3.377 \(\int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=215 \[ \frac {59 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 d^{5/2} f}-\frac {\tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 d^{5/2} f}+\frac {63}{8 a^3 d^2 f \sqrt {d \tan (e+f x)}}+\frac {11}{8 a^3 d f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}-\frac {55}{24 a^3 d f (d \tan (e+f x))^{3/2}}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}} \]

[Out]

59/8*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/d^(5/2)/f-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/d^(5/2)/f*2^(1/2)+63/8/a^3/d^2/f/(d*tan(f*x+e))^(1/2)-55/24/a^3/d/f/(d*tan(f*x+e))^(3/
2)+11/8/a^3/d/f/(d*tan(f*x+e))^(3/2)/(1+tan(f*x+e))+1/4/a/d/f/(d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2

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Rubi [A]  time = 1.03, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3569, 3649, 3650, 3653, 3532, 205, 3634, 63} \[ \frac {59 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 d^{5/2} f}-\frac {\tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 d^{5/2} f}+\frac {63}{8 a^3 d^2 f \sqrt {d \tan (e+f x)}}+\frac {11}{8 a^3 d f (\tan (e+f x)+1) (d \tan (e+f x))^{3/2}}-\frac {55}{24 a^3 d f (d \tan (e+f x))^{3/2}}+\frac {1}{4 a d f (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3),x]

[Out]

(59*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*d^(5/2)*f) - ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sqrt[2]
*Sqrt[d*Tan[e + f*x]])]/(2*Sqrt[2]*a^3*d^(5/2)*f) - 55/(24*a^3*d*f*(d*Tan[e + f*x])^(3/2)) + 63/(8*a^3*d^2*f*S
qrt[d*Tan[e + f*x]]) + 11/(8*a^3*d*f*(d*Tan[e + f*x])^(3/2)*(1 + Tan[e + f*x])) + 1/(4*a*d*f*(d*Tan[e + f*x])^
(3/2)*(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 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 3650

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*t
an[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 + a^2*C)*(a + b*Tan[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)) - a*C*(b*c*(m + 1)
+ a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - b*C)*Tan[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 2)*Tan[e + f*x]^2,
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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3653

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}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3} \, dx &=\frac {1}{4 a d f (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2}+\frac {\int \frac {\frac {11 a^2 d}{2}-2 a^2 d \tan (e+f x)+\frac {7}{2} a^2 d \tan ^2(e+f x)}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^2} \, dx}{4 a^3 d}\\ &=\frac {11}{8 a^3 d f (d \tan (e+f x))^{3/2} (1+\tan (e+f x))}+\frac {1}{4 a d f (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2}+\frac {\int \frac {\frac {55 a^4 d^2}{2}-4 a^4 d^2 \tan (e+f x)+\frac {55}{2} a^4 d^2 \tan ^2(e+f x)}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx}{8 a^6 d^2}\\ &=-\frac {55}{24 a^3 d f (d \tan (e+f x))^{3/2}}+\frac {11}{8 a^3 d f (d \tan (e+f x))^{3/2} (1+\tan (e+f x))}+\frac {1}{4 a d f (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2}-\frac {\int \frac {\frac {189 a^5 d^4}{4}+\frac {165}{4} a^5 d^4 \tan ^2(e+f x)}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx}{12 a^7 d^5}\\ &=-\frac {55}{24 a^3 d f (d \tan (e+f x))^{3/2}}+\frac {63}{8 a^3 d^2 f \sqrt {d \tan (e+f x)}}+\frac {11}{8 a^3 d f (d \tan (e+f x))^{3/2} (1+\tan (e+f x))}+\frac {1}{4 a d f (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2}+\frac {\int \frac {\frac {189 a^6 d^6}{8}+3 a^6 d^6 \tan (e+f x)+\frac {189}{8} a^6 d^6 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{6 a^8 d^8}\\ &=-\frac {55}{24 a^3 d f (d \tan (e+f x))^{3/2}}+\frac {63}{8 a^3 d^2 f \sqrt {d \tan (e+f x)}}+\frac {11}{8 a^3 d f (d \tan (e+f x))^{3/2} (1+\tan (e+f x))}+\frac {1}{4 a d f (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2}+\frac {\int \frac {3 a^7 d^6+3 a^7 d^6 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{12 a^{10} d^8}+\frac {59 \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^2}\\ &=-\frac {55}{24 a^3 d f (d \tan (e+f x))^{3/2}}+\frac {63}{8 a^3 d^2 f \sqrt {d \tan (e+f x)}}+\frac {11}{8 a^3 d f (d \tan (e+f x))^{3/2} (1+\tan (e+f x))}+\frac {1}{4 a d f (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2}+\frac {59 \operatorname {Subst}\left (\int \frac {1}{\sqrt {d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{16 a^2 d^2 f}-\frac {\left (3 a^4 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{18 a^{14} d^{12}+d x^2} \, dx,x,\frac {3 a^7 d^6-3 a^7 d^6 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{2 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 d^{5/2} f}-\frac {55}{24 a^3 d f (d \tan (e+f x))^{3/2}}+\frac {63}{8 a^3 d^2 f \sqrt {d \tan (e+f x)}}+\frac {11}{8 a^3 d f (d \tan (e+f x))^{3/2} (1+\tan (e+f x))}+\frac {1}{4 a d f (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2}+\frac {59 \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^3 f}\\ &=\frac {59 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 d^{5/2} 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 d^{5/2} f}-\frac {55}{24 a^3 d f (d \tan (e+f x))^{3/2}}+\frac {63}{8 a^3 d^2 f \sqrt {d \tan (e+f x)}}+\frac {11}{8 a^3 d f (d \tan (e+f x))^{3/2} (1+\tan (e+f x))}+\frac {1}{4 a d f (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]  time = 6.30, size = 368, normalized size = 1.71 \[ \frac {\tan ^3(e+f x) \sec ^3(e+f x) (\sin (e+f x)+\cos (e+f x))^3 \left (6 \cot (e+f x)-\frac {2}{3} \csc ^2(e+f x)-\frac {17 \sin (e+f x)}{8 (\sin (e+f x)+\cos (e+f x))}+\frac {1}{8 (\sin (e+f x)+\cos (e+f x))^2}+\frac {8}{3}\right )}{f (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}}+\frac {\tan ^{\frac {5}{2}}(e+f x) \sec ^3(e+f x) (\sin (e+f x)+\cos (e+f x))^3 \left (\frac {126 \tan ^{-1}\left (\sqrt {\tan (e+f x)}\right ) (\tan (e+f x)+1) \csc (e+f x) \sec ^3(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 (\cot (e+f x)+1)}+\frac {2 \sin (2 (e+f x)) \left (\sqrt {2} \left (\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )-\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )\right )-2 \tan ^{-1}\left (\sqrt {\tan (e+f x)}\right )\right ) (\tan (e+f x)+1) \csc ^2(e+f x) \sec ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) (\cot (e+f x)+1)}\right )}{16 f (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3),x]

[Out]

(Sec[e + f*x]^3*(Cos[e + f*x] + Sin[e + f*x])^3*(8/3 + 6*Cot[e + f*x] - (2*Csc[e + f*x]^2)/3 + 1/(8*(Cos[e + f
*x] + Sin[e + f*x])^2) - (17*Sin[e + f*x])/(8*(Cos[e + f*x] + Sin[e + f*x])))*Tan[e + f*x]^3)/(f*(d*Tan[e + f*
x])^(5/2)*(a + a*Tan[e + f*x])^3) + (Sec[e + f*x]^3*(Cos[e + f*x] + Sin[e + f*x])^3*Tan[e + f*x]^(5/2)*((126*A
rcTan[Sqrt[Tan[e + f*x]]]*Csc[e + f*x]*Sec[e + f*x]^3*(1 + Tan[e + f*x]))/((1 + Cot[e + f*x])*(1 + Tan[e + f*x
]^2)^2) + (2*(Sqrt[2]*(-ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] + ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]) - 2*A
rcTan[Sqrt[Tan[e + f*x]]])*Csc[e + f*x]^2*Sec[e + f*x]^2*Sin[2*(e + f*x)]*(1 + Tan[e + f*x]))/((1 + Cot[e + f*
x])*(1 + Tan[e + f*x]^2))))/(16*f*(d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3)

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fricas [A]  time = 0.77, size = 486, normalized size = 2.26 \[ \left [-\frac {6 \, \sqrt {2} {\left (\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )^{2}\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 177 \, {\left (\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )^{2}\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) - 2 \, {\left (189 \, \tan \left (f x + e\right )^{3} + 323 \, \tan \left (f x + e\right )^{2} + 112 \, \tan \left (f x + e\right ) - 16\right )} \sqrt {d \tan \left (f x + e\right )}}{48 \, {\left (a^{3} d^{3} f \tan \left (f x + e\right )^{4} + 2 \, a^{3} d^{3} f \tan \left (f x + e\right )^{3} + a^{3} d^{3} f \tan \left (f x + e\right )^{2}\right )}}, \frac {6 \, \sqrt {2} {\left (\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )^{2}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) + 177 \, {\left (\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )^{2}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) + {\left (189 \, \tan \left (f x + e\right )^{3} + 323 \, \tan \left (f x + e\right )^{2} + 112 \, \tan \left (f x + e\right ) - 16\right )} \sqrt {d \tan \left (f x + e\right )}}{24 \, {\left (a^{3} d^{3} f \tan \left (f x + e\right )^{4} + 2 \, a^{3} d^{3} f \tan \left (f x + e\right )^{3} + a^{3} d^{3} f \tan \left (f x + e\right )^{2}\right )}}\right ] \]

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))^3,x, algorithm="fricas")

[Out]

[-1/48*(6*sqrt(2)*(tan(f*x + e)^4 + 2*tan(f*x + e)^3 + tan(f*x + e)^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)) + 177*(tan(
f*x + e)^4 + 2*tan(f*x + e)^3 + tan(f*x + e)^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*(189*tan(f*x + e)^3 + 323*tan(f*x + e)^2 + 112*tan(f*x + e) - 16)*sqrt(d*tan(f*x
+ e)))/(a^3*d^3*f*tan(f*x + e)^4 + 2*a^3*d^3*f*tan(f*x + e)^3 + a^3*d^3*f*tan(f*x + e)^2), 1/24*(6*sqrt(2)*(t
an(f*x + e)^4 + 2*tan(f*x + e)^3 + tan(f*x + e)^2)*sqrt(d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x +
e) - 1)/(sqrt(d)*tan(f*x + e))) + 177*(tan(f*x + e)^4 + 2*tan(f*x + e)^3 + tan(f*x + e)^2)*sqrt(d)*arctan(sqrt
(d*tan(f*x + e))/sqrt(d)) + (189*tan(f*x + e)^3 + 323*tan(f*x + e)^2 + 112*tan(f*x + e) - 16)*sqrt(d*tan(f*x +
e)))/(a^3*d^3*f*tan(f*x + e)^4 + 2*a^3*d^3*f*tan(f*x + e)^3 + a^3*d^3*f*tan(f*x + e)^2)]

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giac [B]  time = 3.08, size = 366, normalized size = 1.70 \[ \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{8 \, a^{3} d^{4} f} + \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{8 \, a^{3} d^{4} f} + \frac {59 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{8 \, a^{3} d^{\frac {5}{2}} f} + \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{16 \, a^{3} d^{4} f} - \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{16 \, a^{3} d^{4} f} + \frac {15 \, \sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + 17 \, \sqrt {d \tan \left (f x + e\right )} d}{8 \, {\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} d^{2} f} + \frac {2 \, {\left (9 \, d \tan \left (f x + e\right ) - d\right )}}{3 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{3} f \tan \left (f x + e\right )} \]

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))^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^4*f) + 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^4*f) + 59/8*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3*d^(5
/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(a
bs(d)) + abs(d))/(a^3*d^4*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^4*f) + 1/8*(15*sqrt(d*tan(f*x + e))*d*tan(f*x + e) + 17*sqrt(d*t
an(f*x + e))*d)/((d*tan(f*x + e) + d)^2*a^3*d^2*f) + 2/3*(9*d*tan(f*x + e) - d)/(sqrt(d*tan(f*x + e))*a^3*d^3*
f*tan(f*x + e))

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maple [B]  time = 0.34, size = 482, normalized size = 2.24 \[ \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^{3}}+\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^{3}}-\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^{3}}+\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^{2} \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^{2} \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^{2} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {15 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8 f \,a^{3} d^{2} \left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {17 \sqrt {d \tan \left (f x +e \right )}}{8 f \,a^{3} d \left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {59 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{8 a^{3} d^{\frac {5}{2}} f}-\frac {2}{3 a^{3} d f \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {6}{a^{3} d^{2} 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))^(5/2)/(a+a*tan(f*x+e))^3,x)

[Out]

1/16/f/a^3/d^3*(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*t
an(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*(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^3*(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*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*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*2^(1/2)/(d^2)^(1/4)*arctan(-2^(1/2)/(d^2
)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+15/8/f/a^3/d^2/(d*tan(f*x+e)+d)^2*(d*tan(f*x+e))^(3/2)+17/8/f/a^3/d/(d*tan(f*x
+e)+d)^2*(d*tan(f*x+e))^(1/2)+59/8*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/d^(5/2)/f-2/3/a^3/d/f/(d*tan(f*x+e
))^(3/2)+6/a^3/d^2/f/(d*tan(f*x+e))^(1/2)

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maxima [A]  time = 0.49, size = 210, normalized size = 0.98 \[ \frac {\frac {189 \, d^{3} \tan \left (f x + e\right )^{3} + 323 \, d^{3} \tan \left (f x + e\right )^{2} + 112 \, d^{3} \tan \left (f x + e\right ) - 16 \, d^{3}}{\left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} a^{3} d + 2 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} d^{2} + \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d^{3}} + \frac {6 \, {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )}}{a^{3} d} + \frac {177 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3} d^{\frac {3}{2}}}}{24 \, d f} \]

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))^3,x, algorithm="maxima")

[Out]

1/24*((189*d^3*tan(f*x + e)^3 + 323*d^3*tan(f*x + e)^2 + 112*d^3*tan(f*x + e) - 16*d^3)/((d*tan(f*x + e))^(7/2
)*a^3*d + 2*(d*tan(f*x + e))^(5/2)*a^3*d^2 + (d*tan(f*x + e))^(3/2)*a^3*d^3) + 6*(sqrt(2)*arctan(1/2*sqrt(2)*(
sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*
sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/(a^3*d) + 177*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3*d^(3/2)))/(d*f
)

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mupad [B]  time = 5.02, size = 192, normalized size = 0.89 \[ \frac {\frac {63\,d\,{\mathrm {tan}\left (e+f\,x\right )}^3}{8}+\frac {323\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{24}+\frac {14\,d\,\mathrm {tan}\left (e+f\,x\right )}{3}-\frac {2\,d}{3}}{a^3\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}+2\,a^3\,d\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}+a^3\,d^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {59\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,d^{5/2}\,f}+\frac {\sqrt {2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{8\,a^3\,d^{5/2}\,f} \]

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))^3),x)

[Out]

((14*d*tan(e + f*x))/3 - (2*d)/3 + (323*d*tan(e + f*x)^2)/24 + (63*d*tan(e + f*x)^3)/8)/(a^3*f*(d*tan(e + f*x)
)^(7/2) + 2*a^3*d*f*(d*tan(e + f*x))^(5/2) + a^3*d^2*f*(d*tan(e + f*x))^(3/2)) + (59*atan((d*tan(e + f*x))^(1/
2)/d^(1/2)))/(8*a^3*d^(5/2)*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^(5/2)*f)

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

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))**3,x)

[Out]

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

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