3.1.41 \(\int (e+f x)^m (a+b \text {ArcTan}(c+d x)) \, dx\) [41]

Optimal. Leaf size=177 \[ \frac {(e+f x)^{1+m} (a+b \text {ArcTan}(c+d x))}{f (1+m)}-\frac {i b d (e+f x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac {d (e+f x)}{d e+i f-c f}\right )}{2 f (d e+(i-c) f) (1+m) (2+m)}+\frac {i b d (e+f x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac {d (e+f x)}{d e-(i+c) f}\right )}{2 f (d e-(i+c) f) (1+m) (2+m)} \]

[Out]

(f*x+e)^(1+m)*(a+b*arctan(d*x+c))/f/(1+m)-1/2*I*b*d*(f*x+e)^(2+m)*hypergeom([1, 2+m],[3+m],d*(f*x+e)/(d*e+I*f-
c*f))/f/(d*e+(I-c)*f)/(1+m)/(2+m)+1/2*I*b*d*(f*x+e)^(2+m)*hypergeom([1, 2+m],[3+m],d*(f*x+e)/(d*e-(I+c)*f))/f/
(d*e-(I+c)*f)/(1+m)/(2+m)

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Rubi [A]
time = 0.22, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5155, 4972, 726, 70} \begin {gather*} \frac {(e+f x)^{m+1} (a+b \text {ArcTan}(c+d x))}{f (m+1)}-\frac {i b d (e+f x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac {d (e+f x)}{d e-c f+i f}\right )}{2 f (m+1) (m+2) (d e+(-c+i) f)}+\frac {i b d (e+f x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac {d (e+f x)}{d e-(c+i) f}\right )}{2 f (m+1) (m+2) (d e-(c+i) f)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^m*(a + b*ArcTan[c + d*x]),x]

[Out]

((e + f*x)^(1 + m)*(a + b*ArcTan[c + d*x]))/(f*(1 + m)) - ((I/2)*b*d*(e + f*x)^(2 + m)*Hypergeometric2F1[1, 2
+ m, 3 + m, (d*(e + f*x))/(d*e + I*f - c*f)])/(f*(d*e + (I - c)*f)*(1 + m)*(2 + m)) + ((I/2)*b*d*(e + f*x)^(2
+ m)*Hypergeometric2F1[1, 2 + m, 3 + m, (d*(e + f*x))/(d*e - (I + c)*f)])/(f*(d*e - (I + c)*f)*(1 + m)*(2 + m)
)

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 726

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 4972

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((a + b*
ArcTan[c*x])/(e*(q + 1))), x] - Dist[b*(c/(e*(q + 1))), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 5155

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x]
&& IGtQ[p, 0]

Rubi steps

\begin {align*} \int (e+f x)^m \left (a+b \tan ^{-1}(c+d x)\right ) \, dx &=\frac {\text {Subst}\left (\int \left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^m \left (a+b \tan ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {(e+f x)^{1+m} \left (a+b \tan ^{-1}(c+d x)\right )}{f (1+m)}-\frac {b \text {Subst}\left (\int \frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{1+x^2} \, dx,x,c+d x\right )}{f (1+m)}\\ &=\frac {(e+f x)^{1+m} \left (a+b \tan ^{-1}(c+d x)\right )}{f (1+m)}-\frac {b \text {Subst}\left (\int \left (\frac {i \left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{2 (i-x)}+\frac {i \left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{2 (i+x)}\right ) \, dx,x,c+d x\right )}{f (1+m)}\\ &=\frac {(e+f x)^{1+m} \left (a+b \tan ^{-1}(c+d x)\right )}{f (1+m)}-\frac {(i b) \text {Subst}\left (\int \frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{i-x} \, dx,x,c+d x\right )}{2 f (1+m)}-\frac {(i b) \text {Subst}\left (\int \frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{i+x} \, dx,x,c+d x\right )}{2 f (1+m)}\\ &=\frac {(e+f x)^{1+m} \left (a+b \tan ^{-1}(c+d x)\right )}{f (1+m)}-\frac {i b d (e+f x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac {d (e+f x)}{d e+i f-c f}\right )}{2 f (d e+(i-c) f) (1+m) (2+m)}+\frac {i b d (e+f x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac {d (e+f x)}{d e-(i+c) f}\right )}{2 f (d e-(i+c) f) (1+m) (2+m)}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 162, normalized size = 0.92 \begin {gather*} \frac {(e+f x)^{1+m} \left (2 (a+b \text {ArcTan}(c+d x))+\frac {b d (e+f x) \left ((d e-(i+c) f) \, _2F_1\left (1,2+m;3+m;\frac {d (e+f x)}{d e-(-i+c) f}\right )+(-d e+(-i+c) f) \, _2F_1\left (1,2+m;3+m;\frac {d (e+f x)}{d e-(i+c) f}\right )\right )}{(i d e+f-i c f) (d e-(-i+c) f) (2+m)}\right )}{2 f (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^m*(a + b*ArcTan[c + d*x]),x]

[Out]

((e + f*x)^(1 + m)*(2*(a + b*ArcTan[c + d*x]) + (b*d*(e + f*x)*((d*e - (I + c)*f)*Hypergeometric2F1[1, 2 + m,
3 + m, (d*(e + f*x))/(d*e - (-I + c)*f)] + (-(d*e) + (-I + c)*f)*Hypergeometric2F1[1, 2 + m, 3 + m, (d*(e + f*
x))/(d*e - (I + c)*f)]))/((I*d*e + f - I*c*f)*(d*e - (-I + c)*f)*(2 + m))))/(2*f*(1 + m))

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \left (f x +e \right )^{m} \left (a +b \arctan \left (d x +c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^m*(a+b*arctan(d*x+c)),x)

[Out]

int((f*x+e)^m*(a+b*arctan(d*x+c)),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^m*(a+b*arctan(d*x+c)),x, algorithm="maxima")

[Out]

1/2*((3*m^2*e + (3*f*m^2 + 2*f*m + f)*x + 2*m*e + e)*(f*x + e)^m*arctan(d*x + c) + ((f*m + f)*x + m*e + e)*(f*
x + e)^m*log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 2*(f*m^3 + f*m^2 + f*m + f)*integrate(-1/2*(2*((c^2 + 1)*f*m^3 + 2
*(c^2 + 1)*f*m^2 + (c^2 + 1)*f*m + (d^2*f*m^3 + 2*d^2*f*m^2 + d^2*f*m)*x^2 + 2*(c*d*f*m^3 + 2*c*d*f*m^2 + c*d*
f*m)*x)*(f*x + e)^m*arctan(d*x + c) - ((c^2 + 1)*f*m^3 - (c^2 + 1)*f*m + (d^2*f*m^3 - d^2*f*m)*x^2 + 2*(c*d*f*
m^3 - c*d*f*m)*x)*(f*x + e)^m*log(d^2*x^2 + 2*c*d*x + c^2 + 1) - 2*((c*e - e)*d*m^2 - (c*e + e)*d*m + (d^2*f*m
^2 - d^2*f*m)*x^2 + (((c - 1)*d*f + d^2*e)*m^2 - ((c + 1)*d*f + d^2*e)*m)*x)*(f*x + e)^m)/((c^2 + 1)*f*m^3 + (
c^2 + 1)*f*m^2 + (c^2 + 1)*f*m + (d^2*f*m^3 + d^2*f*m^2 + d^2*f*m + d^2*f)*x^2 + (c^2 + 1)*f + 2*(c*d*f*m^3 +
c*d*f*m^2 + c*d*f*m + c*d*f)*x), x))*b/(f*m^3 + f*m^2 + f*m + f) + (f*x + e)^(m + 1)*a/(f*(m + 1))

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^m*(a+b*arctan(d*x+c)),x, algorithm="fricas")

[Out]

integral((b*arctan(d*x + c) + a)*(f*x + e)^m, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**m*(a+b*atan(d*x+c)),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^m*(a+b*arctan(d*x+c)),x, algorithm="giac")

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e+f\,x\right )}^m\,\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)^m*(a + b*atan(c + d*x)),x)

[Out]

int((e + f*x)^m*(a + b*atan(c + d*x)), x)

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