3.4.75 \(\int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3} \, dx\) [375]

Optimal. Leaf size=165 \[ \frac {11 \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 \sqrt {d} f}+\frac {\text {ArcTan}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 \sqrt {d} f}+\frac {7 \sqrt {d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2} \]

[Out]

11/8*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f/d^(1/2)+1/4*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)/d^(1/2)+7/8*(d*tan(f*x+e))^(1/2)/a^3/d/f/(1+tan(f*x+e))+1/4*(d*tan(f*x+e))^(
1/2)/a/d/f/(a+a*tan(f*x+e))^2

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Rubi [A]
time = 0.37, antiderivative size = 165, 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 = {3650, 3730, 3734, 3613, 211, 3715, 65} \begin {gather*} \frac {11 \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 \sqrt {d} f}+\frac {\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 \sqrt {d} f}+\frac {7 \sqrt {d \tan (e+f x)}}{8 a^3 d f (\tan (e+f x)+1)}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])^3),x]

[Out]

(11*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*Sqrt[d]*f) + ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sqrt[2]
*Sqrt[d*Tan[e + f*x]])]/(2*Sqrt[2]*a^3*Sqrt[d]*f) + (7*Sqrt[d*Tan[e + f*x]])/(8*a^3*d*f*(1 + Tan[e + f*x])) +
Sqrt[d*Tan[e + f*x]]/(4*a*d*f*(a + a*Tan[e + f*x])^2)

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 3650

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

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 {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3} \, dx &=\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}+\frac {\int \frac {\frac {7 a^2 d}{2}-2 a^2 d \tan (e+f x)+\frac {3}{2} a^2 d \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2} \, dx}{4 a^3 d}\\ &=\frac {7 \sqrt {d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}+\frac {\int \frac {\frac {7 a^4 d^2}{2}-4 a^4 d^2 \tan (e+f x)+\frac {7}{2} a^4 d^2 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{8 a^6 d^2}\\ &=\frac {7 \sqrt {d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}+\frac {11 \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 {\int \frac {-4 a^5 d^2-4 a^5 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{16 a^8 d^2}\\ &=\frac {7 \sqrt {d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}+\frac {11 \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^2 d^2\right ) \text {Subst}\left (\int \frac {1}{32 a^{10} d^4+d x^2} \, dx,x,\frac {-4 a^5 d^2+4 a^5 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f}\\ &=\frac {\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 \sqrt {d} f}+\frac {7 \sqrt {d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}+\frac {11 \text {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^2 d f}\\ &=\frac {11 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 \sqrt {d} f}+\frac {\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 \sqrt {d} f}+\frac {7 \sqrt {d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]
time = 1.08, size = 217, normalized size = 1.32 \begin {gather*} \frac {\left (22 \text {ArcTan}\left (\sqrt {\tan (e+f x)}\right )+4 \sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) (\cos (e+f x)+\sin (e+f x))^2-4 \sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right ) (\cos (e+f x)+\sin (e+f x))^2+22 \text {ArcTan}\left (\sqrt {\tan (e+f x)}\right ) \sin (2 (e+f x))+9 \sqrt {\tan (e+f x)}+9 \cos (2 (e+f x)) \sqrt {\tan (e+f x)}+7 \sin (2 (e+f x)) \sqrt {\tan (e+f x)}\right ) \sqrt {\tan (e+f x)}}{16 a^3 f (\cos (e+f x)+\sin (e+f x))^2 \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])^3),x]

[Out]

((22*ArcTan[Sqrt[Tan[e + f*x]]] + 4*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]*(Cos[e + f*x] + Sin[e + f*x
])^2 - 4*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*(Cos[e + f*x] + Sin[e + f*x])^2 + 22*ArcTan[Sqrt[Tan[e
+ f*x]]]*Sin[2*(e + f*x)] + 9*Sqrt[Tan[e + f*x]] + 9*Cos[2*(e + f*x)]*Sqrt[Tan[e + f*x]] + 7*Sin[2*(e + f*x)]
*Sqrt[Tan[e + f*x]])*Sqrt[Tan[e + f*x]])/(16*a^3*f*(Cos[e + f*x] + Sin[e + f*x])^2*Sqrt[d*Tan[e + f*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(348\) vs. \(2(136)=272\).
time = 0.24, size = 349, normalized size = 2.12

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\frac {\frac {\frac {7 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {9 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {11 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{4}}+\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{4}}\right )}{f \,a^{3}}\)	\(349\)
default	\(\frac {2 d^{4} \left (\frac {\frac {\frac {7 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {9 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {11 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{4}}+\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{4}}\right )}{f \,a^{3}}\)	\(349\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2/f/a^3*d^4*(1/4/d^4*((7/4*(d*tan(f*x+e))^(3/2)+9/4*d*(d*tan(f*x+e))^(1/2))/(d*tan(f*x+e)+d)^2+11/4/d^(1/2)*ar
ctan((d*tan(f*x+e))^(1/2)/d^(1/2)))+1/4/d^4*(-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*arcta
n(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 = 0.53, size = 187, normalized size = 1.13 \begin {gather*} \frac {\frac {7 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d + 9 \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{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}} - \frac {2 \, d {\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 {11 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3}}}{8 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

1/8*((7*(d*tan(f*x + e))^(3/2)*d + 9*sqrt(d*tan(f*x + e))*d^2)/(a^3*d^2*tan(f*x + e)^2 + 2*a^3*d^2*tan(f*x + e
) + a^3*d^2) - 2*d*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + s
qrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/a^3 + 11*sqrt(d)*arcta
n(sqrt(d*tan(f*x + e))/sqrt(d))/a^3)/(d*f)

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Fricas [A]
time = 0.93, size = 407, normalized size = 2.47 \begin {gather*} \left [-\frac {2 \, \sqrt {2} {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 11 \, {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\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 \, \sqrt {d \tan \left (f x + e\right )} {\left (7 \, \tan \left (f x + e\right ) + 9\right )}}{16 \, {\left (a^{3} d f \tan \left (f x + e\right )^{2} + 2 \, a^{3} d f \tan \left (f x + e\right ) + a^{3} d f\right )}}, -\frac {2 \, \sqrt {2} {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) - 11 \, {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) - \sqrt {d \tan \left (f x + e\right )} {\left (7 \, \tan \left (f x + e\right ) + 9\right )}}{8 \, {\left (a^{3} d f \tan \left (f x + e\right )^{2} + 2 \, a^{3} d f \tan \left (f x + e\right ) + a^{3} d f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/16*(2*sqrt(2)*(tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(-d)*log((d*tan(f*x + e)^2 + 2*sqrt(2)*sqrt(d*tan(
f*x + e))*sqrt(-d)*(tan(f*x + e) - 1) - 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) + 11*(tan(f*x + e)^2 + 2*t
an(f*x + e) + 1)*sqrt(-d)*log((d*tan(f*x + e) - 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)) - 2*s
qrt(d*tan(f*x + e))*(7*tan(f*x + e) + 9))/(a^3*d*f*tan(f*x + e)^2 + 2*a^3*d*f*tan(f*x + e) + a^3*d*f), -1/8*(2
*sqrt(2)*(tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) -
1)/(sqrt(d)*tan(f*x + e))) - 11*(tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqr
t(d)) - sqrt(d*tan(f*x + e))*(7*tan(f*x + e) + 9))/(a^3*d*f*tan(f*x + e)^2 + 2*a^3*d*f*tan(f*x + e) + a^3*d*f)
]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (144) = 288\).
time = 0.75, size = 318, normalized size = 1.93 \begin {gather*} -\frac {\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 )}{8 \, a^{3} d^{2} f} - \frac {\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 )}{8 \, a^{3} d^{2} f} + \frac {11 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{8 \, a^{3} \sqrt {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 )}{16 \, a^{3} d^{2} 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 )}{16 \, a^{3} d^{2} f} + \frac {7 \, \sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + 9 \, \sqrt {d \tan \left (f x + e\right )} d}{8 \, {\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/8*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^2*f) - 1/8*sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(ab
s(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a^3*d^2*f) + 11/8*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3*sqr
t(d)*f) - 1/16*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^2*f) + 1/16*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^2*f) + 1/8*(7*sqrt(d*tan(f*x + e))*d*tan(f*x + e) + 9*sqrt(d*ta
n(f*x + e))*d)/((d*tan(f*x + e) + d)^2*a^3*f)

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Mupad [B]
time = 4.78, size = 172, normalized size = 1.04 \begin {gather*} \frac {\frac {9\,d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8}+\frac {7\,{\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 {11\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,\sqrt {d}\,f}-\frac {\sqrt {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\,\sqrt {d}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((9*d*(d*tan(e + f*x))^(1/2))/8 + (7*(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)) + (11*atan((d*tan(e + f*x))^(1/2)/d^(1/2)))/(8*a^3*d^(1/2)*f) - (2^(1/2)*(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*d^(1/2)*f)

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