3.4.47 \(\int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{3/2}} \, dx\) [347]

Optimal. Leaf size=222 \[ -\frac {\sqrt {2} a^2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{3/2} f}+\frac {\sqrt {2} a^2 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{3/2} f}-\frac {a^2 \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} d^{3/2} f}+\frac {a^2 \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} d^{3/2} f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}} \]

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

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Rubi [A]
time = 0.14, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3623, 12, 3557, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {\sqrt {2} a^2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{3/2} f}+\frac {\sqrt {2} a^2 \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{d^{3/2} 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^{3/2} 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^{3/2} f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*a^2*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(3/2)*f)) + (Sqrt[2]*a^2*ArcTan[1 + (Sqr
t[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(3/2)*f) - (a^2*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Ta
n[e + f*x]]])/(Sqrt[2]*d^(3/2)*f) + (a^2*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(
Sqrt[2]*d^(3/2)*f) - (2*a^2)/(d*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 217

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 1.87, size = 232, normalized size = 1.05 \begin {gather*} -\frac {a^2 (1+\tan (e+f x))^2 \left (6 \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\tan ^2(e+f x)\right ) \sin (2 (e+f x))-4 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(e+f x)\right ) \sin ^2(e+f x) \tan (e+f x)+3 \sqrt {2} \cos ^2(e+f x) \left (2 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-2 \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right )+\log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )\right ) \tan ^{\frac {3}{2}}(e+f x)\right )}{6 f (\cos (e+f x)+\sin (e+f x))^2 (d \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(a^2*(1 + Tan[e + f*x])^2*(6*Hypergeometric2F1[-1/4, 1, 3/4, -Tan[e + f*x]^2]*Sin[2*(e + f*x)] - 4*Hyperg
eometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2]*Sin[e + f*x]^2*Tan[e + f*x] + 3*Sqrt[2]*Cos[e + f*x]^2*(2*ArcTan[1 -
Sqrt[2]*Sqrt[Tan[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]^(3/2)))/(f*(Cos[e + f*x] + S
in[e + f*x])^2*(d*Tan[e + f*x])^(3/2))

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Maple [A]
time = 0.19, size = 159, normalized size = 0.72

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (-\frac {1}{\sqrt {d \tan \left (f x +e \right )}}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}\right )}{f d}\)	\(159\)
default	\(\frac {2 a^{2} \left (-\frac {1}{\sqrt {d \tan \left (f x +e \right )}}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}\right )}{f d}\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*a^2/d*(-1/(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)))

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Maxima [A]
time = 0.52, size = 182, normalized size = 0.82 \begin {gather*} \frac {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 )} - \frac {4 \, a^{2}}{\sqrt {d \tan \left (f x + e\right )}}}{2 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 790 vs. \(2 (184) = 368\).
time = 1.14, size = 790, normalized size = 3.56 \begin {gather*} \frac {4 \, a^{2} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 4 \, {\left (\sqrt {2} d^{2} f \cos \left (f x + e\right )^{2} - \sqrt {2} d^{2} f\right )} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} a^{2} d^{4} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {3}{4}} - \sqrt {2} d^{4} f^{3} \sqrt {\frac {d^{4} f^{2} \sqrt {\frac {a^{8}}{d^{6} f^{4}}} \cos \left (f x + e\right ) + \sqrt {2} a^{2} d^{2} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {1}{4}} \cos \left (f x + e\right ) + a^{4} d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {3}{4}} + a^{8}}{a^{8}}\right ) - 4 \, {\left (\sqrt {2} d^{2} f \cos \left (f x + e\right )^{2} - \sqrt {2} d^{2} f\right )} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} a^{2} d^{4} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {3}{4}} - \sqrt {2} d^{4} f^{3} \sqrt {\frac {d^{4} f^{2} \sqrt {\frac {a^{8}}{d^{6} f^{4}}} \cos \left (f x + e\right ) - \sqrt {2} a^{2} d^{2} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {1}{4}} \cos \left (f x + e\right ) + a^{4} d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {3}{4}} - a^{8}}{a^{8}}\right ) + {\left (\sqrt {2} d^{2} f \cos \left (f x + e\right )^{2} - \sqrt {2} d^{2} f\right )} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {1}{4}} \log \left (\frac {d^{4} f^{2} \sqrt {\frac {a^{8}}{d^{6} f^{4}}} \cos \left (f x + e\right ) + \sqrt {2} a^{2} d^{2} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {1}{4}} \cos \left (f x + e\right ) + a^{4} d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) - {\left (\sqrt {2} d^{2} f \cos \left (f x + e\right )^{2} - \sqrt {2} d^{2} f\right )} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {1}{4}} \log \left (\frac {d^{4} f^{2} \sqrt {\frac {a^{8}}{d^{6} f^{4}}} \cos \left (f x + e\right ) - \sqrt {2} a^{2} d^{2} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {a^{8}}{d^{6} f^{4}}\right )^{\frac {1}{4}} \cos \left (f x + e\right ) + a^{4} d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right )}{2 \, {\left (d^{2} f \cos \left (f x + e\right )^{2} - d^{2} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a^2*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - 4*(sqrt(2)*d^2*f*cos(f*x + e)^2 - sqr
t(2)*d^2*f)*(a^8/(d^6*f^4))^(1/4)*arctan(-(sqrt(2)*a^2*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(a^8/(d^6*f^4
))^(3/4) - sqrt(2)*d^4*f^3*sqrt((d^4*f^2*sqrt(a^8/(d^6*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x +
e)/cos(f*x + e))*(a^8/(d^6*f^4))^(1/4)*cos(f*x + e) + a^4*d*sin(f*x + e))/cos(f*x + e))*(a^8/(d^6*f^4))^(3/4)
+ a^8)/a^8) - 4*(sqrt(2)*d^2*f*cos(f*x + e)^2 - sqrt(2)*d^2*f)*(a^8/(d^6*f^4))^(1/4)*arctan(-(sqrt(2)*a^2*d^4
*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(a^8/(d^6*f^4))^(3/4) - sqrt(2)*d^4*f^3*sqrt((d^4*f^2*sqrt(a^8/(d^6*f^4
))*cos(f*x + e) - sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(a^8/(d^6*f^4))^(1/4)*cos(f*x + e) + a^4
*d*sin(f*x + e))/cos(f*x + e))*(a^8/(d^6*f^4))^(3/4) - a^8)/a^8) + (sqrt(2)*d^2*f*cos(f*x + e)^2 - sqrt(2)*d^2
*f)*(a^8/(d^6*f^4))^(1/4)*log((d^4*f^2*sqrt(a^8/(d^6*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e
)/cos(f*x + e))*(a^8/(d^6*f^4))^(1/4)*cos(f*x + e) + a^4*d*sin(f*x + e))/cos(f*x + e)) - (sqrt(2)*d^2*f*cos(f*
x + e)^2 - sqrt(2)*d^2*f)*(a^8/(d^6*f^4))^(1/4)*log((d^4*f^2*sqrt(a^8/(d^6*f^4))*cos(f*x + e) - sqrt(2)*a^2*d^
2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(a^8/(d^6*f^4))^(1/4)*cos(f*x + e) + a^4*d*sin(f*x + e))/cos(f*x + e)))/
(d^2*f*cos(f*x + e)^2 - d^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {2 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.73, size = 225, normalized size = 1.01 \begin {gather*} \frac {\frac {2 \, \sqrt {2} a^{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 )}{d f} + \frac {2 \, \sqrt {2} a^{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 )}{d f} + \frac {\sqrt {2} a^{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 )}{d f} - \frac {\sqrt {2} a^{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 )}{d f} - \frac {4 \, a^{2}}{\sqrt {d \tan \left (f x + e\right )} f}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*sqrt(2)*a^2*sqrt(abs(d))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)
))/(d*f) + 2*sqrt(2)*a^2*sqrt(abs(d))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt
(abs(d)))/(d*f) + sqrt(2)*a^2*sqrt(abs(d))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + ab
s(d))/(d*f) - sqrt(2)*a^2*sqrt(abs(d))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d)
)/(d*f) - 4*a^2/(sqrt(d*tan(f*x + e))*f))/d

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Mupad [B]
time = 4.25, size = 86, normalized size = 0.39 \begin {gather*} -\frac {2\,a^2}{d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {{\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 )\,2{}\mathrm {i}}{d^{3/2}\,f}-\frac {{\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 )\,2{}\mathrm {i}}{d^{3/2}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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