3.4.63 \(\int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx\) [363]

Optimal. Leaf size=135 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a d^{5/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a d^{5/2} f}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}+\frac {2}{a d^2 f \sqrt {d \tan (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.33, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3650, 3730, 12, 16, 3654, 3613, 214, 3715, 65, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a d^{5/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a d^{5/2} f}+\frac {2}{a d^2 f \sqrt {d \tan (e+f x)}}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])),x]

[Out]

ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]]/(a*d^(5/2)*f) - ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*
Tan[e + f*x]])]/(Sqrt[2]*a*d^(5/2)*f) - 2/(3*a*d*f*(d*Tan[e + f*x])^(3/2)) + 2/(a*d^2*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 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[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 3654

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1
/(c^2 + d^2), Int[Simp[a^2*c - b^2*c + 2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e +
f*x]], x], x] + Dist[(b*c - a*d)^2/(c^2 + d^2), Int[(1 + Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan
[e + f*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]

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])))

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 1.47, size = 130, normalized size = 0.96 \begin {gather*} \frac {-8+24 \tan (e+f x)+12 \text {ArcTan}\left (\sqrt {\tan (e+f x)}\right ) \tan ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} \left (\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)}{12 a d f (d \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])),x]

[Out]

(-8 + 24*Tan[e + f*x] + 12*ArcTan[Sqrt[Tan[e + f*x]]]*Tan[e + f*x]^(3/2) + 3*Sqrt[2]*(Log[-1 + Sqrt[2]*Sqrt[Ta
n[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))/(12*a*d*
f*(d*Tan[e + f*x])^(3/2))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(332\) vs. \(2(112)=224\).
time = 0.16, size = 333, normalized size = 2.47

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (-\frac {1}{3 d^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 d^{\frac {9}{2}}}+\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{2 d^{4}}\right )}{f a}\)	\(333\)
default	\(\frac {2 d^{2} \left (-\frac {1}{3 d^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 d^{\frac {9}{2}}}+\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{2 d^{4}}\right )}{f a}\)	\(333\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

2/f/a*d^2*(-1/3/d^3/(d*tan(f*x+e))^(3/2)+1/d^4/(d*tan(f*x+e))^(1/2)+1/2/d^(9/2)*arctan((d*tan(f*x+e))^(1/2)/d^
(1/2))+1/2/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*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))))

________________________________________________________________________________________

Maxima [A]
time = 0.51, size = 151, normalized size = 1.12 \begin {gather*} -\frac {\frac {3 \, {\left (\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 )}}{a d} - \frac {12 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a d^{\frac {3}{2}}} - \frac {8 \, {\left (3 \, d \tan \left (f x + e\right ) - d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a d}}{12 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x, algorithm="maxima")

[Out]

-1/12*(3*(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))/(a*d) - 12*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(
a*d^(3/2)) - 8*(3*d*tan(f*x + e) - d)/((d*tan(f*x + e))^(3/2)*a*d))/(d*f)

________________________________________________________________________________________

Fricas [A]
time = 0.98, size = 326, normalized size = 2.41 \begin {gather*} \left [\frac {3 \, \sqrt {2} \sqrt {-d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{2} - 3 \, \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 ) \tan \left (f x + e\right )^{2} + 4 \, \sqrt {d \tan \left (f x + e\right )} {\left (3 \, \tan \left (f x + e\right ) - 1\right )}}{6 \, a d^{3} f \tan \left (f x + e\right )^{2}}, \frac {3 \, \sqrt {2} \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 ) \tan \left (f x + e\right )^{2} + 12 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) \tan \left (f x + e\right )^{2} + 8 \, \sqrt {d \tan \left (f x + e\right )} {\left (3 \, \tan \left (f x + e\right ) - 1\right )}}{12 \, a d^{3} f \tan \left (f x + e\right )^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x, algorithm="fricas")

[Out]

[1/6*(3*sqrt(2)*sqrt(-d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan(f*x + e) + 1)/(d*tan(f*x + e)))
*tan(f*x + e)^2 - 3*sqrt(-d)*log((d*tan(f*x + e) - 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1))*ta
n(f*x + e)^2 + 4*sqrt(d*tan(f*x + e))*(3*tan(f*x + e) - 1))/(a*d^3*f*tan(f*x + e)^2), 1/12*(3*sqrt(2)*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))*tan(f*x + e)^2 + 12*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))*tan(f*x + e)^2 + 8*sqrt(d*t
an(f*x + e))*(3*tan(f*x + e) - 1))/(a*d^3*f*tan(f*x + e)^2)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )} + \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (117) = 234\).
time = 0.75, size = 305, normalized size = 2.26 \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 )}{4 \, a d^{4} 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 )}{4 \, a d^{4} f} + \frac {\arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a d^{\frac {5}{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 )}{8 \, a d^{4} 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 )}{8 \, a d^{4} f} + \frac {2 \, {\left (3 \, d \tan \left (f x + e\right ) - d\right )}}{3 \, \sqrt {d \tan \left (f x + e\right )} a d^{3} f \tan \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 4.82, size = 130, normalized size = 0.96 \begin {gather*} \frac {\frac {2\,\mathrm {tan}\left (e+f\,x\right )}{d}-\frac {2}{3\,d}}{a\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,d^{5/2}\,f}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {12\,\sqrt {2}\,a^3\,d^{21/2}\,f^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{12\,a^3\,d^{11}\,f^3+12\,a^3\,d^{11}\,f^3\,\mathrm {tan}\left (e+f\,x\right )}\right )}{2\,a\,d^{5/2}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*tan(e + f*x))/d - 2/(3*d))/(a*f*(d*tan(e + f*x))^(3/2)) + atan((d*tan(e + f*x))^(1/2)/d^(1/2))/(a*d^(5/2)*
f) - (2^(1/2)*atanh((12*2^(1/2)*a^3*d^(21/2)*f^3*(d*tan(e + f*x))^(1/2))/(12*a^3*d^11*f^3 + 12*a^3*d^11*f^3*ta
n(e + f*x))))/(2*a*d^(5/2)*f)

________________________________________________________________________________________