\(\int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx\) [348]

Optimal result

Integrand size = 25, antiderivative size = 247 \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {\sqrt {2} a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {a^2 \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} d^{5/2} f}+\frac {a^2 \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} d^{5/2} f}-\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 a^2}{d^2 f \sqrt {d \tan (e+f x)}} \]

[Out]

-1/2*a^2*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/d^(5/2)/f*2^(1/2)+1/2*a^2*ln(d^(1/2)+2^(1
/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/d^(5/2)/f*2^(1/2)+a^2*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/
2))*2^(1/2)/d^(5/2)/f-a^2*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/d^(5/2)/f-4*a^2/d^2/f/(d*tan(
f*x+e))^(1/2)-2/3*a^2/d/f/(d*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3623, 12, 3555, 3557, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {\sqrt {2} a^2 \arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{d^{5/2} f}-\frac {a^2 \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} d^{5/2} f}+\frac {a^2 \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} d^{5/2} f}-\frac {4 a^2}{d^2 f \sqrt {d \tan (e+f x)}}-\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}} \]

[In]

Int[(a + a*Tan[e + f*x])^2/(d*Tan[e + f*x])^(5/2),x]

[Out]

(Sqrt[2]*a^2*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(5/2)*f) - (Sqrt[2]*a^2*ArcTan[1 + (Sqrt[2
]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(5/2)*f) - (a^2*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e
+ f*x]]])/(Sqrt[2]*d^(5/2)*f) + (a^2*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqr
t[2]*d^(5/2)*f) - (2*a^2)/(3*d*f*(d*Tan[e + f*x])^(3/2)) - (4*a^2)/(d^2*f*Sqrt[d*Tan[e + f*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

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

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3555

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3623

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(b*c - a*d)^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Ta
n[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Tan[e + f*x], x], x], x] /; FreeQ
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}+\frac {\int \frac {2 a^2 d}{(d \tan (e+f x))^{3/2}} \, dx}{d^2} \\ & = -\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}+\frac {\left (2 a^2\right ) \int \frac {1}{(d \tan (e+f x))^{3/2}} \, dx}{d} \\ & = -\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 a^2}{d^2 f \sqrt {d \tan (e+f x)}}-\frac {\left (2 a^2\right ) \int \sqrt {d \tan (e+f x)} \, dx}{d^3} \\ & = -\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 a^2}{d^2 f \sqrt {d \tan (e+f x)}}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{d^2 f} \\ & = -\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 a^2}{d^2 f \sqrt {d \tan (e+f x)}}-\frac {\left (4 a^2\right ) \text {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{d^2 f} \\ & = -\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 a^2}{d^2 f \sqrt {d \tan (e+f x)}}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{d^2 f}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{d^2 f} \\ & = -\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 a^2}{d^2 f \sqrt {d \tan (e+f x)}}-\frac {a^2 \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} d^{5/2} f}-\frac {a^2 \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} d^{5/2} f}-\frac {a^2 \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{d^2 f}-\frac {a^2 \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{d^2 f} \\ & = -\frac {a^2 \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} d^{5/2} f}+\frac {a^2 \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} d^{5/2} f}-\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 a^2}{d^2 f \sqrt {d \tan (e+f x)}}-\frac {\left (\sqrt {2} a^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}+\frac {\left (\sqrt {2} a^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f} \\ & = \frac {\sqrt {2} a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {\sqrt {2} a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {a^2 \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} d^{5/2} f}+\frac {a^2 \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} d^{5/2} f}-\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 a^2}{d^2 f \sqrt {d \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 5.55 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.62 \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=\frac {a^2 (1+\cot (e+f x))^2 \left (-48 \sin ^2(e+f x)-4 \sin (2 (e+f x))-6 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^2(e+f x) \tan ^{\frac {5}{2}}(e+f x)+6 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^2(e+f x) \tan ^{\frac {5}{2}}(e+f x)-3 \sqrt {2} \cos ^2(e+f x) \log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \tan ^{\frac {5}{2}}(e+f x)+3 \sqrt {2} \cos ^2(e+f x) \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \tan ^{\frac {5}{2}}(e+f x)+12 \text {arctanh}\left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \left (\cos ^2(e+f x) (-\tan (e+f x))^{3/4} \tan ^{\frac {7}{4}}(e+f x)+2 \sin ^2(e+f x) \sqrt [4]{-\tan ^2(e+f x)}\right )+12 \arctan \left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \cos ^2(e+f x) \left ((-\tan (e+f x))^{3/4} \tan ^{\frac {7}{4}}(e+f x)+2 \left (-\tan ^2(e+f x)\right )^{5/4}\right )\right )}{12 d^2 f (\cos (e+f x)+\sin (e+f x))^2 \sqrt {d \tan (e+f x)}} \]

[In]

Integrate[(a + a*Tan[e + f*x])^2/(d*Tan[e + f*x])^(5/2),x]

[Out]

(a^2*(1 + Cot[e + f*x])^2*(-48*Sin[e + f*x]^2 - 4*Sin[2*(e + f*x)] - 6*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e +
f*x]]]*Cos[e + f*x]^2*Tan[e + f*x]^(5/2) + 6*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^2*Ta
n[e + f*x]^(5/2) - 3*Sqrt[2]*Cos[e + f*x]^2*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*Tan[e + f*x]^(5
/2) + 3*Sqrt[2]*Cos[e + f*x]^2*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*Tan[e + f*x]^(5/2) + 12*ArcT
anh[(-Tan[e + f*x]^2)^(1/4)]*(Cos[e + f*x]^2*(-Tan[e + f*x])^(3/4)*Tan[e + f*x]^(7/4) + 2*Sin[e + f*x]^2*(-Tan
[e + f*x]^2)^(1/4)) + 12*ArcTan[(-Tan[e + f*x]^2)^(1/4)]*Cos[e + f*x]^2*((-Tan[e + f*x])^(3/4)*Tan[e + f*x]^(7
/4) + 2*(-Tan[e + f*x]^2)^(5/4))))/(12*d^2*f*(Cos[e + f*x] + Sin[e + f*x])^2*Sqrt[d*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (-\frac {1}{3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2}{d \sqrt {d \tan \left (f x +e \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f d}\)	\(174\)
default	\(\frac {2 a^{2} \left (-\frac {1}{3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2}{d \sqrt {d \tan \left (f x +e \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f d}\)	\(174\)
parts	\(\frac {2 a^{2} d \left (-\frac {1}{3 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{4}}\right )}{f}+\frac {a^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 f \,d^{3}}+\frac {2 a^{2} \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d^{2} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {2}{d^{2} \sqrt {d \tan \left (f x +e \right )}}\right )}{f}\)	\(459\)

[In]

int((a+a*tan(f*x+e))^2/(d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/f*a^2/d*(-1/3/(d*tan(f*x+e))^(3/2)-2/d/(d*tan(f*x+e))^(1/2)-1/4/d/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)-(d^2
)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)
^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/
2)+1)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.21 \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {3 \, d^{3} f \left (-\frac {a^{8}}{d^{10} f^{4}}\right )^{\frac {1}{4}} \log \left (d^{8} f^{3} \left (-\frac {a^{8}}{d^{10} f^{4}}\right )^{\frac {3}{4}} + \sqrt {d \tan \left (f x + e\right )} a^{6}\right ) \tan \left (f x + e\right )^{2} - 3 i \, d^{3} f \left (-\frac {a^{8}}{d^{10} f^{4}}\right )^{\frac {1}{4}} \log \left (i \, d^{8} f^{3} \left (-\frac {a^{8}}{d^{10} f^{4}}\right )^{\frac {3}{4}} + \sqrt {d \tan \left (f x + e\right )} a^{6}\right ) \tan \left (f x + e\right )^{2} + 3 i \, d^{3} f \left (-\frac {a^{8}}{d^{10} f^{4}}\right )^{\frac {1}{4}} \log \left (-i \, d^{8} f^{3} \left (-\frac {a^{8}}{d^{10} f^{4}}\right )^{\frac {3}{4}} + \sqrt {d \tan \left (f x + e\right )} a^{6}\right ) \tan \left (f x + e\right )^{2} - 3 \, d^{3} f \left (-\frac {a^{8}}{d^{10} f^{4}}\right )^{\frac {1}{4}} \log \left (-d^{8} f^{3} \left (-\frac {a^{8}}{d^{10} f^{4}}\right )^{\frac {3}{4}} + \sqrt {d \tan \left (f x + e\right )} a^{6}\right ) \tan \left (f x + e\right )^{2} + 2 \, {\left (6 \, a^{2} \tan \left (f x + e\right ) + a^{2}\right )} \sqrt {d \tan \left (f x + e\right )}}{3 \, d^{3} f \tan \left (f x + e\right )^{2}} \]

[In]

integrate((a+a*tan(f*x+e))^2/(d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/3*(3*d^3*f*(-a^8/(d^10*f^4))^(1/4)*log(d^8*f^3*(-a^8/(d^10*f^4))^(3/4) + sqrt(d*tan(f*x + e))*a^6)*tan(f*x
+ e)^2 - 3*I*d^3*f*(-a^8/(d^10*f^4))^(1/4)*log(I*d^8*f^3*(-a^8/(d^10*f^4))^(3/4) + sqrt(d*tan(f*x + e))*a^6)*t
an(f*x + e)^2 + 3*I*d^3*f*(-a^8/(d^10*f^4))^(1/4)*log(-I*d^8*f^3*(-a^8/(d^10*f^4))^(3/4) + sqrt(d*tan(f*x + e)
)*a^6)*tan(f*x + e)^2 - 3*d^3*f*(-a^8/(d^10*f^4))^(1/4)*log(-d^8*f^3*(-a^8/(d^10*f^4))^(3/4) + sqrt(d*tan(f*x
+ e))*a^6)*tan(f*x + e)^2 + 2*(6*a^2*tan(f*x + e) + a^2)*sqrt(d*tan(f*x + e)))/(d^3*f*tan(f*x + e)^2)

Sympy [F]

\[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=a^{2} \left (\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {2 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx\right ) \]

[In]

integrate((a+a*tan(f*x+e))**2/(d*tan(f*x+e))**(5/2),x)

[Out]

a**2*(Integral((d*tan(e + f*x))**(-5/2), x) + Integral(2*tan(e + f*x)/(d*tan(e + f*x))**(5/2), x) + Integral(t
an(e + f*x)**2/(d*tan(e + f*x))**(5/2), x))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.80 \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {\frac {3 \, a^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d} + \frac {4 \, {\left (6 \, a^{2} d \tan \left (f x + e\right ) + a^{2} d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d}}{6 \, d f} \]

[In]

integrate((a+a*tan(f*x+e))^2/(d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-1/6*(3*a^2*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + 2*sqrt
(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - sqrt(2)*log(d*tan(f*x +
e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x +
e))*sqrt(d) + d)/sqrt(d))/d + 4*(6*a^2*d*tan(f*x + e) + a^2*d)/((d*tan(f*x + e))^(3/2)*d))/(d*f)

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*tan(f*x+e))^2/(d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.62 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.40 \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{d^{5/2}\,f}-\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{d^{5/2}\,f}-\frac {4\,a^2\,\mathrm {tan}\left (e+f\,x\right )+\frac {2\,a^2}{3}}{d\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \]

[In]

int((a + a*tan(e + f*x))^2/(d*tan(e + f*x))^(5/2),x)

[Out]

(2*(-1)^(1/4)*a^2*atanh(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/(d^(5/2)*f) - (2*(-1)^(1/4)*a^2*atan(((-
1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/(d^(5/2)*f) - (4*a^2*tan(e + f*x) + (2*a^2)/3)/(d*f*(d*tan(e + f*x)
)^(3/2))