3.751 \(\int \frac {x^3 \sqrt {\tan ^{-1}(a x)}}{(c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=215 \[ \frac {3 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{4 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} C\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{12 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {3 \sqrt {\tan ^{-1}(a x)}}{4 a^4 c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 x^2+1} \sqrt {\tan ^{-1}(a x)} \cos \left (3 \tan ^{-1}(a x)\right )}{12 a^4 c^2 \sqrt {a^2 c x^2+c}} \]

[Out]

-1/72*FresnelC(6^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*6^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)/a^4/c^2/(a^2*c*x^2+c)^(1
/2)+3/8*FresnelC(2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)/a^4/c^2/(a^2*c*x^2+c)^
(1/2)-3/4*arctan(a*x)^(1/2)/a^4/c^2/(a^2*c*x^2+c)^(1/2)+1/12*cos(3*arctan(a*x))*(a^2*x^2+1)^(1/2)*arctan(a*x)^
(1/2)/a^4/c^2/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4971, 4970, 3312, 3296, 3304, 3352} \[ \frac {3 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{4 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} \text {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{12 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {3 \sqrt {\tan ^{-1}(a x)}}{4 a^4 c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 x^2+1} \sqrt {\tan ^{-1}(a x)} \cos \left (3 \tan ^{-1}(a x)\right )}{12 a^4 c^2 \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Sqrt[ArcTan[a*x]])/(c + a^2*c*x^2)^(5/2),x]

[Out]

(-3*Sqrt[ArcTan[a*x]])/(4*a^4*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[1 + a^2*x^2]*Sqrt[ArcTan[a*x]]*Cos[3*ArcTan[a*x
]])/(12*a^4*c^2*Sqrt[c + a^2*c*x^2]) + (3*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])
/(4*a^4*c^2*Sqrt[c + a^2*c*x^2]) - (Sqrt[Pi/6]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/(12*a
^4*c^2*Sqrt[c + a^2*c*x^2])

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(f*x^2)/d],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 4970

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c^(m
+ 1), Subst[Int[((a + b*x)^p*Sin[x]^m)/Cos[x]^(m + 2*(q + 1)), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d,
e, p}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q] || GtQ[d, 0])

Rule 4971

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[(d^(q +
1/2)*Sqrt[1 + c^2*x^2])/Sqrt[d + e*x^2], Int[x^m*(1 + c^2*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] &&  !(IntegerQ[q] || GtQ[d, 0])

Rubi steps

\begin {align*} \int \frac {x^3 \sqrt {\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=\frac {\sqrt {1+a^2 x^2} \int \frac {x^3 \sqrt {\tan ^{-1}(a x)}}{\left (1+a^2 x^2\right )^{5/2}} \, dx}{c^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \sqrt {x} \sin ^3(x) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \left (\frac {3}{4} \sqrt {x} \sin (x)-\frac {1}{4} \sqrt {x} \sin (3 x)\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \sqrt {x} \sin (3 x) \, dx,x,\tan ^{-1}(a x)\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \sqrt {x} \sin (x) \, dx,x,\tan ^{-1}(a x)\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {3 \sqrt {\tan ^{-1}(a x)}}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \sqrt {\tan ^{-1}(a x)} \cos \left (3 \tan ^{-1}(a x)\right )}{12 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{24 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{8 a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {3 \sqrt {\tan ^{-1}(a x)}}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \sqrt {\tan ^{-1}(a x)} \cos \left (3 \tan ^{-1}(a x)\right )}{12 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \cos \left (3 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{12 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {3 \sqrt {\tan ^{-1}(a x)}}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \sqrt {\tan ^{-1}(a x)} \cos \left (3 \tan ^{-1}(a x)\right )}{12 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} C\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{12 a^4 c^2 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.53, size = 324, normalized size = 1.51 \[ \frac {-144 a^2 x^2 \tan ^{-1}(a x)+i a^2 x^2 \sqrt {3 a^2 x^2+3} \sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},-3 i \tan ^{-1}(a x)\right )-i a^2 x^2 \sqrt {3 a^2 x^2+3} \sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},3 i \tan ^{-1}(a x)\right )-27 i \left (a^2 x^2+1\right )^{3/2} \sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},-i \tan ^{-1}(a x)\right )+27 i \left (a^2 x^2+1\right )^{3/2} \sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},i \tan ^{-1}(a x)\right )+i \sqrt {3 a^2 x^2+3} \sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},-3 i \tan ^{-1}(a x)\right )-i \sqrt {3 a^2 x^2+3} \sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},3 i \tan ^{-1}(a x)\right )-96 \tan ^{-1}(a x)}{144 a^4 c^2 \left (a^2 x^2+1\right ) \sqrt {a^2 c x^2+c} \sqrt {\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*Sqrt[ArcTan[a*x]])/(c + a^2*c*x^2)^(5/2),x]

[Out]

(-96*ArcTan[a*x] - 144*a^2*x^2*ArcTan[a*x] - (27*I)*(1 + a^2*x^2)^(3/2)*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)
*ArcTan[a*x]] + (27*I)*(1 + a^2*x^2)^(3/2)*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, I*ArcTan[a*x]] + I*Sqrt[3 + 3*a^2*x^
2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-3*I)*ArcTan[a*x]] + I*a^2*x^2*Sqrt[3 + 3*a^2*x^2]*Sqrt[(-I)*ArcTan[a*x]
]*Gamma[1/2, (-3*I)*ArcTan[a*x]] - I*Sqrt[3 + 3*a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (3*I)*ArcTan[a*x]] - I
*a^2*x^2*Sqrt[3 + 3*a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (3*I)*ArcTan[a*x]])/(144*a^4*c^2*(1 + a^2*x^2)*Sqr
t[c + a^2*c*x^2]*Sqrt[ArcTan[a*x]])

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fricas [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^3*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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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^3*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),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 [F]  time = 9.94, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \sqrt {\arctan \left (a x \right )}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x)

[Out]

int(x^3*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,\sqrt {\mathrm {atan}\left (a\,x\right )}}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*atan(a*x)^(1/2))/(c + a^2*c*x^2)^(5/2),x)

[Out]

int((x^3*atan(a*x)^(1/2))/(c + a^2*c*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \sqrt {\operatorname {atan}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atan(a*x)**(1/2)/(a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(x**3*sqrt(atan(a*x))/(c*(a**2*x**2 + 1))**(5/2), x)

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