3.4.73 \(\int \frac {(d \tan (e+f x))^{3/2}}{(a+a \tan (e+f x))^3} \, dx\) [373]

Optimal. Leaf size=164 \[ -\frac {5 d^{3/2} \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}-\frac {d^{3/2} \text {ArcTan}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}+\frac {d \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )} \]

[Out]

-5/8*d^(3/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f-1/4*d^(3/2)*arctan(1/2*(d^(1/2)-d^(1/2)*tan(f*x+e))*2^
(1/2)/(d*tan(f*x+e))^(1/2))/a^3/f*2^(1/2)+1/4*d*(d*tan(f*x+e))^(1/2)/a/f/(a+a*tan(f*x+e))^2-1/8*d*(d*tan(f*x+e
))^(1/2)/f/(a^3+a^3*tan(f*x+e))

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Rubi [A]
time = 0.39, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3648, 3730, 3734, 3613, 211, 3715, 65} \begin {gather*} -\frac {5 d^{3/2} \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}-\frac {d^{3/2} \text {ArcTan}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}-\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3 \tan (e+f x)+a^3\right )}+\frac {d \sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*f) - (d^(3/2)*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])
/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(2*Sqrt[2]*a^3*f) + (d*Sqrt[d*Tan[e + f*x]])/(4*a*f*(a + a*Tan[e + f*x])^2)
- (d*Sqrt[d*Tan[e + f*x]])/(8*f*(a^3 + a^3*Tan[e + f*x]))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 211

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

Rule 3613

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2*(d^2/f),
Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x
] && EqQ[c^2 - d^2, 0]

Rule 3648

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

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3730

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3734

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[(c + d*Tan[e + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rubi steps

\begin {align*} \int \frac {(d \tan (e+f x))^{3/2}}{(a+a \tan (e+f x))^3} \, dx &=\frac {d \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {\int \frac {\frac {a d^2}{2}-2 a d^2 \tan (e+f x)-\frac {3}{2} a d^2 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2} \, dx}{4 a^2}\\ &=\frac {d \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}-\frac {\int \frac {\frac {a^3 d^3}{2}-4 a^3 d^3 \tan (e+f x)+\frac {1}{2} a^3 d^3 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{8 a^5 d}\\ &=\frac {d \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}-\frac {\int \frac {-4 a^4 d^3-4 a^4 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{16 a^7 d}-\frac {\left (5 d^2\right ) \int \frac {1+\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{16 a^2}\\ &=\frac {d \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}-\frac {\left (5 d^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{16 a^2 f}+\frac {\left (2 a d^5\right ) \text {Subst}\left (\int \frac {1}{32 a^8 d^6+d x^2} \, dx,x,\frac {-4 a^4 d^3+4 a^4 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f}\\ &=-\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}+\frac {d \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}-\frac {(5 d) \text {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^2 f}\\ &=-\frac {5 d^{3/2} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}-\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}+\frac {d \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 2.50, size = 127, normalized size = 0.77 \begin {gather*} \frac {d \left (\frac {(-1+\cot (e+f x)) \cot (e+f x)}{(1+\cot (e+f x))^2}+\frac {-2 \sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )+2 \sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right )-5 \text {ArcTan}\left (\sqrt {\tan (e+f x)}\right )}{\sqrt {\tan (e+f x)}}\right ) \sqrt {d \tan (e+f x)}}{8 a^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(((-1 + Cot[e + f*x])*Cot[e + f*x])/(1 + Cot[e + f*x])^2 + (-2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]
]] + 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] - 5*ArcTan[Sqrt[Tan[e + f*x]]])/Sqrt[Tan[e + f*x]])*Sqrt
[d*Tan[e + f*x]])/(8*a^3*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(348\) vs. \(2(135)=270\).
time = 0.18, size = 349, normalized size = 2.13

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (-\frac {\frac {\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}-\frac {d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {5 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{2}}+\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{2}}\right )}{f \,a^{3}}\)	\(349\)
default	\(\frac {2 d^{4} \left (-\frac {\frac {\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}-\frac {d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {5 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{2}}+\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{2}}\right )}{f \,a^{3}}\)	\(349\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

2/f/a^3*d^4*(-1/4/d^2*((1/4*(d*tan(f*x+e))^(3/2)-1/4*d*(d*tan(f*x+e))^(1/2))/(d*tan(f*x+e)+d)^2+5/4/d^(1/2)*ar
ctan((d*tan(f*x+e))^(1/2)/d^(1/2)))+1/4/d^2*(1/8/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))+1/8/(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 = 0.54, size = 191, normalized size = 1.16 \begin {gather*} \frac {\frac {2 \, d^{3} {\left (\frac {\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} \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}}\right )}}{a^{3}} - \frac {5 \, d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3}} - \frac {\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{3} - \sqrt {d \tan \left (f x + e\right )} d^{4}}{a^{3} d^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}}}{8 \, d f} \end {gather*}

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))^3,x, algorithm="maxima")

[Out]

1/8*(2*d^3*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*a
rctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/a^3 - 5*d^(5/2)*arctan(sqrt(d*
tan(f*x + e))/sqrt(d))/a^3 - ((d*tan(f*x + e))^(3/2)*d^3 - sqrt(d*tan(f*x + e))*d^4)/(a^3*d^2*tan(f*x + e)^2 +
2*a^3*d^2*tan(f*x + e) + a^3*d^2))/(d*f)

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Fricas [A]
time = 0.88, size = 440, normalized size = 2.68 \begin {gather*} \left [\frac {2 \, {\left (\sqrt {2} d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} d \tan \left (f x + e\right ) + \sqrt {2} d\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) - \sqrt {2}\right )} \sqrt {-d} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 5 \, {\left (d \tan \left (f x + e\right )^{2} + 2 \, d \tan \left (f x + e\right ) + d\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) - 2 \, {\left (d \tan \left (f x + e\right ) - d\right )} \sqrt {d \tan \left (f x + e\right )}}{16 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}, -\frac {5 \, {\left (d \tan \left (f x + e\right )^{2} + 2 \, d \tan \left (f x + e\right ) + d\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) - 2 \, {\left (\sqrt {2} d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} d \tan \left (f x + e\right ) + \sqrt {2} d\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) - \sqrt {2}\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) + {\left (d \tan \left (f x + e\right ) - d\right )} \sqrt {d \tan \left (f x + e\right )}}{8 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}\right ] \end {gather*}

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))^3,x, algorithm="fricas")

[Out]

[1/16*(2*(sqrt(2)*d*tan(f*x + e)^2 + 2*sqrt(2)*d*tan(f*x + e) + sqrt(2)*d)*sqrt(-d)*log((d*tan(f*x + e)^2 + 2*
sqrt(d*tan(f*x + e))*(sqrt(2)*tan(f*x + e) - sqrt(2))*sqrt(-d) - 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) +
5*(d*tan(f*x + e)^2 + 2*d*tan(f*x + e) + d)*sqrt(-d)*log((d*tan(f*x + e) - 2*sqrt(d*tan(f*x + e))*sqrt(-d) -
d)/(tan(f*x + e) + 1)) - 2*(d*tan(f*x + e) - d)*sqrt(d*tan(f*x + e)))/(a^3*f*tan(f*x + e)^2 + 2*a^3*f*tan(f*x
+ e) + a^3*f), -1/8*(5*(d*tan(f*x + e)^2 + 2*d*tan(f*x + e) + d)*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))
- 2*(sqrt(2)*d*tan(f*x + e)^2 + 2*sqrt(2)*d*tan(f*x + e) + sqrt(2)*d)*sqrt(d)*arctan(1/2*sqrt(d*tan(f*x + e))*
(sqrt(2)*tan(f*x + e) - sqrt(2))/(sqrt(d)*tan(f*x + e))) + (d*tan(f*x + e) - d)*sqrt(d*tan(f*x + e)))/(a^3*f*t
an(f*x + e)^2 + 2*a^3*f*tan(f*x + e) + a^3*f)]

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

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))**3,x)

[Out]

Integral((d*tan(e + f*x))**(3/2)/(tan(e + f*x)**3 + 3*tan(e + f*x)**2 + 3*tan(e + f*x) + 1), x)/a**3

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (143) = 286\).
time = 0.77, size = 323, normalized size = 1.97 \begin {gather*} \frac {1}{16} \, d {\left (\frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\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 )}{a^{3} d f} + \frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\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 )}{a^{3} d f} - \frac {10 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3} f} + \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\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 )}{a^{3} d f} - \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\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 )}{a^{3} d f} - \frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - \sqrt {d \tan \left (f x + e\right )} d^{2}\right )}}{{\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} f}\right )} \end {gather*}

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))^3,x, algorithm="giac")

[Out]

1/16*d*(2*sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x
+ e)))/sqrt(abs(d)))/(a^3*d*f) + 2*sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(a
bs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a^3*d*f) - 10*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^
3*f) + sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d))
+ abs(d))/(a^3*d*f) - sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e)
)*sqrt(abs(d)) + abs(d))/(a^3*d*f) - 2*(sqrt(d*tan(f*x + e))*d^2*tan(f*x + e) - sqrt(d*tan(f*x + e))*d^2)/((d*
tan(f*x + e) + d)^2*a^3*f))

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Mupad [B]
time = 4.77, size = 177, normalized size = 1.08 \begin {gather*} \frac {\frac {d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8}-\frac {d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8}}{f\,a^3\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,f\,a^3\,d^2\,\mathrm {tan}\left (e+f\,x\right )+f\,a^3\,d^2}-\frac {5\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}+\frac {\sqrt {2}\,d^{3/2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{8\,a^3\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d^3*(d*tan(e + f*x))^(1/2))/8 - (d^2*(d*tan(e + f*x))^(3/2))/8)/(a^3*d^2*f + a^3*d^2*f*tan(e + f*x)^2 + 2*a^
3*d^2*f*tan(e + f*x)) - (5*d^(3/2)*atan((d*tan(e + f*x))^(1/2)/d^(1/2)))/(8*a^3*f) + (2^(1/2)*d^(3/2)*(2*atan(
(2^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/2))) + 2*atan((2^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/2)) + (2^(1/2)
*(d*tan(e + f*x))^(3/2))/(2*d^(3/2)))))/(8*a^3*f)

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