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

Optimal. Leaf size=365 \[ -\frac {2 b^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{f \left (a^2+b^2\right ) (b c-a d)^{5/2}}+\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^2 d^2 (3 A-C)+A d^4-2 B c^3 d+c^4 C\right )-a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )\right )}{f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}+\frac {(A-i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (b+i a) (c-i d)^{5/2}}+\frac {(i A-B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (a+i b) (c+i d)^{5/2}} \]

[Out]

(A-I*B-C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(I*a+b)/(c-I*d)^(5/2)/f+(I*A-B-I*C)*arctanh((c+d*tan(f
*x+e))^(1/2)/(c+I*d)^(1/2))/(a+I*b)/(c+I*d)^(5/2)/f-2*b^(3/2)*(A*b^2-a*(B*b-C*a))*arctanh(b^(1/2)*(c+d*tan(f*x
+e))^(1/2)/(-a*d+b*c)^(1/2))/(a^2+b^2)/(-a*d+b*c)^(5/2)/f+2*(b*(c^4*C-2*B*c^3*d+c^2*(3*A-C)*d^2+A*d^4)-a*d^2*(
2*c*(A-C)*d-B*(c^2-d^2)))/(-a*d+b*c)^2/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)+2/3*(A*d^2-B*c*d+C*c^2)/(-a*d+b*c)
/(c^2+d^2)/f/(c+d*tan(f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 2.47, antiderivative size = 365, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {3649, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac {2 b^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{f \left (a^2+b^2\right ) (b c-a d)^{5/2}}+\frac {2 \left (b \left (c^2 d^2 (3 A-C)+A d^4-2 B c^3 d+c^4 C\right )-a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )\right )}{f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}+\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}+\frac {(A-i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (b+i a) (c-i d)^{5/2}}+\frac {(i A-B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (a+i b) (c+i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 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 {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}} \, dx &=\frac {2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \int \frac {-\frac {3}{2} \left (a A c d-a d (c C-B d)-A b \left (c^2+d^2\right )\right )+\frac {3}{2} (b c-a d) (B c-(A-C) d) \tan (e+f x)+\frac {3}{2} b \left (c^2 C-B c d+A d^2\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx}{3 (b c-a d) \left (c^2+d^2\right )}\\ &=\frac {2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {4 \int \frac {-\frac {3}{4} \left (A \left (2 a b c^3 d-a^2 d^2 \left (c^2-d^2\right )-b^2 \left (c^2+d^2\right )^2\right )+a d \left (a d \left (c^2 C-2 B c d-C d^2\right )-b \left (2 c^3 C-3 B c^2 d-B d^3\right )\right )\right )-\frac {3}{4} (b c-a d)^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right ) \tan (e+f x)+\frac {3}{4} b \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{3 (b c-a d)^2 \left (c^2+d^2\right )^2}\\ &=\frac {2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\left (b^2 \left (A b^2-a (b B-a C)\right )\right ) \int \frac {1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{\left (a^2+b^2\right ) (b c-a d)^2}+\frac {4 \int \frac {-\frac {3}{4} (b c-a d)^2 \left (a \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )-\frac {3}{4} (b c-a d)^2 \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )-a B \left (c^2-d^2\right )-b \left (c^2 C-2 B c d-C d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right )^2}\\ &=\frac {2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {(A-i B-C) \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b) (c-i d)^2}+\frac {(A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b) (c+i d)^2}+\frac {\left (b^2 \left (A b^2-a (b B-a C)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{\left (a^2+b^2\right ) (b c-a d)^2 f}\\ &=\frac {2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {(i A+B-i C) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (a-i b) (c-i d)^2 f}-\frac {(i (A+i B-C)) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (a+i b) (c+i d)^2 f}+\frac {\left (2 b^2 \left (A b^2-a (b B-a C)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{\left (a^2+b^2\right ) d (b c-a d)^2 f}\\ &=-\frac {2 b^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right ) (b c-a d)^{5/2} f}+\frac {2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}-\frac {(A+i B-C) \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 )}{(a+i b) (c+i d)^2 d f}+\frac {(A-i B-C) \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 )}{(a-i b) d (i c+d)^2 f}\\ &=\frac {(A-i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(i a+b) (c-i d)^{5/2} f}-\frac {(A+i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b) (c+i d)^{5/2} f}-\frac {2 b^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right ) (b c-a d)^{5/2} f}+\frac {2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 6.28, size = 1948, normalized size = 5.34 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A*d^2 - c*(-(c*C) + B*d)))/(3*(-(b*c) + a*d)*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) - (2*((-2*(((I*Sqr
t[c - I*d]*((b*(-(b*c) + a*d)*((-3*c*(b*c - a*d)*(B*c - (A - C)*d))/2 - (3*b*d*(c^2*C - B*c*d + A*d^2))/2 - (3
*d*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2))/2 + a*((-3*((b*d^2)/2 - (c*(-(b*c) + a*d))/2)*(a*A*c*d -
a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2 - (a*d*((3*d*(b*c - a*d)*(B*c - (A - C)*d))/2 - (3*b*c*(c^2*C - B*c*d +
A*d^2))/2))/2 - (b*((-3*d^2*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2 - c*((3*d*(b*c - a*d)*(B*c - (A
- C)*d))/2 - (3*b*c*(c^2*C - B*c*d + A*d^2))/2)))/2) - I*((a*(-(b*c) + a*d)*((-3*c*(b*c - a*d)*(B*c - (A - C)*
d))/2 - (3*b*d*(c^2*C - B*c*d + A*d^2))/2 - (3*d*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2))/2 - b*((-3
*((b*d^2)/2 - (c*(-(b*c) + a*d))/2)*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2 - (a*d*((3*d*(b*c - a*d)*
(B*c - (A - C)*d))/2 - (3*b*c*(c^2*C - B*c*d + A*d^2))/2))/2 - (b*((-3*d^2*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c
^2 + d^2)))/2 - c*((3*d*(b*c - a*d)*(B*c - (A - C)*d))/2 - (3*b*c*(c^2*C - B*c*d + A*d^2))/2)))/2)))*ArcTanh[S
qrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((-c + I*d)*f) - (I*Sqrt[c + I*d]*((b*(-(b*c) + a*d)*((-3*c*(b*c - a*d
)*(B*c - (A - C)*d))/2 - (3*b*d*(c^2*C - B*c*d + A*d^2))/2 - (3*d*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d^2)
))/2))/2 + a*((-3*((b*d^2)/2 - (c*(-(b*c) + a*d))/2)*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2 - (a*d*(
(3*d*(b*c - a*d)*(B*c - (A - C)*d))/2 - (3*b*c*(c^2*C - B*c*d + A*d^2))/2))/2 - (b*((-3*d^2*(a*A*c*d - a*d*(c*
C - B*d) - A*b*(c^2 + d^2)))/2 - c*((3*d*(b*c - a*d)*(B*c - (A - C)*d))/2 - (3*b*c*(c^2*C - B*c*d + A*d^2))/2)
))/2) + I*((a*(-(b*c) + a*d)*((-3*c*(b*c - a*d)*(B*c - (A - C)*d))/2 - (3*b*d*(c^2*C - B*c*d + A*d^2))/2 - (3*
d*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2))/2 - b*((-3*((b*d^2)/2 - (c*(-(b*c) + a*d))/2)*(a*A*c*d -
a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2 - (a*d*((3*d*(b*c - a*d)*(B*c - (A - C)*d))/2 - (3*b*c*(c^2*C - B*c*d +
A*d^2))/2))/2 - (b*((-3*d^2*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2 - c*((3*d*(b*c - a*d)*(B*c - (A -
C)*d))/2 - (3*b*c*(c^2*C - B*c*d + A*d^2))/2)))/2)))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((-c -
I*d)*f))/(a^2 + b^2) + (2*Sqrt[b*c - a*d]*(-1/2*(a*b*(-(b*c) + a*d)*((-3*c*(b*c - a*d)*(B*c - (A - C)*d))/2 -
(3*b*d*(c^2*C - B*c*d + A*d^2))/2 - (3*d*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2)) + (a^2*b*((-3*d^2*
(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2 - c*((3*d*(b*c - a*d)*(B*c - (A - C)*d))/2 - (3*b*c*(c^2*C -
B*c*d + A*d^2))/2)))/2 + b^2*((-3*((b*d^2)/2 - (c*(-(b*c) + a*d))/2)*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d
^2)))/2 - (a*d*((3*d*(b*c - a*d)*(B*c - (A - C)*d))/2 - (3*b*c*(c^2*C - B*c*d + A*d^2))/2))/2))*ArcTanh[(Sqrt[
b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(Sqrt[b]*(a^2 + b^2)*(-(b*c) + a*d)*f)))/((-(b*c) + a*d)*(c^2 +
d^2)) - (2*((-3*d^2*(a*A*c*d - a*d*(c*C - B*d) - A*b*(c^2 + d^2)))/2 - c*((3*d*(b*c - a*d)*(B*c - (A - C)*d))
/2 - (3*b*c*(c^2*C - B*c*d + A*d^2))/2)))/((-(b*c) + a*d)*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]])))/(3*(-(b*c)
+ a*d)*(c^2 + d^2))

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.67, size = 45119, normalized size = 123.61 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

________________________________________________________________________________________

mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________