3.376 \(\int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=189 \[ -\frac {31 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 d^{3/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 d^{3/2} f}-\frac {27}{8 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {9}{8 a^3 d f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}} \]

[Out]

-31/8*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/d^(3/2)/f-1/4*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/d^(3/2)/f*2^(1/2)-27/8/a^3/d/f/(d*tan(f*x+e))^(1/2)+9/8/a^3/d/f/(d*tan(f*x+e))^(1/2)
/(1+tan(f*x+e))+1/4/a/d/f/(d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e))^2

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Rubi [A]  time = 0.80, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3569, 3649, 3654, 3532, 208, 3634, 63, 205} \[ -\frac {31 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 d^{3/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 d^{3/2} f}-\frac {27}{8 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {9}{8 a^3 d f (\tan (e+f x)+1) \sqrt {d \tan (e+f x)}}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^3),x]

[Out]

(-31*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*d^(3/2)*f) - ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[
2]*Sqrt[d*Tan[e + f*x]])]/(2*Sqrt[2]*a^3*d^(3/2)*f) - 27/(8*a^3*d*f*Sqrt[d*Tan[e + f*x]]) + 9/(8*a^3*d*f*Sqrt[
d*Tan[e + f*x]]*(1 + Tan[e + f*x])) + 1/(4*a*d*f*Sqrt[d*Tan[e + f*x]]*(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 3569

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 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 {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3} \, dx &=\frac {1}{4 a d f \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2}+\frac {\int \frac {\frac {9 a^2 d}{2}-2 a^2 d \tan (e+f x)+\frac {5}{2} a^2 d \tan ^2(e+f x)}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \, dx}{4 a^3 d}\\ &=\frac {9}{8 a^3 d f \sqrt {d \tan (e+f x)} (1+\tan (e+f x))}+\frac {1}{4 a d f \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2}+\frac {\int \frac {\frac {27 a^4 d^2}{2}-4 a^4 d^2 \tan (e+f x)+\frac {27}{2} a^4 d^2 \tan ^2(e+f x)}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx}{8 a^6 d^2}\\ &=-\frac {27}{8 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {9}{8 a^3 d f \sqrt {d \tan (e+f x)} (1+\tan (e+f x))}+\frac {1}{4 a d f \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2}-\frac {\int \frac {\frac {35 a^5 d^4}{4}+\frac {27}{4} a^5 d^4 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{4 a^7 d^5}\\ &=-\frac {27}{8 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {9}{8 a^3 d f \sqrt {d \tan (e+f x)} (1+\tan (e+f x))}+\frac {1}{4 a d f \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2}-\frac {\int \frac {2 a^6 d^4-2 a^6 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^9 d^5}-\frac {31 \int \frac {1+\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{16 a^2 d}\\ &=-\frac {27}{8 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {9}{8 a^3 d f \sqrt {d \tan (e+f x)} (1+\tan (e+f x))}+\frac {1}{4 a d f \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2}-\frac {31 \operatorname {Subst}\left (\int \frac {1}{\sqrt {d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{16 a^2 d f}+\frac {\left (a^3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{-8 a^{12} d^8+d x^2} \, dx,x,\frac {2 a^6 d^4+2 a^6 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f}\\ &=-\frac {\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 d^{3/2} f}-\frac {27}{8 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {9}{8 a^3 d f \sqrt {d \tan (e+f x)} (1+\tan (e+f x))}+\frac {1}{4 a d f \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2}-\frac {31 \operatorname {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^2 d^2 f}\\ &=-\frac {31 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 d^{3/2} f}-\frac {\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 d^{3/2} f}-\frac {27}{8 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {9}{8 a^3 d f \sqrt {d \tan (e+f x)} (1+\tan (e+f x))}+\frac {1}{4 a d f \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]  time = 1.43, size = 173, normalized size = 0.92 \[ \frac {\tan ^{\frac {3}{2}}(e+f x) \left (-62 \tan ^{-1}\left (\sqrt {\tan (e+f x)}\right )+2 \sqrt {2} \left (\log \left (-\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}-1\right )-\log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )\right )-\frac {\sqrt {\tan (e+f x)} (90 \cos (2 (e+f x))+75 \cot (e+f x)-11 \cos (3 (e+f x)) \csc (e+f x)+90)}{2 (\sin (e+f x)+\cos (e+f x))^2}\right )}{16 a^3 f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^3),x]

[Out]

((-62*ArcTan[Sqrt[Tan[e + f*x]]] + 2*Sqrt[2]*(Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[e + f*x]] - Log[1 + Sq
rt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]) - ((90 + 90*Cos[2*(e + f*x)] + 75*Cot[e + f*x] - 11*Cos[3*(e + f*x)]
*Csc[e + f*x])*Sqrt[Tan[e + f*x]])/(2*(Cos[e + f*x] + Sin[e + f*x])^2))*Tan[e + f*x]^(3/2))/(16*a^3*f*(d*Tan[e
+ f*x])^(3/2))

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fricas [A]  time = 0.58, size = 457, normalized size = 2.42 \[ \left [\frac {4 \, \sqrt {2} {\left (\tan \left (f x + e\right )^{3} + 2 \, \tan \left (f x + e\right )^{2} + \tan \left (f x + e\right )\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) - 31 \, {\left (\tan \left (f x + e\right )^{3} + 2 \, \tan \left (f x + e\right )^{2} + \tan \left (f x + e\right )\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (27 \, \tan \left (f x + e\right )^{2} + 45 \, \tan \left (f x + e\right ) + 16\right )}}{16 \, {\left (a^{3} d^{2} f \tan \left (f x + e\right )^{3} + 2 \, a^{3} d^{2} f \tan \left (f x + e\right )^{2} + a^{3} d^{2} f \tan \left (f x + e\right )\right )}}, \frac {\sqrt {2} {\left (\tan \left (f x + e\right )^{3} + 2 \, \tan \left (f x + e\right )^{2} + \tan \left (f x + e\right )\right )} \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) - 31 \, {\left (\tan \left (f x + e\right )^{3} + 2 \, \tan \left (f x + e\right )^{2} + \tan \left (f x + e\right )\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) - \sqrt {d \tan \left (f x + e\right )} {\left (27 \, \tan \left (f x + e\right )^{2} + 45 \, \tan \left (f x + e\right ) + 16\right )}}{8 \, {\left (a^{3} d^{2} f \tan \left (f x + e\right )^{3} + 2 \, a^{3} d^{2} f \tan \left (f x + e\right )^{2} + a^{3} d^{2} f \tan \left (f x + e\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(4*sqrt(2)*(tan(f*x + e)^3 + 2*tan(f*x + e)^2 + tan(f*x + e))*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))) - 31*(tan(f*x + e)^3 + 2*tan(f*x + e)^2 + tan(f*x + e))*s
qrt(-d)*log((d*tan(f*x + e) + 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)) - 2*sqrt(d*tan(f*x + e)
)*(27*tan(f*x + e)^2 + 45*tan(f*x + e) + 16))/(a^3*d^2*f*tan(f*x + e)^3 + 2*a^3*d^2*f*tan(f*x + e)^2 + a^3*d^2
*f*tan(f*x + e)), 1/8*(sqrt(2)*(tan(f*x + e)^3 + 2*tan(f*x + e)^2 + tan(f*x + e))*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)) -
31*(tan(f*x + e)^3 + 2*tan(f*x + e)^2 + tan(f*x + e))*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d)) - sqrt(d*ta
n(f*x + e))*(27*tan(f*x + e)^2 + 45*tan(f*x + e) + 16))/(a^3*d^2*f*tan(f*x + e)^3 + 2*a^3*d^2*f*tan(f*x + e)^2
+ a^3*d^2*f*tan(f*x + e))]

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giac [B]  time = 2.43, size = 341, normalized size = 1.80 \[ -\frac {\frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a^{3} d^{2} f} + \frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a^{3} d^{2} f} + \frac {62 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3} \sqrt {d} f} + \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a^{3} d^{2} f} - \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a^{3} d^{2} f} + \frac {32}{\sqrt {d \tan \left (f x + e\right )} a^{3} f} + \frac {2 \, {\left (11 \, \sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + 13 \, \sqrt {d \tan \left (f x + e\right )} d\right )}}{{\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} f}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(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^2*f) + 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^2*f) + 62*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3*sqr
t(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^2*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^2*f) + 32/(sqrt(d*tan(f*x + e))*a^3*f) + 2*(11*sqrt(d*tan(f*x + e))*d*tan
(f*x + e) + 13*sqrt(d*tan(f*x + e))*d)/((d*tan(f*x + e) + d)^2*a^3*f))/d

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maple [B]  time = 0.33, size = 458, normalized size = 2.42 \[ -\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{16 f \,a^{3} d^{2}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3} d^{2}}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3} d^{2}}+\frac {\sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{16 f \,a^{3} d \left (d^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3} d \left (d^{2}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3} d \left (d^{2}\right )^{\frac {1}{4}}}-\frac {11 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8 f \,a^{3} d \left (d \tan \left (f x +e \right )+d \right )^{2}}-\frac {13 \sqrt {d \tan \left (f x +e \right )}}{8 f \,a^{3} \left (d \tan \left (f x +e \right )+d \right )^{2}}-\frac {31 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{8 a^{3} d^{\frac {3}{2}} f}-\frac {2}{a^{3} d f \sqrt {d \tan \left (f x +e \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(f*x+e))^(3/2)/(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*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*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*2^(1/2)/(d^2)^(1/4)*arctan(-2^(1/2)/(d^2)^(1/
4)*(d*tan(f*x+e))^(1/2)+1)-11/8/f/a^3/d/(d*tan(f*x+e)+d)^2*(d*tan(f*x+e))^(3/2)-13/8/f/a^3/(d*tan(f*x+e)+d)^2*
(d*tan(f*x+e))^(1/2)-31/8*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/d^(3/2)/f-2/a^3/d/f/(d*tan(f*x+e))^(1/2)

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maxima [A]  time = 0.67, size = 190, normalized size = 1.01 \[ -\frac {\frac {27 \, d^{2} \tan \left (f x + e\right )^{2} + 45 \, d^{2} \tan \left (f x + e\right ) + 16 \, d^{2}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} + 2 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d + \sqrt {d \tan \left (f x + e\right )} a^{3} d^{2}} + \frac {\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}}{a^{3}} + \frac {31 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3} \sqrt {d}}}{8 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((27*d^2*tan(f*x + e)^2 + 45*d^2*tan(f*x + e) + 16*d^2)/((d*tan(f*x + e))^(5/2)*a^3 + 2*(d*tan(f*x + e))^
(3/2)*a^3*d + sqrt(d*tan(f*x + e))*a^3*d^2) + (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 + 31*arc
tan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3*sqrt(d)))/(d*f)

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mupad [B]  time = 4.82, size = 176, normalized size = 0.93 \[ -\frac {\frac {27\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8}+\frac {45\,d\,\mathrm {tan}\left (e+f\,x\right )}{8}+2\,d}{a^3\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}+2\,a^3\,d\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}+a^3\,d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {31\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,d^{3/2}\,f}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {63504384\,\sqrt {2}\,a^9\,d^{15/2}\,f^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{63504384\,a^9\,d^8\,f^3+63504384\,a^9\,d^8\,f^3\,\mathrm {tan}\left (e+f\,x\right )}\right )}{4\,a^3\,d^{3/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*d + (45*d*tan(e + f*x))/8 + (27*d*tan(e + f*x)^2)/8)/(a^3*f*(d*tan(e + f*x))^(5/2) + 2*a^3*d*f*(d*tan(e +
f*x))^(3/2) + a^3*d^2*f*(d*tan(e + f*x))^(1/2)) - (31*atan((d*tan(e + f*x))^(1/2)/d^(1/2)))/(8*a^3*d^(3/2)*f)
- (2^(1/2)*atanh((63504384*2^(1/2)*a^9*d^(15/2)*f^3*(d*tan(e + f*x))^(1/2))/(63504384*a^9*d^8*f^3 + 63504384*
a^9*d^8*f^3*tan(e + f*x))))/(4*a^3*d^(3/2)*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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