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*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)

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Rubi [A]  time = 1.02621, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, 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 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 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]

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 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 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 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]

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*}

Mathematica [A]  time = 6.27509, size = 368, normalized size = 1.71 \[ \frac{\tan ^{\frac{5}{2}}(e+f x) \sec ^3(e+f x) (\sin (e+f x)+\cos (e+f x))^3 \left (\frac{126 \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right ) (\tan (e+f x)+1) \csc (e+f x) \sec ^3(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 (\cot (e+f x)+1)}+\frac{2 \sin (2 (e+f x)) \left (\sqrt{2} \left (\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right )-\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right )\right )-2 \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right )\right ) (\tan (e+f x)+1) \csc ^2(e+f x) \sec ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) (\cot (e+f x)+1)}\right )}{16 f (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}}+\frac{\tan ^3(e+f x) \sec ^3(e+f x) (\sin (e+f x)+\cos (e+f x))^3 \left (6 \cot (e+f x)-\frac{2}{3} \csc ^2(e+f x)-\frac{17 \sin (e+f x)}{8 (\sin (e+f x)+\cos (e+f x))}+\frac{1}{8 (\sin (e+f x)+\cos (e+f x))^2}+\frac{8}{3}\right )}{f (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}} \]

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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Maple [B]  time = 0.039, size = 482, normalized size = 2.2 \begin{align*}{\frac{\sqrt{2}}{16\,f{a}^{3}{d}^{3}}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }+{\frac{\sqrt{2}}{8\,f{a}^{3}{d}^{3}}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{\sqrt{2}}{8\,f{a}^{3}{d}^{3}}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{\sqrt{2}}{16\,f{a}^{3}{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{\sqrt{2}}{8\,f{a}^{3}{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{\sqrt{2}}{8\,f{a}^{3}{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{2}{3\,{a}^{3}df} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}+6\,{\frac{1}{f{a}^{3}{d}^{2}\sqrt{d\tan \left ( fx+e \right ) }}}+{\frac{15}{8\,f{a}^{3}{d}^{2} \left ( d\tan \left ( fx+e \right ) +d \right ) ^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{17}{8\,{a}^{3}df \left ( d\tan \left ( fx+e \right ) +d \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{59}{8\,f{a}^{3}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{d}}}} \right ){d}^{-{\frac{5}{2}}}} \end{align*}

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/(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)-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)+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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.82072, size = 1314, normalized size = 6.11 \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}} \end{align*}

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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Giac [B]  time = 1.29321, size = 501, normalized size = 2.33 \begin{align*} \frac{1}{48} \, d^{4}{\left (\frac{6 \, \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^{8} f} + \frac{6 \, \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^{8} f} + \frac{354 \, \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right )}{a^{3} d^{\frac{13}{2}} f} + \frac{3 \, \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^{8} f} - \frac{3 \, \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^{8} f} + \frac{6 \,{\left (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\right )}}{{\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} d^{6} f} + \frac{32 \,{\left (9 \, d \tan \left (f x + e\right ) - d\right )}}{\sqrt{d \tan \left (f x + e\right )} a^{3} d^{7} f \tan \left (f x + e\right )}\right )} \end{align*}

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]

1/48*d^4*(6*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^8*f) + 6*sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sq
rt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a^3*d^8*f) + 354*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3
*d^(13/2)*f) + 3*sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqr
t(abs(d)) + abs(d))/(a^3*d^8*f) - 3*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^8*f) + 6*(15*sqrt(d*tan(f*x + e))*d*tan(f*x + e) + 17*sqrt(d*tan
(f*x + e))*d)/((d*tan(f*x + e) + d)^2*a^3*d^6*f) + 32*(9*d*tan(f*x + e) - d)/(sqrt(d*tan(f*x + e))*a^3*d^7*f*t
an(f*x + e)))