\(\int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx\) [358]

Optimal result

Integrand size = 25, antiderivative size = 111 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=-\frac {d^{5/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a f}+\frac {d^{5/2} \arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f}+\frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3647, 3734, 3613, 211, 3715, 65} \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=-\frac {d^{5/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a f}+\frac {d^{5/2} \arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f}+\frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f} \]

[In]

Int[(d*Tan[e + f*x])^(5/2)/(a + a*Tan[e + f*x]),x]

[Out]

-((d^(5/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(a*f)) + (d^(5/2)*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sq
rt[2]*Sqrt[d*Tan[e + f*x]])])/(Sqrt[2]*a*f) + (2*d^2*Sqrt[d*Tan[e + f*x]])/(a*f)

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 3647

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 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !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 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*} \text {integral}& = \frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}+\frac {2 \int \frac {-\frac {a d^3}{2}-\frac {1}{2} a d^3 \tan (e+f x)-\frac {1}{2} a d^3 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{a} \\ & = \frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}+\frac {\int \frac {-\frac {1}{2} a^2 d^3-\frac {1}{2} a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{a^3}-\frac {1}{2} d^3 \int \frac {1+\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx \\ & = \frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}-\frac {d^3 \text {Subst}\left (\int \frac {1}{\sqrt {d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{2 f}-\frac {\left (a d^6\right ) \text {Subst}\left (\int \frac {1}{\frac {a^4 d^6}{2}+d x^2} \, dx,x,\frac {-\frac {1}{2} a^2 d^3+\frac {1}{2} a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{2 f} \\ & = \frac {d^{5/2} \arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f}+\frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f}-\frac {d^2 \text {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f} \\ & = -\frac {d^{5/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a f}+\frac {d^{5/2} \arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f}+\frac {2 d^2 \sqrt {d \tan (e+f x)}}{a f} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.17 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=\frac {-2 d^{5/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-(1+i) (-1)^{3/4} d^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+(1+i) \sqrt [4]{-1} d^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+4 d^2 \sqrt {d \tan (e+f x)}}{2 a f} \]

[In]

Integrate[(d*Tan[e + f*x])^(5/2)/(a + a*Tan[e + f*x]),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(311\) vs. \(2(93)=186\).

Time = 0.85 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.81

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{2}-\frac {\sqrt {d}\, \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2}\right )}{f a}\)	\(312\)
default	\(\frac {2 d^{2} \left (\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{2}-\frac {\sqrt {d}\, \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2}\right )}{f a}\)	\(312\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.23 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=\left [\frac {\sqrt {2} \sqrt {-d} d^{2} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) - \sqrt {2}\right )} \sqrt {-d} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, \sqrt {-d} d^{2} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) + 8 \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{4 \, a f}, -\frac {\sqrt {2} d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) - \sqrt {2}\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) + 2 \, d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) - 4 \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{2 \, a f}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=\frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\tan {\left (e + f x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

Integral((d*tan(e + f*x))**(5/2)/(tan(e + f*x) + 1), x)/a

Maxima [A] (verification not implemented)

none

Time = 0.55 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.17 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=-\frac {\frac {d^{4} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )}}{a} + \frac {2 \, d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a} - \frac {4 \, \sqrt {d \tan \left (f x + e\right )} d^{3}}{a}}{2 \, d f} \]

[In]

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

[Out]

-1/2*(d^4*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*ar
ctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/a + 2*d^(7/2)*arctan(sqrt(d*tan
(f*x + e))/sqrt(d))/a - 4*sqrt(d*tan(f*x + e))*d^3/a)/(d*f)

Giac [F(-1)]

Timed out. \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 4.37 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.12 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+a \tan (e+f x)} \, dx=\frac {2\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{a\,f}-\frac {d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,f}-\frac {\sqrt {2}\,d^{5/2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{4\,a\,f} \]

[In]

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

[Out]

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