3.343 \(\int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=269 \[ -\frac {\sqrt {2} a^2 d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {\sqrt {2} a^2 d^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{f}-\frac {a^2 d^{5/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} f}+\frac {a^2 d^{5/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} f}-\frac {4 a^2 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac {4 a^2 (d \tan (e+f x))^{5/2}}{5 f} \]

[Out]

-1/2*a^2*d^(5/2)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/f*2^(1/2)+1/2*a^2*d^(5/2)*ln(d^(1
/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/f*2^(1/2)-a^2*d^(5/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/
2)/d^(1/2))*2^(1/2)/f+a^2*d^(5/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/f-4*a^2*d^2*(d*tan(f*
x+e))^(1/2)/f+4/5*a^2*(d*tan(f*x+e))^(5/2)/f+2/7*a^2*(d*tan(f*x+e))^(7/2)/d/f

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3543, 12, 16, 3473, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {\sqrt {2} a^2 d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {\sqrt {2} a^2 d^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{f}-\frac {4 a^2 d^2 \sqrt {d \tan (e+f x)}}{f}-\frac {a^2 d^{5/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} f}+\frac {a^2 d^{5/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} f}+\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac {4 a^2 (d \tan (e+f x))^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*a^2*d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f) + (Sqrt[2]*a^2*d^(5/2)*ArcTan[1
+ (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f - (a^2*d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[
d*Tan[e + f*x]]])/(Sqrt[2]*f) + (a^2*d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]
])/(Sqrt[2]*f) - (4*a^2*d^2*Sqrt[d*Tan[e + f*x]])/f + (4*a^2*(d*Tan[e + f*x])^(5/2))/(5*f) + (2*a^2*(d*Tan[e +
f*x])^(7/2))/(7*d*f)

Rule 12

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 204

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

Rule 211

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

Rule 329

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 617

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 628

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 1162

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 1165

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 3473

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

Rule 3476

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 3543

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

Rubi steps

\begin {align*} \int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^2 \, dx &=\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\int 2 a^2 \tan (e+f x) (d \tan (e+f x))^{5/2} \, dx\\ &=\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\left (2 a^2\right ) \int \tan (e+f x) (d \tan (e+f x))^{5/2} \, dx\\ &=\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac {\left (2 a^2\right ) \int (d \tan (e+f x))^{7/2} \, dx}{d}\\ &=\frac {4 a^2 (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}-\left (2 a^2 d\right ) \int (d \tan (e+f x))^{3/2} \, dx\\ &=-\frac {4 a^2 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^2 (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\left (2 a^2 d^3\right ) \int \frac {1}{\sqrt {d \tan (e+f x)}} \, dx\\ &=-\frac {4 a^2 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^2 (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac {\left (2 a^2 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=-\frac {4 a^2 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^2 (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac {\left (4 a^2 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=-\frac {4 a^2 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^2 (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac {\left (2 a^2 d^3\right ) \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}+\frac {\left (2 a^2 d^3\right ) \operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=-\frac {4 a^2 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^2 (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}-\frac {\left (a^2 d^{5/2}\right ) \operatorname {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} f}-\frac {\left (a^2 d^{5/2}\right ) \operatorname {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} f}+\frac {\left (a^2 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}+\frac {\left (a^2 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=-\frac {a^2 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}+\frac {a^2 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}-\frac {4 a^2 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^2 (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac {\left (\sqrt {2} a^2 d^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {\left (\sqrt {2} a^2 d^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}\\ &=-\frac {\sqrt {2} a^2 d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {\sqrt {2} a^2 d^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {a^2 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}+\frac {a^2 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}-\frac {4 a^2 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^2 (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.27, size = 187, normalized size = 0.70 \[ \frac {a^2 (d \tan (e+f x))^{5/2} \left (-70 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )+70 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )+20 \tan ^{\frac {7}{2}}(e+f x)+56 \tan ^{\frac {5}{2}}(e+f x)-280 \sqrt {\tan (e+f x)}-35 \sqrt {2} \log \left (\tan (e+f x)-\sqrt {2} \sqrt {\tan (e+f x)}+1\right )+35 \sqrt {2} \log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )\right )}{70 f \tan ^{\frac {5}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(d*Tan[e + f*x])^(5/2)*(-70*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] + 70*Sqrt[2]*ArcTan[1 + Sqrt[2
]*Sqrt[Tan[e + f*x]]] - 35*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] + 35*Sqrt[2]*Log[1 + Sqr
t[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - 280*Sqrt[Tan[e + f*x]] + 56*Tan[e + f*x]^(5/2) + 20*Tan[e + f*x]^(7/
2)))/(70*f*Tan[e + f*x]^(5/2))

________________________________________________________________________________________

fricas [B]  time = 0.46, size = 736, normalized size = 2.74 \[ -\frac {140 \, \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {1}{4}} f \arctan \left (-\frac {a^{8} d^{10} + \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {3}{4}} a^{2} d^{2} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} - \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {3}{4}} f^{3} \sqrt {\frac {a^{4} d^{5} \sin \left (f x + e\right ) + \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {1}{4}} a^{2} d^{2} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{10}}{f^{4}}} f^{2} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}}}{a^{8} d^{10}}\right ) \cos \left (f x + e\right )^{3} + 140 \, \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {1}{4}} f \arctan \left (\frac {a^{8} d^{10} - \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {3}{4}} a^{2} d^{2} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} + \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {3}{4}} f^{3} \sqrt {\frac {a^{4} d^{5} \sin \left (f x + e\right ) - \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {1}{4}} a^{2} d^{2} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{10}}{f^{4}}} f^{2} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}}}{a^{8} d^{10}}\right ) \cos \left (f x + e\right )^{3} - 35 \, \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {1}{4}} f \cos \left (f x + e\right )^{3} \log \left (\frac {a^{4} d^{5} \sin \left (f x + e\right ) + \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {1}{4}} a^{2} d^{2} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{10}}{f^{4}}} f^{2} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) + 35 \, \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {1}{4}} f \cos \left (f x + e\right )^{3} \log \left (\frac {a^{4} d^{5} \sin \left (f x + e\right ) - \sqrt {2} \left (\frac {a^{8} d^{10}}{f^{4}}\right )^{\frac {1}{4}} a^{2} d^{2} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{10}}{f^{4}}} f^{2} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) + 4 \, {\left (84 \, a^{2} d^{2} \cos \left (f x + e\right )^{3} - 14 \, a^{2} d^{2} \cos \left (f x + e\right ) + 5 \, {\left (a^{2} d^{2} \cos \left (f x + e\right )^{2} - a^{2} d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{70 \, f \cos \left (f x + e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/70*(140*sqrt(2)*(a^8*d^10/f^4)^(1/4)*f*arctan(-(a^8*d^10 + sqrt(2)*(a^8*d^10/f^4)^(3/4)*a^2*d^2*f^3*sqrt(d*
sin(f*x + e)/cos(f*x + e)) - sqrt(2)*(a^8*d^10/f^4)^(3/4)*f^3*sqrt((a^4*d^5*sin(f*x + e) + sqrt(2)*(a^8*d^10/f
^4)^(1/4)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e) + sqrt(a^8*d^10/f^4)*f^2*cos(f*x + e))/cos(
f*x + e)))/(a^8*d^10))*cos(f*x + e)^3 + 140*sqrt(2)*(a^8*d^10/f^4)^(1/4)*f*arctan((a^8*d^10 - sqrt(2)*(a^8*d^1
0/f^4)^(3/4)*a^2*d^2*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e)) + sqrt(2)*(a^8*d^10/f^4)^(3/4)*f^3*sqrt((a^4*d^5*si
n(f*x + e) - sqrt(2)*(a^8*d^10/f^4)^(1/4)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e) + sqrt(a^8*
d^10/f^4)*f^2*cos(f*x + e))/cos(f*x + e)))/(a^8*d^10))*cos(f*x + e)^3 - 35*sqrt(2)*(a^8*d^10/f^4)^(1/4)*f*cos(
f*x + e)^3*log((a^4*d^5*sin(f*x + e) + sqrt(2)*(a^8*d^10/f^4)^(1/4)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e)
)*cos(f*x + e) + sqrt(a^8*d^10/f^4)*f^2*cos(f*x + e))/cos(f*x + e)) + 35*sqrt(2)*(a^8*d^10/f^4)^(1/4)*f*cos(f*
x + e)^3*log((a^4*d^5*sin(f*x + e) - sqrt(2)*(a^8*d^10/f^4)^(1/4)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*
cos(f*x + e) + sqrt(a^8*d^10/f^4)*f^2*cos(f*x + e))/cos(f*x + e)) + 4*(84*a^2*d^2*cos(f*x + e)^3 - 14*a^2*d^2*
cos(f*x + e) + 5*(a^2*d^2*cos(f*x + e)^2 - a^2*d^2)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(f*cos(f*
x + e)^3)

________________________________________________________________________________________

giac [A]  time = 2.03, size = 293, normalized size = 1.09 \[ \frac {\sqrt {2} a^{2} d^{2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{f} + \frac {\sqrt {2} a^{2} d^{2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{f} + \frac {\sqrt {2} a^{2} d^{2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{2 \, f} - \frac {\sqrt {2} a^{2} d^{2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{2 \, f} + \frac {2 \, {\left (5 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{9} f^{6} \tan \left (f x + e\right )^{3} + 14 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{9} f^{6} \tan \left (f x + e\right )^{2} - 70 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{9} f^{6}\right )}}{35 \, d^{7} f^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*a^2*d^2*sqrt(abs(d))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/
f + sqrt(2)*a^2*d^2*sqrt(abs(d))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(
d)))/f + 1/2*sqrt(2)*a^2*d^2*sqrt(abs(d))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs
(d))/f - 1/2*sqrt(2)*a^2*d^2*sqrt(abs(d))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs
(d))/f + 2/35*(5*sqrt(d*tan(f*x + e))*a^2*d^9*f^6*tan(f*x + e)^3 + 14*sqrt(d*tan(f*x + e))*a^2*d^9*f^6*tan(f*x
+ e)^2 - 70*sqrt(d*tan(f*x + e))*a^2*d^9*f^6)/(d^7*f^7)

________________________________________________________________________________________

maple [A]  time = 0.23, size = 234, normalized size = 0.87 \[ \frac {2 a^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7 d f}+\frac {4 a^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}-\frac {4 a^{2} d^{2} \sqrt {d \tan \left (f x +e \right )}}{f}+\frac {a^{2} d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f}-\frac {a^{2} d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f}+\frac {a^{2} d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*a^2*(d*tan(f*x+e))^(7/2)/d/f+4/5*a^2*(d*tan(f*x+e))^(5/2)/f-4*a^2*d^2*(d*tan(f*x+e))^(1/2)/f+1/2/f*a^2*d^2
*(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)))-1/f*a^2*d^2*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)
*(d*tan(f*x+e))^(1/2)+1)+1/f*a^2*d^2*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 210, normalized size = 0.78 \[ \frac {20 \, \left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} a^{2} + 56 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{2} d - 280 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{3} + 35 \, {\left (2 \, \sqrt {2} d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right ) + 2 \, \sqrt {2} d^{\frac {7}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right ) + \sqrt {2} d^{\frac {7}{2}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) - \sqrt {2} d^{\frac {7}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )\right )} a^{2}}{70 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/70*(20*(d*tan(f*x + e))^(7/2)*a^2 + 56*(d*tan(f*x + e))^(5/2)*a^2*d - 280*sqrt(d*tan(f*x + e))*a^2*d^3 + 35*
(2*sqrt(2)*d^(7/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d)) + 2*sqrt(2)*d^(7/2)*
arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d)) + sqrt(2)*d^(7/2)*log(d*tan(f*x + e) +
sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d) - sqrt(2)*d^(7/2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))
*sqrt(d) + d))*a^2)/(d*f)

________________________________________________________________________________________

mupad [B]  time = 5.30, size = 125, normalized size = 0.46 \[ \frac {4\,a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,f}-\frac {4\,a^2\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}+\frac {2\,a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}}{7\,d\,f}-\frac {{\left (-1\right )}^{1/4}\,a^2\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{f}-\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a^2*(d*tan(e + f*x))^(5/2))/(5*f) - (4*a^2*d^2*(d*tan(e + f*x))^(1/2))/f + (2*a^2*(d*tan(e + f*x))^(7/2))/(
7*d*f) - ((-1)^(1/4)*a^2*d^(5/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2))*2i)/f - (2*(-1)^(1/4)*a^2*d
^(5/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2)*1i)/d^(1/2)))/f

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________