3.372 \(\int \frac {(d \tan (e+f x))^{5/2}}{(a+a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=164 \[ \frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}-\frac {d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}+\frac {5 d^2 \sqrt {d \tan (e+f x)}}{8 a^3 f (\tan (e+f x)+1)}-\frac {d^2 \sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.56, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3565, 3649, 3654, 3532, 208, 3634, 63, 205} \[ \frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {5 d^2 \sqrt {d \tan (e+f x)}}{8 a^3 f (\tan (e+f x)+1)}-\frac {d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}-\frac {d^2 \sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(5/2)/(a + a*Tan[e + f*x])^3,x]

[Out]

(d^(5/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*f) - (d^(5/2)*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(
Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(2*Sqrt[2]*a^3*f) + (5*d^2*Sqrt[d*Tan[e + f*x]])/(8*a^3*f*(1 + Tan[e + f*x]))
- (d^2*Sqrt[d*Tan[e + f*x]])/(4*a*f*(a + a*Tan[e + f*x])^2)

Rule 63

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 205

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

Rule 208

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

Rule 3532

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 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*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] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3634

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 3649

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*T
an[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 3654

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*ta
n[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*x])^n*Simp[a*(A - C) - (A*b - b*
C)*Tan[e + f*x], x], x], x] + Dist[(A*b^2 + 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, 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))^{5/2}}{(a+a \tan (e+f x))^3} \, dx &=-\frac {d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}+\frac {\int \frac {\frac {a^2 d^3}{2}-2 a^2 d^3 \tan (e+f x)+\frac {5}{2} a^2 d^3 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2} \, dx}{4 a^3}\\ &=\frac {5 d^2 \sqrt {d \tan (e+f x)}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}+\frac {\int \frac {-\frac {3}{2} a^4 d^4+\frac {5}{2} a^4 d^4 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{8 a^6 d}\\ &=\frac {5 d^2 \sqrt {d \tan (e+f x)}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}+\frac {\int \frac {-4 a^5 d^4+4 a^5 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{16 a^8 d}+\frac {d^3 \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 {5 d^2 \sqrt {d \tan (e+f x)}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}+\frac {d^3 \operatorname {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^7\right ) \operatorname {Subst}\left (\int \frac {1}{-32 a^{10} d^8+d x^2} \, dx,x,\frac {-4 a^5 d^4-4 a^5 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f}\\ &=-\frac {d^{5/2} \tanh ^{-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 {5 d^2 \sqrt {d \tan (e+f x)}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}+\frac {d^2 \operatorname {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 {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}-\frac {d^{5/2} \tanh ^{-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 {5 d^2 \sqrt {d \tan (e+f x)}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]  time = 3.05, size = 192, normalized size = 1.17 \[ \frac {\sec (e+f x) (d \tan (e+f x))^{5/2} (\sin (e+f x)+\cos (e+f x))^3 \left (\frac {\csc ^4(e+f x) (5 \sin (2 (e+f x))+3 \cos (2 (e+f x))+3)}{(\cot (e+f x)+1)^2}+\frac {2 \csc (e+f x) \sec (e+f x) \left (\tan ^{-1}\left (\sqrt {\tan (e+f x)}\right )+\sqrt {2} \left (\log \left (-\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}-1\right )-\log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )\right )\right )}{\tan ^{\frac {3}{2}}(e+f x)}\right )}{16 a^3 f (\tan (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(5/2)/(a + a*Tan[e + f*x])^3,x]

[Out]

(Sec[e + f*x]*(Cos[e + f*x] + Sin[e + f*x])^3*((Csc[e + f*x]^4*(3 + 3*Cos[2*(e + f*x)] + 5*Sin[2*(e + f*x)]))/
(1 + Cot[e + f*x])^2 + (2*Csc[e + f*x]*(ArcTan[Sqrt[Tan[e + f*x]]] + Sqrt[2]*(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]]))*Sec[e + f*x])/Tan[e + f*x]^(3/2))*(
d*Tan[e + f*x])^(5/2))/(16*a^3*f*(1 + Tan[e + f*x])^3)

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fricas [A]  time = 0.59, size = 442, normalized size = 2.70 \[ \left [\frac {4 \, {\left (\sqrt {2} d^{2} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} d^{2} \tan \left (f x + e\right ) + \sqrt {2} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {-d}}{2 \, d \tan \left (f x + e\right )}\right ) + {\left (d^{2} \tan \left (f x + e\right )^{2} + 2 \, d^{2} \tan \left (f x + e\right ) + d^{2}\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) + 2 \, {\left (5 \, d^{2} \tan \left (f x + e\right ) + 3 \, d^{2}\right )} \sqrt {d \tan \left (f x + e\right )}}{16 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}, \frac {{\left (d^{2} \tan \left (f x + e\right )^{2} + 2 \, d^{2} \tan \left (f x + e\right ) + d^{2}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) + {\left (\sqrt {2} d^{2} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} d^{2} \tan \left (f x + e\right ) + \sqrt {2} d^{2}\right )} \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {d} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + {\left (5 \, d^{2} \tan \left (f x + e\right ) + 3 \, d^{2}\right )} \sqrt {d \tan \left (f x + e\right )}}{8 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/16*(4*(sqrt(2)*d^2*tan(f*x + e)^2 + 2*sqrt(2)*d^2*tan(f*x + e) + sqrt(2)*d^2)*sqrt(-d)*arctan(1/2*sqrt(d*ta
n(f*x + e))*(sqrt(2)*tan(f*x + e) + sqrt(2))*sqrt(-d)/(d*tan(f*x + e))) + (d^2*tan(f*x + e)^2 + 2*d^2*tan(f*x
+ e) + d^2)*sqrt(-d)*log((d*tan(f*x + e) + 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)) + 2*(5*d^2
*tan(f*x + e) + 3*d^2)*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*((d^2*
tan(f*x + e)^2 + 2*d^2*tan(f*x + e) + d^2)*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d)) + (sqrt(2)*d^2*tan(f*x
+ e)^2 + 2*sqrt(2)*d^2*tan(f*x + e) + sqrt(2)*d^2)*sqrt(d)*log((d*tan(f*x + e)^2 - 2*sqrt(d*tan(f*x + e))*(sq
rt(2)*tan(f*x + e) + sqrt(2))*sqrt(d) + 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) + (5*d^2*tan(f*x + e) + 3*
d^2)*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)]

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giac [B]  time = 3.94, size = 326, normalized size = 1.99 \[ -\frac {1}{16} \, d^{2} {\left (\frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a^{3} d f} + \frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a^{3} d f} - \frac {2 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3} f} + \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a^{3} d f} - \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a^{3} d f} - \frac {2 \, {\left (5 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) + 3 \, \sqrt {d \tan \left (f x + e\right )} d^{2}\right )}}{{\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} f}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/16*d^2*(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)*sqr
t(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a^3*d*f) - 2*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*(5*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e) + 3*sqrt(d*tan(f*x + e))*d^2
)/((d*tan(f*x + e) + d)^2*a^3*f))

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maple [B]  time = 0.33, size = 440, normalized size = 2.68 \[ -\frac {d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{16 f \,a^{3}}-\frac {d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3}}+\frac {d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3}}+\frac {d^{3} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{16 f \,a^{3} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {d^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {d^{3} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {5 d^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8 f \,a^{3} \left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {3 d^{4} \sqrt {d \tan \left (f x +e \right )}}{8 f \,a^{3} \left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{8 a^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^3,x)

[Out]

-1/16/f/a^3*d^2*(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)))-1/8/f/a^3*d^2*(d^2)^(1/4)*2^(1/2)*arctan(2^(
1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/8/f/a^3*d^2*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(
f*x+e))^(1/2)+1)+1/16/f/a^3*d^3*2^(1/2)/(d^2)^(1/4)*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)))+1/8/f/a^3*d^3*2^(1/2)/(d^2)^
(1/4)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/8/f/a^3*d^3*2^(1/2)/(d^2)^(1/4)*arctan(-2^(1/2)/(d^
2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+5/8/f/a^3*d^3/(d*tan(f*x+e)+d)^2*(d*tan(f*x+e))^(3/2)+3/8/f/a^3*d^4/(d*tan(f*
x+e)+d)^2*(d*tan(f*x+e))^(1/2)+1/8*d^(5/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f

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maxima [A]  time = 0.54, size = 184, normalized size = 1.12 \[ -\frac {\frac {d^{4} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a^{3}} - \frac {d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3}} - \frac {5 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{4} + 3 \, \sqrt {d \tan \left (f x + e\right )} d^{5}}{a^{3} d^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}}}{8 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/8*(d^4*(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^3 - d^(7/2)*arctan(sqrt(d*tan(f*x + e))/sqrt(d
))/a^3 - (5*(d*tan(f*x + e))^(3/2)*d^4 + 3*sqrt(d*tan(f*x + e))*d^5)/(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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mupad [B]  time = 4.80, size = 153, normalized size = 0.93 \[ \frac {\frac {3\,d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8}+\frac {5\,d^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8}}{f\,a^3\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,f\,a^3\,d^2\,\mathrm {tan}\left (e+f\,x\right )+f\,a^3\,d^2}+\frac {d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {\sqrt {2}\,d^{5/2}\,\mathrm {atanh}\left (\frac {9\,\sqrt {2}\,d^{33/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{32\,\left (\frac {9\,d^{17}\,\mathrm {tan}\left (e+f\,x\right )}{32}+\frac {9\,d^{17}}{32}\right )}\right )}{4\,a^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(5/2)/(a + a*tan(e + f*x))^3,x)

[Out]

((3*d^4*(d*tan(e + f*x))^(1/2))/8 + (5*d^3*(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)) + (d^(5/2)*atan((d*tan(e + f*x))^(1/2)/d^(1/2)))/(8*a^3*f) - (2^(1/2)*d^(5/2)*atanh(
(9*2^(1/2)*d^(33/2)*(d*tan(e + f*x))^(1/2))/(32*((9*d^17*tan(e + f*x))/32 + (9*d^17)/32))))/(4*a^3*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\tan ^{3}{\left (e + f x \right )} + 3 \tan ^{2}{\left (e + f x \right )} + 3 \tan {\left (e + f x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(5/2)/(a+a*tan(f*x+e))**3,x)

[Out]

Integral((d*tan(e + f*x))**(5/2)/(tan(e + f*x)**3 + 3*tan(e + f*x)**2 + 3*tan(e + f*x) + 1), x)/a**3

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