3.1280 \(\int \frac{(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=273 \[ -\frac{2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) \sqrt{a+b \tan (e+f x)}}+\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{b^{3/2} f}-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{3/2}}+\frac{i (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{3/2}} \]

[Out]

((-I)*(c - I*d)^(5/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]]
)])/((a - I*b)^(3/2)*f) + (I*(c + I*d)^(5/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*S
qrt[c + d*Tan[e + f*x]])])/((a + I*b)^(3/2)*f) + (2*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b
]*Sqrt[c + d*Tan[e + f*x]])])/(b^(3/2)*f) - (2*(b*c - a*d)^2*Sqrt[c + d*Tan[e + f*x]])/(b*(a^2 + b^2)*f*Sqrt[a
+ b*Tan[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 3.07245, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {3565, 3655, 6725, 63, 217, 206, 93, 208} \[ -\frac{2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) \sqrt{a+b \tan (e+f x)}}+\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{b^{3/2} f}-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{3/2}}+\frac{i (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(c - I*d)^(5/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]]
)])/((a - I*b)^(3/2)*f) + (I*(c + I*d)^(5/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*S
qrt[c + d*Tan[e + f*x]])])/((a + I*b)^(3/2)*f) + (2*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b
]*Sqrt[c + d*Tan[e + f*x]])])/(b^(3/2)*f) - (2*(b*c - a*d)^2*Sqrt[c + d*Tan[e + f*x]])/(b*(a^2 + b^2)*f*Sqrt[a
+ b*Tan[e + f*x]])

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 3655

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] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[((a + b*ff*x)^m*(c + d*ff*x)^n*(A + B*ff*x + C*ff^2*x^2))/(1 + ff^2*x^2), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

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]

Rubi steps

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

Mathematica [B]  time = 6.14188, size = 1198, normalized size = 4.39 \[ \frac{i (c+i d) \left ((c+i d) \left (\frac{2 \sqrt{c+d \tan (e+f x)}}{(-a-i b) \sqrt{a+b \tan (e+f x)}}-\frac{2 (c+i d) \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{(-a-i b) \sqrt{a+i b} \sqrt{-c-i d}}\right )-\frac{2 d \sqrt{c+d \tan (e+f x)} \left (\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}+1\right )^{3/2} \left (1-\frac{\sqrt{b} \sqrt{d} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b c-a d} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}}\right ) \sqrt{a+b \tan (e+f x)}}{\sqrt{b c-a d} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}} \sqrt{\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}+1}}\right )}{b \sqrt{\frac{b}{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}} \sqrt{a+b \tan (e+f x)} \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}} \left (-\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}-1\right )}\right )}{2 f}-\frac{i (i d-c) \left (\frac{2 d \sqrt{c+d \tan (e+f x)} \left (\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}+1\right )^{3/2} \left (1-\frac{\sqrt{b} \sqrt{d} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b c-a d} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}}\right ) \sqrt{a+b \tan (e+f x)}}{\sqrt{b c-a d} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}} \sqrt{\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}+1}}\right )}{b \sqrt{\frac{b}{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}} \sqrt{a+b \tan (e+f x)} \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}} \left (-\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}-1\right )}-(i d-c) \left (\frac{2 \sqrt{c+d \tan (e+f x)}}{(a-i b) \sqrt{a+b \tan (e+f x)}}-\frac{2 (i d-c) \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{i b-a} \sqrt{c+d \tan (e+f x)}}\right )}{(a-i b) \sqrt{i b-a} \sqrt{c-i d}}\right )\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(c + I*d)*((c + I*d)*((-2*(c + I*d)*ArcTan[(Sqrt[-c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqr
t[c + d*Tan[e + f*x]])])/((-a - I*b)*Sqrt[a + I*b]*Sqrt[-c - I*d]) + (2*Sqrt[c + d*Tan[e + f*x]])/((-a - I*b)*
Sqrt[a + b*Tan[e + f*x]])) - (2*d*Sqrt[c + d*Tan[e + f*x]]*(1 + (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*
c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))))^(3/2)*(1 - (Sqrt[b]*Sqrt[d]*ArcSinh[(Sqrt[b]*Sqrt[d]*Sqrt[a + b*Tan[e
+ f*x]])/(Sqrt[b*c - a*d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)])]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b*
c - a*d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)]*Sqrt[1 + (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^
2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))])))/(b*Sqrt[b/((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))]*Sqrt[a + b
*Tan[e + f*x]]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)]*(-1 - (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*
c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))))))/f - ((I/2)*(-c + I*d)*(-((-c + I*d)*((-2*(-c + I*d)*ArcTan[(Sqrt[c
- I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/((a - I*b)*Sqrt[-a + I*b]*Sqrt[c
- I*d]) + (2*Sqrt[c + d*Tan[e + f*x]])/((a - I*b)*Sqrt[a + b*Tan[e + f*x]]))) + (2*d*Sqrt[c + d*Tan[e + f*x]]*
(1 + (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))))^(3/2)*(1 - (Sqrt[b]
*Sqrt[d]*ArcSinh[(Sqrt[b]*Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b*c - a*d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d
)/(b*c - a*d)])]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b*c - a*d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)]*Sq
rt[1 + (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))])))/(b*Sqrt[b/((b^
2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))]*Sqrt[a + b*Tan[e + f*x]]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)]*(
-1 - (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))))))/f

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{ \left ( c+d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}} \left ( a+b\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError