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

Optimal. Leaf size=402 \[ -\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (a^4 C d-a^2 b^2 (d (A-3 C)+B c)+2 a b^3 (A c-B d-c C)+b^4 (A d+B c)\right )}{b^2 f \left (a^2+b^2\right )^2 \sqrt {a+b \tan (e+f x)}}-\frac {(c-i d)^{3/2} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2}}-\frac {(c+i d)^{3/2} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2}}+\frac {2 C d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{5/2} f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 7.13, antiderivative size = 402, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3645, 3655, 6725, 63, 217, 206, 93, 208} \[ -\frac {2 \left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{3/2}}{3 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (-a^2 b^2 (d (A-3 C)+B c)+a^4 C d+2 a b^3 (A c-B d-c C)+b^4 (A d+B c)\right )}{b^2 f \left (a^2+b^2\right )^2 \sqrt {a+b \tan (e+f x)}}-\frac {(c-i d)^{3/2} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2}}-\frac {(c+i d)^{3/2} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2}}+\frac {2 C d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{5/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 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 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]

Rubi steps

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

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Mathematica [C]  time = 41.05, size = 1347065, normalized size = 3350.91 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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

[Out]

Timed out

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

[Out]

Timed out

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}} \left (A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )\right )}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*tan(e + f*x))**(3/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)**2)/(a + b*tan(e + f*x))**(5/2), x)

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