3.1.16 \(\int (\frac {x^2}{\tan ^{\frac {3}{2}}(a+b x)}-\frac {4 x}{b \sqrt {\tan (a+b x)}}+x^2 \sqrt {\tan (a+b x)}) \, dx\) [16]
Optimal. Leaf size=18 \[ -\frac {2 x^2}{b \sqrt {\tan (a+b x)}} \]
[Out]
-2*x^2/b/tan(b*x+a)^(1/2)
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Rubi [A]
time = 1.93, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps
used = 76, number of rules used = 12, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3802, 3819,
209, 281, 6857, 6139, 6057, 2449, 2352, 2497, 6131, 6055} \begin {gather*} -\frac {2 x^2}{b \sqrt {\tan (a+b x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[x^2/Tan[a + b*x]^(3/2) - (4*x)/(b*Sqrt[Tan[a + b*x]]) + x^2*Sqrt[Tan[a + b*x]],x]
[Out]
(-2*x^2)/(b*Sqrt[Tan[a + b*x]])
Rule 209
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 281
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Rule 2352
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]
Rule 2449
Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]
Rule 2497
Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]
Rule 3802
Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(c + d*x)^m*((b*Tan[e +
f*x])^(n + 1)/(b*f*(n + 1))), x] + (-Dist[d*(m/(b*f*(n + 1))), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n + 1)
, x], x] - Dist[1/b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n + 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && LtQ[n
, -1] && GtQ[m, 0]
Rule 3819
Int[((c_.) + (d_.)*(x_))/Sqrt[(a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-I)*((c + d*x)/(f*Rt[
a - I*b, 2]))*ArcTanh[Sqrt[a + b*Tan[e + f*x]]/Rt[a - I*b, 2]], x] + (Dist[I*(d/(f*Rt[a - I*b, 2])), Int[ArcTa
nh[Sqrt[a + b*Tan[e + f*x]]/Rt[a - I*b, 2]], x], x] - Dist[I*(d/(f*Rt[a + I*b, 2])), Int[ArcTanh[Sqrt[a + b*Ta
n[e + f*x]]/Rt[a + I*b, 2]], x], x] + Simp[I*((c + d*x)/(f*Rt[a + I*b, 2]))*ArcTanh[Sqrt[a + b*Tan[e + f*x]]/R
t[a + I*b, 2]], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2, 0]
Rule 6055
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^p)
*(Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 - c^
2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]
Rule 6057
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x]))*(Log[2/
(1 + c*x)]/e), x] + (Dist[b*(c/e), Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Dist[b*(c/e), Int[Log[2*c*((d
+ e*x)/((c*d + e)*(1 + c*x)))]/(1 - c^2*x^2), x], x] + Simp[(a + b*ArcTanh[c*x])*(Log[2*c*((d + e*x)/((c*d + e
)*(1 + c*x)))]/e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]
Rule 6131
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]
Rule 6139
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[a
+ b*ArcTanh[c*x], x^m/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[m] && !(EqQ[m, 1] && NeQ[
a, 0])
Rule 6857
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]
Rubi steps
\begin {align*} \int \left (\frac {x^2}{\tan ^{\frac {3}{2}}(a+b x)}-\frac {4 x}{b \sqrt {\tan (a+b x)}}+x^2 \sqrt {\tan (a+b x)}\right ) \, dx &=-\frac {4 \int \frac {x}{\sqrt {\tan (a+b x)}} \, dx}{b}+\int \frac {x^2}{\tan ^{\frac {3}{2}}(a+b x)} \, dx+\int x^2 \sqrt {\tan (a+b x)} \, dx\\ &=-\frac {2 x^2}{b \sqrt {\tan (a+b x)}}\\ \end {align*}
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Mathematica [A]
time = 0.70, size = 18, normalized size = 1.00 \begin {gather*} -\frac {2 x^2}{b \sqrt {\tan (a+b x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[x^2/Tan[a + b*x]^(3/2) - (4*x)/(b*Sqrt[Tan[a + b*x]]) + x^2*Sqrt[Tan[a + b*x]],x]
[Out]
(-2*x^2)/(b*Sqrt[Tan[a + b*x]])
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Maple [F]
time = 0.36, size = 0, normalized size = 0.00 \[\int -\frac {4 x}{b \sqrt {\tan \left (b x +a \right )}}+x^{2} \left (\sqrt {\tan }\left (b x +a \right )\right )+\frac {x^{2}}{\tan \left (b x +a \right )^{\frac {3}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(-4*x/b/tan(b*x+a)^(1/2)+x^2*tan(b*x+a)^(1/2)+x^2/tan(b*x+a)^(3/2),x)
[Out]
int(-4*x/b/tan(b*x+a)^(1/2)+x^2*tan(b*x+a)^(1/2)+x^2/tan(b*x+a)^(3/2),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(-4*x/b/tan(b*x+a)^(1/2)+x^2*tan(b*x+a)^(1/2)+x^2/tan(b*x+a)^(3/2),x, algorithm="maxima")
[Out]
integrate(x^2*sqrt(tan(b*x + a)) + x^2/tan(b*x + a)^(3/2) - 4*x/(b*sqrt(tan(b*x + a))), x)
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Fricas [A]
time = 0.41, size = 16, normalized size = 0.89 \begin {gather*} -\frac {2 \, x^{2}}{b \sqrt {\tan \left (b x + a\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(-4*x/b/tan(b*x+a)^(1/2)+x^2*tan(b*x+a)^(1/2)+x^2/tan(b*x+a)^(3/2),x, algorithm="fricas")
[Out]
-2*x^2/(b*sqrt(tan(b*x + a)))
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \left (- \frac {4 x}{\sqrt {\tan {\left (a + b x \right )}}}\right )\, dx + \int \frac {b x^{2}}{\tan ^{\frac {3}{2}}{\left (a + b x \right )}}\, dx + \int b x^{2} \sqrt {\tan {\left (a + b x \right )}}\, dx}{b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(-4*x/b/tan(b*x+a)**(1/2)+x**2*tan(b*x+a)**(1/2)+x**2/tan(b*x+a)**(3/2),x)
[Out]
(Integral(-4*x/sqrt(tan(a + b*x)), x) + Integral(b*x**2/tan(a + b*x)**(3/2), x) + Integral(b*x**2*sqrt(tan(a +
b*x)), x))/b
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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(-4*x/b/tan(b*x+a)^(1/2)+x^2*tan(b*x+a)^(1/2)+x^2/tan(b*x+a)^(3/2),x, algorithm="giac")
[Out]
integrate(x^2*sqrt(tan(b*x + a)) + x^2/tan(b*x + a)^(3/2) - 4*x/(b*sqrt(tan(b*x + a))), x)
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Mupad [B]
time = 3.12, size = 50, normalized size = 2.78 \begin {gather*} -\frac {x^2\,\sin \left (2\,a+2\,b\,x\right )\,\sqrt {\frac {\sin \left (2\,a+2\,b\,x\right )}{\cos \left (2\,a+2\,b\,x\right )+1}}}{b\,{\sin \left (a+b\,x\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*tan(a + b*x)^(1/2) + x^2/tan(a + b*x)^(3/2) - (4*x)/(b*tan(a + b*x)^(1/2)),x)
[Out]
-(x^2*sin(2*a + 2*b*x)*(sin(2*a + 2*b*x)/(cos(2*a + 2*b*x) + 1))^(1/2))/(b*sin(a + b*x)^2)
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