3.1195 \(\int x (d+e x^2)^{5/2} (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=233 \[ \frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e}-\frac {b \left (c^2 d-e\right )^{7/2} \tan ^{-1}\left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{7 c^7 e}-\frac {b x \left (11 c^2 d-6 e\right ) \left (d+e x^2\right )^{3/2}}{168 c^3}-\frac {b x \left (19 c^4 d^2-22 c^2 d e+8 e^2\right ) \sqrt {d+e x^2}}{112 c^5}-\frac {b \left (35 c^6 d^3-70 c^4 d^2 e+56 c^2 d e^2-16 e^3\right ) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{112 c^7 \sqrt {e}}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c} \]

[Out]

-1/168*b*(11*c^2*d-6*e)*x*(e*x^2+d)^(3/2)/c^3-1/42*b*x*(e*x^2+d)^(5/2)/c+1/7*(e*x^2+d)^(7/2)*(a+b*arctan(c*x))
/e-1/7*b*(c^2*d-e)^(7/2)*arctan(x*(c^2*d-e)^(1/2)/(e*x^2+d)^(1/2))/c^7/e-1/112*b*(35*c^6*d^3-70*c^4*d^2*e+56*c
^2*d*e^2-16*e^3)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/c^7/e^(1/2)-1/112*b*(19*c^4*d^2-22*c^2*d*e+8*e^2)*x*(e*x^2
+d)^(1/2)/c^5

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Rubi [A]  time = 0.33, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {4974, 416, 528, 523, 217, 206, 377, 203} \[ \frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e}-\frac {b x \left (19 c^4 d^2-22 c^2 d e+8 e^2\right ) \sqrt {d+e x^2}}{112 c^5}-\frac {b \left (-70 c^4 d^2 e+35 c^6 d^3+56 c^2 d e^2-16 e^3\right ) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{112 c^7 \sqrt {e}}-\frac {b x \left (11 c^2 d-6 e\right ) \left (d+e x^2\right )^{3/2}}{168 c^3}-\frac {b \left (c^2 d-e\right )^{7/2} \tan ^{-1}\left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{7 c^7 e}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(19*c^4*d^2 - 22*c^2*d*e + 8*e^2)*x*Sqrt[d + e*x^2])/(112*c^5) - (b*(11*c^2*d - 6*e)*x*(d + e*x^2)^(3/2))/
(168*c^3) - (b*x*(d + e*x^2)^(5/2))/(42*c) + ((d + e*x^2)^(7/2)*(a + b*ArcTan[c*x]))/(7*e) - (b*(c^2*d - e)^(7
/2)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(7*c^7*e) - (b*(35*c^6*d^3 - 70*c^4*d^2*e + 56*c^2*d*e^2 - 16
*e^3)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(112*c^7*Sqrt[e])

Rule 203

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

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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 528

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

Rule 4974

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

Rubi steps

\begin {align*} \int x \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e}-\frac {(b c) \int \frac {\left (d+e x^2\right )^{7/2}}{1+c^2 x^2} \, dx}{7 e}\\ &=-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e}-\frac {b \int \frac {\left (d+e x^2\right )^{3/2} \left (d \left (6 c^2 d-e\right )+\left (11 c^2 d-6 e\right ) e x^2\right )}{1+c^2 x^2} \, dx}{42 c e}\\ &=-\frac {b \left (11 c^2 d-6 e\right ) x \left (d+e x^2\right )^{3/2}}{168 c^3}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e}-\frac {b \int \frac {\sqrt {d+e x^2} \left (3 d \left (8 c^4 d^2-5 c^2 d e+2 e^2\right )+3 e \left (19 c^4 d^2-22 c^2 d e+8 e^2\right ) x^2\right )}{1+c^2 x^2} \, dx}{168 c^3 e}\\ &=-\frac {b \left (19 c^4 d^2-22 c^2 d e+8 e^2\right ) x \sqrt {d+e x^2}}{112 c^5}-\frac {b \left (11 c^2 d-6 e\right ) x \left (d+e x^2\right )^{3/2}}{168 c^3}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e}-\frac {b \int \frac {3 d \left (16 c^6 d^3-29 c^4 d^2 e+26 c^2 d e^2-8 e^3\right )+3 e \left (35 c^6 d^3-70 c^4 d^2 e+56 c^2 d e^2-16 e^3\right ) x^2}{\left (1+c^2 x^2\right ) \sqrt {d+e x^2}} \, dx}{336 c^5 e}\\ &=-\frac {b \left (19 c^4 d^2-22 c^2 d e+8 e^2\right ) x \sqrt {d+e x^2}}{112 c^5}-\frac {b \left (11 c^2 d-6 e\right ) x \left (d+e x^2\right )^{3/2}}{168 c^3}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e}-\frac {\left (b \left (c^2 d-e\right )^4\right ) \int \frac {1}{\left (1+c^2 x^2\right ) \sqrt {d+e x^2}} \, dx}{7 c^7 e}-\frac {\left (b \left (35 c^6 d^3-70 c^4 d^2 e+56 c^2 d e^2-16 e^3\right )\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{112 c^7}\\ &=-\frac {b \left (19 c^4 d^2-22 c^2 d e+8 e^2\right ) x \sqrt {d+e x^2}}{112 c^5}-\frac {b \left (11 c^2 d-6 e\right ) x \left (d+e x^2\right )^{3/2}}{168 c^3}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e}-\frac {\left (b \left (c^2 d-e\right )^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-\left (-c^2 d+e\right ) x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{7 c^7 e}-\frac {\left (b \left (35 c^6 d^3-70 c^4 d^2 e+56 c^2 d e^2-16 e^3\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{112 c^7}\\ &=-\frac {b \left (19 c^4 d^2-22 c^2 d e+8 e^2\right ) x \sqrt {d+e x^2}}{112 c^5}-\frac {b \left (11 c^2 d-6 e\right ) x \left (d+e x^2\right )^{3/2}}{168 c^3}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e}-\frac {b \left (c^2 d-e\right )^{7/2} \tan ^{-1}\left (\frac {\sqrt {c^2 d-e} x}{\sqrt {d+e x^2}}\right )}{7 c^7 e}-\frac {b \left (35 c^6 d^3-70 c^4 d^2 e+56 c^2 d e^2-16 e^3\right ) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{112 c^7 \sqrt {e}}\\ \end {align*}

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Mathematica [C]  time = 0.58, size = 353, normalized size = 1.52 \[ \frac {c^2 \sqrt {d+e x^2} \left (48 a c^5 \left (d+e x^2\right )^3-b e x \left (c^4 \left (87 d^2+38 d e x^2+8 e^2 x^4\right )-6 c^2 e \left (13 d+2 e x^2\right )+24 e^2\right )\right )+48 b c^7 \tan ^{-1}(c x) \left (d+e x^2\right )^{7/2}-24 i b \left (c^2 d-e\right )^{7/2} \log \left (\frac {28 c^8 e \left (-i \sqrt {c^2 d-e} \sqrt {d+e x^2}-i c d+e x\right )}{b (c x-i) \left (c^2 d-e\right )^{9/2}}\right )+24 i b \left (c^2 d-e\right )^{7/2} \log \left (\frac {28 c^8 e \left (i \sqrt {c^2 d-e} \sqrt {d+e x^2}+i c d+e x\right )}{b (c x+i) \left (c^2 d-e\right )^{9/2}}\right )+3 b \sqrt {e} \left (-35 c^6 d^3+70 c^4 d^2 e-56 c^2 d e^2+16 e^3\right ) \log \left (\sqrt {e} \sqrt {d+e x^2}+e x\right )}{336 c^7 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Sqrt[d + e*x^2]*(48*a*c^5*(d + e*x^2)^3 - b*e*x*(24*e^2 - 6*c^2*e*(13*d + 2*e*x^2) + c^4*(87*d^2 + 38*d*e
*x^2 + 8*e^2*x^4))) + 48*b*c^7*(d + e*x^2)^(7/2)*ArcTan[c*x] - (24*I)*b*(c^2*d - e)^(7/2)*Log[(28*c^8*e*((-I)*
c*d + e*x - I*Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(9/2)*(-I + c*x))] + (24*I)*b*(c^2*d - e)^(7/2)
*Log[(28*c^8*e*(I*c*d + e*x + I*Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(9/2)*(I + c*x))] + 3*b*Sqrt[
e]*(-35*c^6*d^3 + 70*c^4*d^2*e - 56*c^2*d*e^2 + 16*e^3)*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(336*c^7*e)

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fricas [A]  time = 15.58, size = 1562, normalized size = 6.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/672*(3*(35*b*c^6*d^3 - 70*b*c^4*d^2*e + 56*b*c^2*d*e^2 - 16*b*e^3)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d
)*sqrt(e)*x - d) + 24*(b*c^6*d^3 - 3*b*c^4*d^2*e + 3*b*c^2*d*e^2 - b*e^3)*sqrt(-c^2*d + e)*log(((c^4*d^2 - 8*c
^2*d*e + 8*e^2)*x^4 - 2*(3*c^2*d^2 - 4*d*e)*x^2 + 4*((c^2*d - 2*e)*x^3 - d*x)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d)
+ d^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) - 2*(48*a*c^7*e^3*x^6 + 144*a*c^7*d*e^2*x^4 - 8*b*c^6*e^3*x^5 + 144*a*c^7*d
^2*e*x^2 + 48*a*c^7*d^3 - 2*(19*b*c^6*d*e^2 - 6*b*c^4*e^3)*x^3 - 3*(29*b*c^6*d^2*e - 26*b*c^4*d*e^2 + 8*b*c^2*
e^3)*x + 48*(b*c^7*e^3*x^6 + 3*b*c^7*d*e^2*x^4 + 3*b*c^7*d^2*e*x^2 + b*c^7*d^3)*arctan(c*x))*sqrt(e*x^2 + d))/
(c^7*e), -1/672*(48*(b*c^6*d^3 - 3*b*c^4*d^2*e + 3*b*c^2*d*e^2 - b*e^3)*sqrt(c^2*d - e)*arctan(1/2*sqrt(c^2*d
- e)*((c^2*d - 2*e)*x^2 - d)*sqrt(e*x^2 + d)/((c^2*d*e - e^2)*x^3 + (c^2*d^2 - d*e)*x)) + 3*(35*b*c^6*d^3 - 70
*b*c^4*d^2*e + 56*b*c^2*d*e^2 - 16*b*e^3)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - 2*(48*a*c^
7*e^3*x^6 + 144*a*c^7*d*e^2*x^4 - 8*b*c^6*e^3*x^5 + 144*a*c^7*d^2*e*x^2 + 48*a*c^7*d^3 - 2*(19*b*c^6*d*e^2 - 6
*b*c^4*e^3)*x^3 - 3*(29*b*c^6*d^2*e - 26*b*c^4*d*e^2 + 8*b*c^2*e^3)*x + 48*(b*c^7*e^3*x^6 + 3*b*c^7*d*e^2*x^4
+ 3*b*c^7*d^2*e*x^2 + b*c^7*d^3)*arctan(c*x))*sqrt(e*x^2 + d))/(c^7*e), 1/336*(3*(35*b*c^6*d^3 - 70*b*c^4*d^2*
e + 56*b*c^2*d*e^2 - 16*b*e^3)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - 12*(b*c^6*d^3 - 3*b*c^4*d^2*e + 3
*b*c^2*d*e^2 - b*e^3)*sqrt(-c^2*d + e)*log(((c^4*d^2 - 8*c^2*d*e + 8*e^2)*x^4 - 2*(3*c^2*d^2 - 4*d*e)*x^2 + 4*
((c^2*d - 2*e)*x^3 - d*x)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d) + d^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) + (48*a*c^7*e^3*x
^6 + 144*a*c^7*d*e^2*x^4 - 8*b*c^6*e^3*x^5 + 144*a*c^7*d^2*e*x^2 + 48*a*c^7*d^3 - 2*(19*b*c^6*d*e^2 - 6*b*c^4*
e^3)*x^3 - 3*(29*b*c^6*d^2*e - 26*b*c^4*d*e^2 + 8*b*c^2*e^3)*x + 48*(b*c^7*e^3*x^6 + 3*b*c^7*d*e^2*x^4 + 3*b*c
^7*d^2*e*x^2 + b*c^7*d^3)*arctan(c*x))*sqrt(e*x^2 + d))/(c^7*e), -1/336*(24*(b*c^6*d^3 - 3*b*c^4*d^2*e + 3*b*c
^2*d*e^2 - b*e^3)*sqrt(c^2*d - e)*arctan(1/2*sqrt(c^2*d - e)*((c^2*d - 2*e)*x^2 - d)*sqrt(e*x^2 + d)/((c^2*d*e
- e^2)*x^3 + (c^2*d^2 - d*e)*x)) - 3*(35*b*c^6*d^3 - 70*b*c^4*d^2*e + 56*b*c^2*d*e^2 - 16*b*e^3)*sqrt(-e)*arc
tan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (48*a*c^7*e^3*x^6 + 144*a*c^7*d*e^2*x^4 - 8*b*c^6*e^3*x^5 + 144*a*c^7*d^2*e*
x^2 + 48*a*c^7*d^3 - 2*(19*b*c^6*d*e^2 - 6*b*c^4*e^3)*x^3 - 3*(29*b*c^6*d^2*e - 26*b*c^4*d*e^2 + 8*b*c^2*e^3)*
x + 48*(b*c^7*e^3*x^6 + 3*b*c^7*d*e^2*x^4 + 3*b*c^7*d^2*e*x^2 + b*c^7*d^3)*arctan(c*x))*sqrt(e*x^2 + d))/(c^7*
e)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [F]  time = 0.97, size = 0, normalized size = 0.00 \[ \int x \left (e \,x^{2}+d \right )^{\frac {5}{2}} \left (a +b \arctan \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(x*(e*x^2+d)^(5/2)*(a+b*arctan(c*x)),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(e-c^2*d>0)', see `assume?` for
more details)Is e-c^2*d zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(a + b*atan(c*x))*(d + e*x**2)**(5/2), x)

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