3} \left (\frac {5 \sqrt [3]{2} \sqrt {\pi } \Gamma \left (\frac {5}{3}\right ) \, _3F_2\left (1,\frac {4}{3},\frac {4}{3};\frac {11}{6},\frac {7}{3};\frac {1}{(a+b x)^2+1}\right )}{\left ((a+b x)^2+1\right )^2}+\Gamma \left (\frac {11}{6}\right ) \Gamma \left (\frac {7}{3}\right ) \left (\frac {24 (a+b x) \, _2F_1\left (1,\frac {4}{3};\frac {11}{6};\frac {1}{(a+b x)^2+1}\right ) \tan ^{-1}(a+b x)}{\left ((a+b x)^2+1\right )^2}+\frac {90}{(a+b x)^2+1}+5 \tan ^{-1}(a+b x) \left (6 \sin \left (2 \tan ^{-1}(a+b x)\right )-4 (a+b x)\right )+15\right )\right )}{140 b \Gamma \left (\frac {11}{6}\right ) \Gamma \left (\frac {7}{3}\right ) \sqrt [3]{c \left (a^2+2 a b x+b^2 x^2+1\right )}}$

[In]

Integrate[((a + b*x)^2*ArcTan[a + b*x])/((1 + a^2)*c + 2*a*b*c*x + b^2*c*x^2)^(1/3),x]

[Out]

(-3*(1 + a^2 + 2*a*b*x + b^2*x^2)^(1/3)*(1 + (a + b*x)^2)^(2/3)*((5*2^(1/3)*Sqrt[Pi]*Gamma[5/3]*Hypergeometric
PFQ[{1, 4/3, 4/3}, {11/6, 7/3}, (1 + (a + b*x)^2)^(-1)])/(1 + (a + b*x)^2)^2 + Gamma[11/6]*Gamma[7/3]*(15 + 90
/(1 + (a + b*x)^2) + (24*(a + b*x)*ArcTan[a + b*x]*Hypergeometric2F1[1, 4/3, 11/6, (1 + (a + b*x)^2)^(-1)])/(1
+ (a + b*x)^2)^2 + 5*ArcTan[a + b*x]*(-4*(a + b*x) + 6*Sin[2*ArcTan[a + b*x]]))))/(140*b*(c*(1 + a^2 + 2*a*b*
x + b^2*x^2))^(1/3)*Gamma[11/6]*Gamma[7/3])