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

Optimal. Leaf size=247 \[ \frac{\sqrt{2} a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{5/2} f}-\frac{\sqrt{2} a^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{d^{5/2} f}-\frac{4 a^2}{d^2 f \sqrt{d \tan (e+f x)}}-\frac{a^2 \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} d^{5/2} f}+\frac{a^2 \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} d^{5/2} f}-\frac{2 a^2}{3 d f (d \tan (e+f x))^{3/2}} \]

[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]])

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Rubi [A]  time = 0.235596, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {3542, 12, 3474, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{\sqrt{2} a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{5/2} f}-\frac{\sqrt{2} a^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{d^{5/2} f}-\frac{4 a^2}{d^2 f \sqrt{d \tan (e+f x)}}-\frac{a^2 \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} d^{5/2} f}+\frac{a^2 \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} d^{5/2} f}-\frac{2 a^2}{3 d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[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 3542

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]

Rule 12

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

Rule 3474

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 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 \frac{(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx &=-\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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 \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} d^{5/2} f}-\frac{a^2 \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} d^{5/2} f}-\frac{a^2 \operatorname{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 \operatorname{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 ) \operatorname{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 ) \operatorname{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 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{5/2} f}-\frac{\sqrt{2} a^2 \tan ^{-1}\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 [C]  time = 1.32303, size = 229, normalized size = 0.93 \[ -\frac{a^2 (\cot (e+f x)+1)^2 \left (48 \sin ^2(e+f x) \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\tan ^2(e+f x)\right )+4 \sin (2 (e+f x)) \, _2F_1\left (-\frac{3}{4},1;\frac{1}{4};-\tan ^2(e+f x)\right )+3 \sqrt{2} \cos ^2(e+f x) \tan ^{\frac{5}{2}}(e+f x) \left (2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right )-2 \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right )+\log \left (\tan (e+f x)-\sqrt{2} \sqrt{\tan (e+f x)}+1\right )-\log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )\right )\right )}{12 d^2 f \sqrt{d \tan (e+f x)} (\sin (e+f x)+\cos (e+f x))^2} \]

Antiderivative was successfully verified.

[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*Hypergeometric2F1[-1/4, 1, 3/4, -Tan[e + f*x]^2]*Sin[e + f*x]^2 + 4*Hypergeomet
ric2F1[-3/4, 1, 1/4, -Tan[e + f*x]^2]*Sin[2*(e + f*x)] + 3*Sqrt[2]*Cos[e + f*x]^2*(2*ArcTan[1 - Sqrt[2]*Sqrt[T
an[e + f*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] + Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]
- Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]])*Tan[e + f*x]^(5/2)))/(12*d^2*f*(Cos[e + f*x] + Sin[e + f
*x])^2*Sqrt[d*Tan[e + f*x]])

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Maple [A]  time = 0.021, size = 216, normalized size = 0.9 \begin{align*} -{\frac{{a}^{2}\sqrt{2}}{2\,f{d}^{2}}\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}\sqrt{2}}{f{d}^{2}}\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}\sqrt{2}}{f{d}^{2}}\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{2\,{a}^{2}}{3\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-4\,{\frac{{a}^{2}}{f{d}^{2}\sqrt{d\tan \left ( fx+e \right ) }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.97227, size = 1918, normalized size = 7.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(12*(sqrt(2)*d^3*f*cos(f*x + e)^2 - sqrt(2)*d^3*f)*(a^8/(d^10*f^4))^(1/4)*arctan(-(sqrt(2)*a^6*d^2*f*sqrt(
d*sin(f*x + e)/cos(f*x + e))*(a^8/(d^10*f^4))^(1/4) + a^8 - sqrt(2)*d^2*f*sqrt((sqrt(2)*a^6*d^8*f^3*sqrt(d*sin
(f*x + e)/cos(f*x + e))*(a^8/(d^10*f^4))^(3/4)*cos(f*x + e) + a^8*d^6*f^2*sqrt(a^8/(d^10*f^4))*cos(f*x + e) +
a^12*d*sin(f*x + e))/cos(f*x + e))*(a^8/(d^10*f^4))^(1/4))/a^8) + 12*(sqrt(2)*d^3*f*cos(f*x + e)^2 - sqrt(2)*d
^3*f)*(a^8/(d^10*f^4))^(1/4)*arctan(-(sqrt(2)*a^6*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(a^8/(d^10*f^4))^(1/
4) - a^8 - sqrt(2)*d^2*f*sqrt(-(sqrt(2)*a^6*d^8*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(a^8/(d^10*f^4))^(3/4)*c
os(f*x + e) - a^8*d^6*f^2*sqrt(a^8/(d^10*f^4))*cos(f*x + e) - a^12*d*sin(f*x + e))/cos(f*x + e))*(a^8/(d^10*f^
4))^(1/4))/a^8) + 3*(sqrt(2)*d^3*f*cos(f*x + e)^2 - sqrt(2)*d^3*f)*(a^8/(d^10*f^4))^(1/4)*log((sqrt(2)*a^6*d^8
*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(a^8/(d^10*f^4))^(3/4)*cos(f*x + e) + a^8*d^6*f^2*sqrt(a^8/(d^10*f^4))*
cos(f*x + e) + a^12*d*sin(f*x + e))/cos(f*x + e)) - 3*(sqrt(2)*d^3*f*cos(f*x + e)^2 - sqrt(2)*d^3*f)*(a^8/(d^1
0*f^4))^(1/4)*log(-(sqrt(2)*a^6*d^8*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(a^8/(d^10*f^4))^(3/4)*cos(f*x + e)
- a^8*d^6*f^2*sqrt(a^8/(d^10*f^4))*cos(f*x + e) - a^12*d*sin(f*x + e))/cos(f*x + e)) + 4*(a^2*cos(f*x + e)^2 +
6*a^2*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(d^3*f*cos(f*x + e)^2 - d^3*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.25776, size = 336, normalized size = 1.36 \begin{align*} -\frac{\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^{4} f} - \frac{\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^{4} f} + \frac{\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 )}{2 \, d^{4} f} - \frac{\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 )}{2 \, d^{4} f} - \frac{2 \,{\left (6 \, a^{2} d \tan \left (f x + e\right ) + a^{2} d\right )}}{3 \, \sqrt{d \tan \left (f x + e\right )} d^{3} f \tan \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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