3.1282 \(\int \frac{(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=398 \[ \frac{2 \left (-3 a^2 b^2 \left (15 c^2-13 d^2\right )+20 a^3 b c d+2 a^4 d^2-100 a b^3 c d+b^4 \left (15 c^2-23 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 b f \left (a^2+b^2\right )^3 \sqrt{a+b \tan (e+f x)}}-\frac{2 (b c-a d) \left (a^2 d+10 a b c+11 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 b f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))^{3/2}}-\frac{2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{5 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{5/2}}-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{7/2}}+\frac{i (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{7/2}} \]

[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)^(7/2)*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)^(7/2)*f) - (2*(b*c - a*d)^2*Sqrt[c + d*Tan[e + f*x]])/(5*b*(a^2 + b^2)*f
*(a + b*Tan[e + f*x])^(5/2)) - (2*(b*c - a*d)*(10*a*b*c + a^2*d + 11*b^2*d)*Sqrt[c + d*Tan[e + f*x]])/(15*b*(a
^2 + b^2)^2*f*(a + b*Tan[e + f*x])^(3/2)) + (2*(20*a^3*b*c*d - 100*a*b^3*c*d + 2*a^4*d^2 + b^4*(15*c^2 - 23*d^
2) - 3*a^2*b^2*(15*c^2 - 13*d^2))*Sqrt[c + d*Tan[e + f*x]])/(15*b*(a^2 + b^2)^3*f*Sqrt[a + b*Tan[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 2.36332, antiderivative size = 398, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {3565, 3649, 3616, 3615, 93, 208} \[ \frac{2 \left (-3 a^2 b^2 \left (15 c^2-13 d^2\right )+20 a^3 b c d+2 a^4 d^2-100 a b^3 c d+b^4 \left (15 c^2-23 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 b f \left (a^2+b^2\right )^3 \sqrt{a+b \tan (e+f x)}}-\frac{2 (b c-a d) \left (a^2 d+10 a b c+11 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 b f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))^{3/2}}-\frac{2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{5 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{5/2}}-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{7/2}}+\frac{i (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])^(5/2)/(a + b*Tan[e + f*x])^(7/2),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)^(7/2)*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)^(7/2)*f) - (2*(b*c - a*d)^2*Sqrt[c + d*Tan[e + f*x]])/(5*b*(a^2 + b^2)*f
*(a + b*Tan[e + f*x])^(5/2)) - (2*(b*c - a*d)*(10*a*b*c + a^2*d + 11*b^2*d)*Sqrt[c + d*Tan[e + f*x]])/(15*b*(a
^2 + b^2)^2*f*(a + b*Tan[e + f*x])^(3/2)) + (2*(20*a^3*b*c*d - 100*a*b^3*c*d + 2*a^4*d^2 + b^4*(15*c^2 - 23*d^
2) - 3*a^2*b^2*(15*c^2 - 13*d^2))*Sqrt[c + d*Tan[e + f*x]])/(15*b*(a^2 + b^2)^3*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 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 3616

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A + I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 -
I*Tan[e + f*x]), x], x] + Dist[(A - I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 + I*Tan[e +
f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2
+ B^2, 0]

Rule 3615

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[A^2/f, Subst[Int[((a + b*x)^m*(c + d*x)^n)/(A - B*x), x], x, Tan[e
+ f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 +
B^2, 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))^{7/2}} \, dx &=-\frac{2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}+\frac{2 \int \frac{\frac{1}{2} \left (11 b^2 c^2 d+a^2 d^3+a b c \left (5 c^2-7 d^2\right )\right )-\frac{5}{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} d \left (\left (a^2+5 b^2\right ) d^2-4 b c (b c-2 a d)\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}} \, dx}{5 b \left (a^2+b^2\right )}\\ &=-\frac{2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{2 (b c-a d) \left (10 a b c+a^2 d+11 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}-\frac{4 \int \frac{-\frac{1}{4} (b c-a d) \left (15 a^2 b c^3-15 b^3 c^3+70 a b^2 c^2 d-27 a^2 b c d^2+23 b^3 c d^2+2 a^3 d^3-8 a b^2 d^3\right )+\frac{15}{4} b (b c-a d) \left (2 a b c \left (c^2-3 d^2\right )+b^2 d \left (3 c^2-d^2\right )-a^2 \left (3 c^2 d-d^3\right )\right ) \tan (e+f x)+\frac{1}{2} d (b c-a d)^2 \left (10 a b c+a^2 d+11 b^2 d\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}} \, dx}{15 b \left (a^2+b^2\right )^2 (b c-a d)}\\ &=-\frac{2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{2 (b c-a d) \left (10 a b c+a^2 d+11 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac{2 \left (20 a^3 b c d-100 a b^3 c d+2 a^4 d^2+b^4 \left (15 c^2-23 d^2\right )-3 a^2 b^2 \left (15 c^2-13 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^3 f \sqrt{a+b \tan (e+f x)}}+\frac{8 \int \frac{\frac{15}{8} b (b c-a d)^2 (a c+b d) \left (a^2 c^2-3 b^2 c^2+8 a b c d-3 a^2 d^2+b^2 d^2\right )-\frac{15}{8} b (b c-a d)^3 \left (8 a b c d-b^2 \left (c^2-3 d^2\right )+a^2 \left (3 c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{15 b \left (a^2+b^2\right )^3 (b c-a d)^2}\\ &=-\frac{2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{2 (b c-a d) \left (10 a b c+a^2 d+11 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac{2 \left (20 a^3 b c d-100 a b^3 c d+2 a^4 d^2+b^4 \left (15 c^2-23 d^2\right )-3 a^2 b^2 \left (15 c^2-13 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^3 f \sqrt{a+b \tan (e+f x)}}+\frac{(c-i d)^3 \int \frac{1+i \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b)^3}+\frac{(c+i d)^3 \int \frac{1-i \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b)^3}\\ &=-\frac{2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{2 (b c-a d) \left (10 a b c+a^2 d+11 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac{2 \left (20 a^3 b c d-100 a b^3 c d+2 a^4 d^2+b^4 \left (15 c^2-23 d^2\right )-3 a^2 b^2 \left (15 c^2-13 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^3 f \sqrt{a+b \tan (e+f x)}}+\frac{(c-i d)^3 \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a-i b)^3 f}+\frac{(c+i d)^3 \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a+i b)^3 f}\\ &=-\frac{2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{2 (b c-a d) \left (10 a b c+a^2 d+11 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac{2 \left (20 a^3 b c d-100 a b^3 c d+2 a^4 d^2+b^4 \left (15 c^2-23 d^2\right )-3 a^2 b^2 \left (15 c^2-13 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^3 f \sqrt{a+b \tan (e+f x)}}+\frac{(c-i d)^3 \operatorname{Subst}\left (\int \frac{1}{i a+b-(i c+d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{(a-i b)^3 f}+\frac{(c+i d)^3 \operatorname{Subst}\left (\int \frac{1}{-i a+b-(-i c+d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{(a+i b)^3 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)^{7/2} 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)^{7/2} f}-\frac{2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{5 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac{2 (b c-a d) \left (10 a b c+a^2 d+11 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac{2 \left (20 a^3 b c d-100 a b^3 c d+2 a^4 d^2+b^4 \left (15 c^2-23 d^2\right )-3 a^2 b^2 \left (15 c^2-13 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 b \left (a^2+b^2\right )^3 f \sqrt{a+b \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.61735, size = 431, normalized size = 1.08 \[ \frac{\frac{5 (d+i c) \left (\frac{3 (c-i d)^{3/2} \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{-a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{-a+i b}}+\frac{\sqrt{c+d \tan (e+f x)} ((a d+3 b c-4 i b d) \tan (e+f x)+4 a c-3 i a d-i b c)}{(a+b \tan (e+f x))^{3/2}}\right )}{(a-i b)^3}-\frac{5 (-d+i c) \left (\frac{\sqrt{c+d \tan (e+f x)} ((a d+3 b c+4 i b d) \tan (e+f x)+4 a c+3 i a d+i b c)}{(a+i b)^2 (a+b \tan (e+f x))^{3/2}}-\frac{3 (-c-i d)^{3/2} \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2}}\right )}{a+i b}+\frac{3 (c+d \tan (e+f x))^{5/2}}{(-b+i a) (a+b \tan (e+f x))^{5/2}}-\frac{3 (c+d \tan (e+f x))^{5/2}}{(b+i a) (a+b \tan (e+f x))^{5/2}}}{15 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Tan[e + f*x])^(5/2)/(a + b*Tan[e + f*x])^(7/2),x]

[Out]

((3*(c + d*Tan[e + f*x])^(5/2))/((I*a - b)*(a + b*Tan[e + f*x])^(5/2)) - (3*(c + d*Tan[e + f*x])^(5/2))/((I*a
+ b)*(a + b*Tan[e + f*x])^(5/2)) + (5*(I*c + d)*((3*(c - I*d)^(3/2)*ArcTan[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f
*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[-a + I*b] + (Sqrt[c + d*Tan[e + f*x]]*(4*a*c - I*b*c -
(3*I)*a*d + (3*b*c + a*d - (4*I)*b*d)*Tan[e + f*x]))/(a + b*Tan[e + f*x])^(3/2)))/(a - I*b)^3 - (5*(I*c - d)*(
(-3*(-c - I*d)^(3/2)*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)^(5/2) + (Sqrt[c + d*Tan[e + f*x]]*(4*a*c + I*b*c + (3*I)*a*d + (3*b*c + a*d + (4*I)*b*d)*Tan[e +
f*x]))/((a + I*b)^2*(a + b*Tan[e + f*x])^(3/2))))/(a + I*b))/(15*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{7}{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))^(7/2),x)

[Out]

int((c+d*tan(f*x+e))^(5/2)/(a+b*tan(f*x+e))^(7/2),x)

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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((c+d*tan(f*x+e))^(5/2)/(a+b*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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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))^(7/2),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))**(7/2),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))^(7/2),x, algorithm="giac")

[Out]

Exception raised: TypeError