3.218 \(\int x (c+a^2 c x^2)^{5/2} \tan ^{-1}(a x) \, dx\)

Optimal. Leaf size=134 \[ -\frac {5 c^{5/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{112 a^2}-\frac {5 c^2 x \sqrt {a^2 c x^2+c}}{112 a}-\frac {x \left (a^2 c x^2+c\right )^{5/2}}{42 a}-\frac {5 c x \left (a^2 c x^2+c\right )^{3/2}}{168 a}+\frac {\left (a^2 c x^2+c\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c} \]

[Out]

-5/168*c*x*(a^2*c*x^2+c)^(3/2)/a-1/42*x*(a^2*c*x^2+c)^(5/2)/a+1/7*(a^2*c*x^2+c)^(7/2)*arctan(a*x)/a^2/c-5/112*
c^(5/2)*arctanh(a*x*c^(1/2)/(a^2*c*x^2+c)^(1/2))/a^2-5/112*c^2*x*(a^2*c*x^2+c)^(1/2)/a

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Rubi [A]  time = 0.09, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4930, 195, 217, 206} \[ -\frac {5 c^2 x \sqrt {a^2 c x^2+c}}{112 a}-\frac {5 c^{5/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{112 a^2}-\frac {x \left (a^2 c x^2+c\right )^{5/2}}{42 a}-\frac {5 c x \left (a^2 c x^2+c\right )^{3/2}}{168 a}+\frac {\left (a^2 c x^2+c\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(c + a^2*c*x^2)^(5/2)*ArcTan[a*x],x]

[Out]

(-5*c^2*x*Sqrt[c + a^2*c*x^2])/(112*a) - (5*c*x*(c + a^2*c*x^2)^(3/2))/(168*a) - (x*(c + a^2*c*x^2)^(5/2))/(42
*a) + ((c + a^2*c*x^2)^(7/2)*ArcTan[a*x])/(7*a^2*c) - (5*c^(5/2)*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/(
112*a^2)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rubi steps

\begin {align*} \int x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x) \, dx &=\frac {\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac {\int \left (c+a^2 c x^2\right )^{5/2} \, dx}{7 a}\\ &=-\frac {x \left (c+a^2 c x^2\right )^{5/2}}{42 a}+\frac {\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac {(5 c) \int \left (c+a^2 c x^2\right )^{3/2} \, dx}{42 a}\\ &=-\frac {5 c x \left (c+a^2 c x^2\right )^{3/2}}{168 a}-\frac {x \left (c+a^2 c x^2\right )^{5/2}}{42 a}+\frac {\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac {\left (5 c^2\right ) \int \sqrt {c+a^2 c x^2} \, dx}{56 a}\\ &=-\frac {5 c^2 x \sqrt {c+a^2 c x^2}}{112 a}-\frac {5 c x \left (c+a^2 c x^2\right )^{3/2}}{168 a}-\frac {x \left (c+a^2 c x^2\right )^{5/2}}{42 a}+\frac {\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac {\left (5 c^3\right ) \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx}{112 a}\\ &=-\frac {5 c^2 x \sqrt {c+a^2 c x^2}}{112 a}-\frac {5 c x \left (c+a^2 c x^2\right )^{3/2}}{168 a}-\frac {x \left (c+a^2 c x^2\right )^{5/2}}{42 a}+\frac {\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac {\left (5 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )}{112 a}\\ &=-\frac {5 c^2 x \sqrt {c+a^2 c x^2}}{112 a}-\frac {5 c x \left (c+a^2 c x^2\right )^{3/2}}{168 a}-\frac {x \left (c+a^2 c x^2\right )^{5/2}}{42 a}+\frac {\left (c+a^2 c x^2\right )^{7/2} \tan ^{-1}(a x)}{7 a^2 c}-\frac {5 c^{5/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{112 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 111, normalized size = 0.83 \[ \frac {c^2 \left (-15 \sqrt {c} \log \left (\sqrt {c} \sqrt {a^2 c x^2+c}+a c x\right )+48 \left (a^2 x^2+1\right )^3 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)-a x \left (8 a^4 x^4+26 a^2 x^2+33\right ) \sqrt {a^2 c x^2+c}\right )}{336 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(c + a^2*c*x^2)^(5/2)*ArcTan[a*x],x]

[Out]

(c^2*(-(a*x*Sqrt[c + a^2*c*x^2]*(33 + 26*a^2*x^2 + 8*a^4*x^4)) + 48*(1 + a^2*x^2)^3*Sqrt[c + a^2*c*x^2]*ArcTan
[a*x] - 15*Sqrt[c]*Log[a*c*x + Sqrt[c]*Sqrt[c + a^2*c*x^2]]))/(336*a^2)

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fricas [A]  time = 0.62, size = 130, normalized size = 0.97 \[ \frac {15 \, c^{\frac {5}{2}} \log \left (-2 \, a^{2} c x^{2} + 2 \, \sqrt {a^{2} c x^{2} + c} a \sqrt {c} x - c\right ) - 2 \, {\left (8 \, a^{5} c^{2} x^{5} + 26 \, a^{3} c^{2} x^{3} + 33 \, a c^{2} x - 48 \, {\left (a^{6} c^{2} x^{6} + 3 \, a^{4} c^{2} x^{4} + 3 \, a^{2} c^{2} x^{2} + c^{2}\right )} \arctan \left (a x\right )\right )} \sqrt {a^{2} c x^{2} + c}}{672 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a^2*c*x^2+c)^(5/2)*arctan(a*x),x, algorithm="fricas")

[Out]

1/672*(15*c^(5/2)*log(-2*a^2*c*x^2 + 2*sqrt(a^2*c*x^2 + c)*a*sqrt(c)*x - c) - 2*(8*a^5*c^2*x^5 + 26*a^3*c^2*x^
3 + 33*a*c^2*x - 48*(a^6*c^2*x^6 + 3*a^4*c^2*x^4 + 3*a^2*c^2*x^2 + c^2)*arctan(a*x))*sqrt(a^2*c*x^2 + c))/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a^2*c*x^2+c)^(5/2)*arctan(a*x),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.90, size = 205, normalized size = 1.53 \[ \frac {c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (48 \arctan \left (a x \right ) x^{6} a^{6}-8 x^{5} a^{5}+144 \arctan \left (a x \right ) x^{4} a^{4}-26 a^{3} x^{3}+144 \arctan \left (a x \right ) x^{2} a^{2}-33 a x +48 \arctan \left (a x \right )\right )}{336 a^{2}}-\frac {5 c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+i\right )}{112 a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {5 c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-i\right )}{112 a^{2} \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a^2*c*x^2+c)^(5/2)*arctan(a*x),x)

[Out]

1/336*c^2/a^2*(c*(a*x-I)*(I+a*x))^(1/2)*(48*arctan(a*x)*x^6*a^6-8*x^5*a^5+144*arctan(a*x)*x^4*a^4-26*a^3*x^3+1
44*arctan(a*x)*x^2*a^2-33*a*x+48*arctan(a*x))-5/112*c^2/a^2*(c*(a*x-I)*(I+a*x))^(1/2)*ln((1+I*a*x)/(a^2*x^2+1)
^(1/2)+I)/(a^2*x^2+1)^(1/2)+5/112*c^2/a^2*(c*(a*x-I)*(I+a*x))^(1/2)*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-I)/(a^2*x^2
+1)^(1/2)

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maxima [B]  time = 0.69, size = 637, normalized size = 4.75 \[ \frac {560 \, {\left (a^{2} c^{2} x^{2} + c^{2}\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {c} \arctan \left (a x\right ) - 280 \, {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} {\left (a c^{2} x \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right ) + 2 \, c^{2} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right )\right )} \sqrt {c} - {\left ({\left (a {\left (\frac {5 \, {\left (\frac {8 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{3}}{a^{2}} - \frac {6 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{a^{4}} + \frac {3 \, \sqrt {a^{2} x^{2} + 1} x}{a^{4}} + \frac {3 \, \operatorname {arsinh}\left (a x\right )}{a^{5}}\right )}}{a^{2}} - \frac {24 \, {\left (\frac {2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{a^{2}} - \frac {\sqrt {a^{2} x^{2} + 1} x}{a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )}{a^{3}}\right )}}{a^{4}} + \frac {64 \, {\left (\sqrt {a^{2} x^{2} + 1} x + \frac {\operatorname {arsinh}\left (a x\right )}{a}\right )}}{a^{6}}\right )} - 16 \, {\left (\frac {15 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{4}}{a^{2}} - \frac {12 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{a^{4}} + \frac {8 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{6}}\right )} \arctan \left (a x\right )\right )} a^{6} c^{2} + 28 \, {\left (a {\left (\frac {3 \, {\left (\frac {2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{a^{2}} - \frac {\sqrt {a^{2} x^{2} + 1} x}{a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )}{a^{3}}\right )}}{a^{2}} - \frac {8 \, {\left (\sqrt {a^{2} x^{2} + 1} x + \frac {\operatorname {arsinh}\left (a x\right )}{a}\right )}}{a^{4}}\right )} - 8 \, {\left (\frac {3 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{a^{2}} - \frac {2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4}}\right )} \arctan \left (a x\right )\right )} a^{4} c^{2} - 140 \, c^{2} \arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) + 2, a x + {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right ) - 140 \, c^{2} \arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) - 2, -a x + {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right )\right )} \sqrt {c}}{1680 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a^2*c*x^2+c)^(5/2)*arctan(a*x),x, algorithm="maxima")

[Out]

1/1680*(560*(a^2*c^2*x^2 + c^2)*sqrt(a^2*x^2 + 1)*sqrt(c)*arctan(a*x) - 280*(a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*(
a*c^2*x*cos(1/2*arctan2(4*a*x, -a^2*x^2 + 3)) + 2*c^2*sin(1/2*arctan2(4*a*x, -a^2*x^2 + 3)))*sqrt(c) - ((a*(5*
(8*(a^2*x^2 + 1)^(3/2)*x^3/a^2 - 6*(a^2*x^2 + 1)^(3/2)*x/a^4 + 3*sqrt(a^2*x^2 + 1)*x/a^4 + 3*arcsinh(a*x)/a^5)
/a^2 - 24*(2*(a^2*x^2 + 1)^(3/2)*x/a^2 - sqrt(a^2*x^2 + 1)*x/a^2 - arcsinh(a*x)/a^3)/a^4 + 64*(sqrt(a^2*x^2 +
1)*x + arcsinh(a*x)/a)/a^6) - 16*(15*(a^2*x^2 + 1)^(3/2)*x^4/a^2 - 12*(a^2*x^2 + 1)^(3/2)*x^2/a^4 + 8*(a^2*x^2
+ 1)^(3/2)/a^6)*arctan(a*x))*a^6*c^2 + 28*(a*(3*(2*(a^2*x^2 + 1)^(3/2)*x/a^2 - sqrt(a^2*x^2 + 1)*x/a^2 - arcs
inh(a*x)/a^3)/a^2 - 8*(sqrt(a^2*x^2 + 1)*x + arcsinh(a*x)/a)/a^4) - 8*(3*(a^2*x^2 + 1)^(3/2)*x^2/a^2 - 2*(a^2*
x^2 + 1)^(3/2)/a^4)*arctan(a*x))*a^4*c^2 - 140*c^2*arctan2((a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*sin(1/2*arctan2(4*
a*x, a^2*x^2 - 3)) + 2, a*x + (a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*cos(1/2*arctan2(4*a*x, a^2*x^2 - 3))) - 140*c^2
*arctan2((a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*sin(1/2*arctan2(4*a*x, a^2*x^2 - 3)) - 2, -a*x + (a^4*x^4 + 10*a^2*x
^2 + 9)^(1/4)*cos(1/2*arctan2(4*a*x, a^2*x^2 - 3))))*sqrt(c))/a^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atan(a*x)*(c + a^2*c*x^2)^(5/2),x)

[Out]

int(x*atan(a*x)*(c + a^2*c*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a**2*c*x**2+c)**(5/2)*atan(a*x),x)

[Out]

Integral(x*(c*(a**2*x**2 + 1))**(5/2)*atan(a*x), x)

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