3.344 \(\int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=246 \[ \frac{\sqrt{2} a^2 d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{\sqrt{2} a^2 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{f}-\frac{a^2 d^{3/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} f}+\frac{a^2 d^{3/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} f}+\frac{2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac{4 a^2 (d \tan (e+f x))^{3/2}}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.227639, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {3543, 12, 16, 3473, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{\sqrt{2} a^2 d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{\sqrt{2} a^2 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{f}-\frac{a^2 d^{3/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} f}+\frac{a^2 d^{3/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} f}+\frac{2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac{4 a^2 (d \tan (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^2,x]

[Out]

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

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])

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 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 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 297

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 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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.374567, size = 52, normalized size = 0.21 \[ \frac{2 a^2 (d \tan (e+f x))^{3/2} \left (-10 \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\tan ^2(e+f x)\right )+3 \tan (e+f x)+10\right )}{15 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^2,x]

[Out]

(2*a^2*(d*Tan[e + f*x])^(3/2)*(10 - 10*Hypergeometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2] + 3*Tan[e + f*x]))/(15*
f)

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Maple [A]  time = 0.018, size = 213, normalized size = 0.9 \begin{align*}{\frac{2\,{a}^{2}}{5\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{4\,{a}^{2}}{3\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{{a}^{2}{d}^{2}\sqrt{2}}{2\,f}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{{a}^{2}{d}^{2}\sqrt{2}}{f}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{{a}^{2}{d}^{2}\sqrt{2}}{f}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^2,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.89997, size = 1769, normalized size = 7.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

1/30*(60*sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*arctan(-(a^8*d^6 + sqrt(2)*(a^8*d^6/f^4)^(1/4)*a^6*d^4*f*sqrt(d*sin(f*x
+ e)/cos(f*x + e)) - sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*sqrt((a^12*d^9*sin(f*x + e) + sqrt(a^8*d^6/f^4)*a^8*d^6*f^
2*cos(f*x + e) + sqrt(2)*(a^8*d^6/f^4)^(3/4)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/cos(f
*x + e)))/(a^8*d^6))*cos(f*x + e)^2 + 60*sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*arctan((a^8*d^6 - sqrt(2)*(a^8*d^6/f^4)
^(1/4)*a^6*d^4*f*sqrt(d*sin(f*x + e)/cos(f*x + e)) + sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*sqrt((a^12*d^9*sin(f*x + e)
+ sqrt(a^8*d^6/f^4)*a^8*d^6*f^2*cos(f*x + e) - sqrt(2)*(a^8*d^6/f^4)^(3/4)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/co
s(f*x + e))*cos(f*x + e))/cos(f*x + e)))/(a^8*d^6))*cos(f*x + e)^2 + 15*sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*cos(f*x
+ e)^2*log((a^12*d^9*sin(f*x + e) + sqrt(a^8*d^6/f^4)*a^8*d^6*f^2*cos(f*x + e) + sqrt(2)*(a^8*d^6/f^4)^(3/4)*a
^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/cos(f*x + e)) - 15*sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*co
s(f*x + e)^2*log((a^12*d^9*sin(f*x + e) + sqrt(a^8*d^6/f^4)*a^8*d^6*f^2*cos(f*x + e) - sqrt(2)*(a^8*d^6/f^4)^(
3/4)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/cos(f*x + e)) - 4*(3*a^2*d*cos(f*x + e)^2 - 1
0*a^2*d*cos(f*x + e)*sin(f*x + e) - 3*a^2*d)*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(f*cos(f*x + e)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}\, dx + \int 2 \left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}} \tan{\left (e + f x \right )}\, dx + \int \left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}} \tan ^{2}{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(3/2)*(a+a*tan(f*x+e))**2,x)

[Out]

a**2*(Integral((d*tan(e + f*x))**(3/2), x) + Integral(2*(d*tan(e + f*x))**(3/2)*tan(e + f*x), x) + Integral((d
*tan(e + f*x))**(3/2)*tan(e + f*x)**2, x))

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Giac [A]  time = 1.29365, size = 370, normalized size = 1.5 \begin{align*} -\frac{1}{30} \,{\left (\frac{30 \, \sqrt{2} a^{2}{\left | d \right |}^{\frac{3}{2}} \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 )}{d f} + \frac{30 \, \sqrt{2} a^{2}{\left | d \right |}^{\frac{3}{2}} \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 )}{d f} - \frac{15 \, \sqrt{2} a^{2}{\left | d \right |}^{\frac{3}{2}} \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 )}{d f} + \frac{15 \, \sqrt{2} a^{2}{\left | d \right |}^{\frac{3}{2}} \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 )}{d f} - \frac{4 \,{\left (3 \, \sqrt{d \tan \left (f x + e\right )} a^{2} d^{10} f^{4} \tan \left (f x + e\right )^{2} + 10 \, \sqrt{d \tan \left (f x + e\right )} a^{2} d^{10} f^{4} \tan \left (f x + e\right )\right )}}{d^{10} f^{5}}\right )} d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-1/30*(30*sqrt(2)*a^2*abs(d)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs
(d)))/(d*f) + 30*sqrt(2)*a^2*abs(d)^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/
sqrt(abs(d)))/(d*f) - 15*sqrt(2)*a^2*abs(d)^(3/2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d
)) + abs(d))/(d*f) + 15*sqrt(2)*a^2*abs(d)^(3/2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)
) + abs(d))/(d*f) - 4*(3*sqrt(d*tan(f*x + e))*a^2*d^10*f^4*tan(f*x + e)^2 + 10*sqrt(d*tan(f*x + e))*a^2*d^10*f
^4*tan(f*x + e))/(d^10*f^5))*d