3.123 \(\int \frac{(a+b \tan (e+f x))^2 (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=358 \[ \frac{2 (b c-a d) \left (3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (-2 c^2 d^2 (A-5 C)+4 A d^4-B c^3 d-7 B c d^3+4 c^4 C\right )\right )}{3 d^3 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{(a-i b)^2 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}-\frac{(a+i b)^2 (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}}+\frac{2 b^2 \left (d^2 (A+3 C)-B c d+4 c^2 C\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 f \left (c^2+d^2\right )} \]

[Out]

-(((a - I*b)^2*(I*A + B - I*C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(5/2)*f)) - ((a + I
*b)^2*(B - I*(A - C))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((c + I*d)^(5/2)*f) - (2*(c^2*C - B*c*d
 + A*d^2)*(a + b*Tan[e + f*x])^2)/(3*d*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + (2*(b*c - a*d)*(b*(4*c^4*C
- B*c^3*d - 2*c^2*(A - 5*C)*d^2 - 7*B*c*d^3 + 4*A*d^4) + 3*a*d^2*(2*c*(A - C)*d - B*(c^2 - d^2))))/(3*d^3*(c^2
 + d^2)^2*f*Sqrt[c + d*Tan[e + f*x]]) + (2*b^2*(4*c^2*C - B*c*d + (A + 3*C)*d^2)*Sqrt[c + d*Tan[e + f*x]])/(3*
d^3*(c^2 + d^2)*f)

________________________________________________________________________________________

Rubi [A]  time = 1.55107, antiderivative size = 358, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.149, Rules used = {3645, 3635, 3630, 3539, 3537, 63, 208} \[ \frac{2 (b c-a d) \left (3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (-2 c^2 d^2 (A-5 C)+4 A d^4-B c^3 d-7 B c d^3+4 c^4 C\right )\right )}{3 d^3 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{(a-i b)^2 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}-\frac{(a+i b)^2 (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}}+\frac{2 b^2 \left (d^2 (A+3 C)-B c d+4 c^2 C\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 f \left (c^2+d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - I*b)^2*(I*A + B - I*C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(5/2)*f)) - ((a + I
*b)^2*(B - I*(A - C))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((c + I*d)^(5/2)*f) - (2*(c^2*C - B*c*d
 + A*d^2)*(a + b*Tan[e + f*x])^2)/(3*d*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + (2*(b*c - a*d)*(b*(4*c^4*C
- B*c^3*d - 2*c^2*(A - 5*C)*d^2 - 7*B*c*d^3 + 4*A*d^4) + 3*a*d^2*(2*c*(A - C)*d - B*(c^2 - d^2))))/(3*d^3*(c^2
 + d^2)^2*f*Sqrt[c + d*Tan[e + f*x]]) + (2*b^2*(4*c^2*C - B*c*d + (A + 3*C)*d^2)*Sqrt[c + d*Tan[e + f*x]])/(3*
d^3*(c^2 + d^2)*f)

Rule 3645

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*d^2 + c*(c*C - B*d))*(a + b*T
an[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3635

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

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3539

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

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 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 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{(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \int \frac{(a+b \tan (e+f x)) \left (\frac{1}{2} \left (2 A d \left (\frac{3 a c}{2}+2 b d\right )+2 \left (2 b c-\frac{3 a d}{2}\right ) (c C-B d)\right )+\frac{3}{2} d ((A-C) (b c-a d)+B (a c+b d)) \tan (e+f x)+\frac{1}{2} b \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{3 d \left (c^2+d^2\right )}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \left (4 c^4 C-B c^3 d-2 c^2 (A-5 C) d^2-7 B c d^3+4 A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 \int \frac{\frac{1}{2} \left (b^2 \left (4 c^4 C-B c^3 d-2 c^2 (A-5 C) d^2-7 B c d^3+4 A d^4\right )-3 a^2 d^2 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+6 a b d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )-\frac{3}{2} d^2 \left (2 a b \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+a^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )-b^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \tan (e+f x)+\frac{1}{2} b^2 \left (c^2+d^2\right ) \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \tan ^2(e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d^2 \left (c^2+d^2\right )^2}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \left (4 c^4 C-B c^3 d-2 c^2 (A-5 C) d^2-7 B c d^3+4 A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{2 \int \frac{-\frac{3}{2} d^2 \left (a^2 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )-b^2 \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )-2 a b \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )-\frac{3}{2} d^2 \left (2 a b \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+a^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )-b^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d^2 \left (c^2+d^2\right )^2}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \left (4 c^4 C-B c^3 d-2 c^2 (A-5 C) d^2-7 B c d^3+4 A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{\left ((a-i b)^2 (A-i B-C)\right ) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)^2}+\frac{\left ((a+i b)^2 (A+i B-C)\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)^2}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \left (4 c^4 C-B c^3 d-2 c^2 (A-5 C) d^2-7 B c d^3+4 A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{\left ((a-i b)^2 (i A+B-i C)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d)^2 f}-\frac{\left (i (a+i b)^2 (A+i B-C)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d)^2 f}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \left (4 c^4 C-B c^3 d-2 c^2 (A-5 C) d^2-7 B c d^3+4 A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac{\left ((a-i b)^2 (A-i B-C)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(c-i d)^2 d f}-\frac{\left ((a+i b)^2 (A+i B-C)\right ) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(c+i d)^2 d f}\\ &=-\frac{(a-i b)^2 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(c-i d)^{5/2} f}-\frac{(a+i b)^2 (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(c+i d)^{5/2} f}-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 (b c-a d) \left (b \left (4 c^4 C-B c^3 d-2 c^2 (A-5 C) d^2-7 B c d^3+4 A d^4\right )+3 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (4 c^2 C-B c d+(A+3 C) d^2\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}\\ \end{align*}

Mathematica [C]  time = 6.47687, size = 502, normalized size = 1.4 \[ \frac{2 C (a+b \tan (e+f x))^2}{d f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (-\frac{(4 a C d+b B d-4 b c C) (a+b \tan (e+f x))}{d f (c+d \tan (e+f x))^{3/2}}-\frac{-\frac{2 \left (8 a^2 C d^2+a b B d^2-16 a b c C d-A b^2 d^2-2 b^2 B c d+8 b^2 c^2 C+b^2 C d^2\right )}{3 d (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (\frac{\left (\frac{3}{2} c d^3 \left (a^2 B+2 a b (A-C)-b^2 B\right )+\frac{3}{2} d^4 \left (a^2 (-(A-C))+2 a b B+b^2 (A-C)\right )\right ) \left (\frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )}{3 (-d+i c) (c+d \tan (e+f x))^{3/2}}-\frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )}{3 (d+i c) (c+d \tan (e+f x))^{3/2}}\right )}{d}-\frac{3}{2} d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \left (\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )}{(-d+i c) \sqrt{c+d \tan (e+f x)}}-\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )}{(d+i c) \sqrt{c+d \tan (e+f x)}}\right )\right )}{3 d}}{2 d f}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*(a + b*Tan[e + f*x])^2)/(d*f*(c + d*Tan[e + f*x])^(3/2)) + (2*(-(((-4*b*c*C + b*B*d + 4*a*C*d)*(a + b*Tan
[e + f*x]))/(d*f*(c + d*Tan[e + f*x])^(3/2))) - ((-2*(8*b^2*c^2*C - 2*b^2*B*c*d - 16*a*b*c*C*d - A*b^2*d^2 + a
*b*B*d^2 + 8*a^2*C*d^2 + b^2*C*d^2))/(3*d*(c + d*Tan[e + f*x])^(3/2)) + (2*((((3*c*(a^2*B - b^2*B + 2*a*b*(A -
 C))*d^3)/2 + (3*(2*a*b*B - a^2*(A - C) + b^2*(A - C))*d^4)/2)*(-Hypergeometric2F1[-3/2, 1, -1/2, (c + d*Tan[e
 + f*x])/(c - I*d)]/(3*(I*c + d)*(c + d*Tan[e + f*x])^(3/2)) + Hypergeometric2F1[-3/2, 1, -1/2, (c + d*Tan[e +
 f*x])/(c + I*d)]/(3*(I*c - d)*(c + d*Tan[e + f*x])^(3/2))))/d - (3*(a^2*B - b^2*B + 2*a*b*(A - C))*d^2*(-(Hyp
ergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)]/((I*c + d)*Sqrt[c + d*Tan[e + f*x]])) + Hypergeom
etric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)]/((I*c - d)*Sqrt[c + d*Tan[e + f*x]])))/2))/(3*d))/(2*d*
f)))/d

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Maple [B]  time = 0.247, size = 61833, normalized size = 172.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{2} \left (A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\left (c + d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(e + f*x))**2*(A + B*tan(e + f*x) + C*tan(e + f*x)**2)/(c + d*tan(e + f*x))**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )}{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(b*tan(f*x + e) + a)^2/(d*tan(f*x + e) + c)^(5/2), x)