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

Optimal. Leaf size=543 \[ -\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{2 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}-\frac {\sqrt {c+d \tan (e+f x)} \left (a^4 C d+3 a^3 b B d-a^2 b^2 (7 A d+4 B c-9 C d)+a b^3 (8 A c-5 B d-8 c C)+b^4 (A d+4 B c)\right )}{4 b f \left (a^2+b^2\right )^2 (b c-a d) (a+b \tan (e+f x))}+\frac {\left (a^6 C d^2+3 a^5 b B d^2-3 a^4 b^2 d (5 A d+4 B c-6 C d)+2 a^3 b^3 \left (20 c d (A-C)+B \left (4 c^2-13 d^2\right )\right )-3 a^2 b^4 \left (8 A c^2-6 A d^2-16 B c d-8 c^2 C+5 C d^2\right )-3 a b^5 \left (8 c d (A-C)+B \left (8 c^2-d^2\right )\right )-b^6 \left (4 c (B d+2 c C)-A \left (8 c^2+d^2\right )\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{4 b^{3/2} f \left (a^2+b^2\right )^3 (b c-a d)^{3/2}}-\frac {\sqrt {c-i d} (A-i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (b+i a)^3}+\frac {\sqrt {c+i d} (A+i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (-b+i a)^3} \]

[Out]

1/4*(3*a^5*b*B*d^2+a^6*C*d^2-3*a^4*b^2*d*(5*A*d+4*B*c-6*C*d)-3*a^2*b^4*(8*A*c^2-6*A*d^2-16*B*c*d-8*C*c^2+5*C*d
^2)+2*a^3*b^3*(20*c*(A-C)*d+B*(4*c^2-13*d^2))-3*a*b^5*(8*c*(A-C)*d+B*(8*c^2-d^2))-b^6*(4*c*(B*d+2*C*c)-A*(8*c^
2+d^2)))*arctanh(b^(1/2)*(c+d*tan(f*x+e))^(1/2)/(-a*d+b*c)^(1/2))/b^(3/2)/(a^2+b^2)^3/(-a*d+b*c)^(3/2)/f-(A-I*
B-C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c-I*d)^(1/2)/(I*a+b)^3/f+(A+I*B-C)*arctanh((c+d*tan(f*x+e)
)^(1/2)/(c+I*d)^(1/2))*(c+I*d)^(1/2)/(I*a-b)^3/f-1/2*(A*b^2-a*(B*b-C*a))*(c+d*tan(f*x+e))^(1/2)/b/(a^2+b^2)/f/
(a+b*tan(f*x+e))^2-1/4*(3*a^3*b*B*d+a^4*C*d+b^4*(A*d+4*B*c)+a*b^3*(8*A*c-5*B*d-8*C*c)-a^2*b^2*(7*A*d+4*B*c-9*C
*d))*(c+d*tan(f*x+e))^(1/2)/b/(a^2+b^2)^2/(-a*d+b*c)/f/(a+b*tan(f*x+e))

________________________________________________________________________________________

Rubi [A]  time = 4.04, antiderivative size = 543, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {3645, 3649, 3653, 3539, 3537, 63, 208, 3634} \[ \frac {\left (2 a^3 b^3 \left (20 c d (A-C)+B \left (4 c^2-13 d^2\right )\right )-3 a^2 b^4 \left (8 A c^2-6 A d^2-16 B c d-8 c^2 C+5 C d^2\right )-3 a^4 b^2 d (5 A d+4 B c-6 C d)+3 a^5 b B d^2+a^6 C d^2-3 a b^5 \left (8 c d (A-C)+B \left (8 c^2-d^2\right )\right )-b^6 \left (4 c (B d+2 c C)-A \left (8 c^2+d^2\right )\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{4 b^{3/2} f \left (a^2+b^2\right )^3 (b c-a d)^{3/2}}-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{2 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}-\frac {\sqrt {c+d \tan (e+f x)} \left (-a^2 b^2 (7 A d+4 B c-9 C d)+3 a^3 b B d+a^4 C d+a b^3 (8 A c-5 B d-8 c C)+b^4 (A d+4 B c)\right )}{4 b f \left (a^2+b^2\right )^2 (b c-a d) (a+b \tan (e+f x))}-\frac {\sqrt {c-i d} (A-i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (b+i a)^3}+\frac {\sqrt {c+i d} (A+i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (-b+i a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A - I*B - C)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((I*a + b)^3*f)) + ((A + I*B -
C)*Sqrt[c + I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((I*a - b)^3*f) + ((3*a^5*b*B*d^2 + a^6*C*d
^2 - 3*a^4*b^2*d*(4*B*c + 5*A*d - 6*C*d) - 3*a^2*b^4*(8*A*c^2 - 8*c^2*C - 16*B*c*d - 6*A*d^2 + 5*C*d^2) + 2*a^
3*b^3*(20*c*(A - C)*d + B*(4*c^2 - 13*d^2)) - 3*a*b^5*(8*c*(A - C)*d + B*(8*c^2 - d^2)) - b^6*(4*c*(2*c*C + B*
d) - A*(8*c^2 + d^2)))*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(4*b^(3/2)*(a^2 + b^2)^3*(
b*c - a*d)^(3/2)*f) - ((A*b^2 - a*(b*B - a*C))*Sqrt[c + d*Tan[e + f*x]])/(2*b*(a^2 + b^2)*f*(a + b*Tan[e + f*x
])^2) - ((3*a^3*b*B*d + a^4*C*d + b^4*(4*B*c + A*d) + a*b^3*(8*A*c - 8*c*C - 5*B*d) - a^2*b^2*(4*B*c + 7*A*d -
9*C*d))*Sqrt[c + d*Tan[e + f*x]])/(4*b*(a^2 + b^2)^2*(b*c - a*d)*f*(a + b*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 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 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 {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^3} \, dx &=-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac {\int \frac {\frac {1}{2} \left (2 (b B-a C) \left (2 b c-\frac {a d}{2}\right )+A b (4 a c+b d)\right )-2 b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)-\frac {1}{2} \left (3 A b^2-3 a b B-a^2 C-4 b^2 C\right ) d \tan ^2(e+f x)}{(a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx}{2 b \left (a^2+b^2\right )}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac {\left (3 a^3 b B d+a^4 C d+b^4 (4 B c+A d)+a b^3 (8 A c-8 c C-5 B d)-a^2 b^2 (4 B c+7 A d-9 C d)\right ) \sqrt {c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))}-\frac {\int \frac {\frac {1}{4} \left (-\left (2 a b c-2 a^2 d-b^2 d\right ) \left (a^2 C d+b^2 (4 B c+A d)+a b (4 A c-4 c C-B d)\right )+(2 b c-a d) \left (3 a^2 b B d+a^3 C d+A b^2 (4 b c-7 a d)-4 b^3 (c C+B d)-4 a b^2 (B c-2 C d)\right )\right )+2 b (b c-a d) \left (2 a b (A c-c C-B d)-a^2 (B c+(A-C) d)+b^2 (B c+(A-C) d)\right ) \tan (e+f x)+\frac {1}{4} d \left (3 a^3 b B d+a^4 C d+b^4 (4 B c+A d)+a b^3 (8 A c-8 c C-5 B d)-a^2 b^2 (4 B c+7 A d-9 C d)\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{2 b \left (a^2+b^2\right )^2 (b c-a d)}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac {\left (3 a^3 b B d+a^4 C d+b^4 (4 B c+A d)+a b^3 (8 A c-8 c C-5 B d)-a^2 b^2 (4 B c+7 A d-9 C d)\right ) \sqrt {c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))}-\frac {\int \frac {-2 b (b c-a d) \left (a^3 (A c-c C-B d)-3 a b^2 (A c-c C-B d)+3 a^2 b (B c+(A-C) d)-b^3 (B c+(A-C) d)\right )+2 b (b c-a d) \left (3 a^2 b (A c-c C-B d)-b^3 (A c-c C-B d)-a^3 (B c+(A-C) d)+3 a b^2 (B c+(A-C) d)\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 b \left (a^2+b^2\right )^3 (b c-a d)}-\frac {\left (3 a^5 b B d^2+a^6 C d^2-3 a^4 b^2 d (4 B c+5 A d-6 C d)-3 a^2 b^4 \left (8 A c^2-8 c^2 C-16 B c d-6 A d^2+5 C d^2\right )+2 a^3 b^3 \left (20 c (A-C) d+B \left (4 c^2-13 d^2\right )\right )-3 a b^5 \left (8 c (A-C) d+B \left (8 c^2-d^2\right )\right )-b^6 \left (4 c (2 c C+B d)-A \left (8 c^2+d^2\right )\right )\right ) \int \frac {1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{8 b \left (a^2+b^2\right )^3 (b c-a d)}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac {\left (3 a^3 b B d+a^4 C d+b^4 (4 B c+A d)+a b^3 (8 A c-8 c C-5 B d)-a^2 b^2 (4 B c+7 A d-9 C d)\right ) \sqrt {c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))}+\frac {((A-i B-C) (c-i d)) \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b)^3}+\frac {((A+i B-C) (c+i d)) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b)^3}-\frac {\left (3 a^5 b B d^2+a^6 C d^2-3 a^4 b^2 d (4 B c+5 A d-6 C d)-3 a^2 b^4 \left (8 A c^2-8 c^2 C-16 B c d-6 A d^2+5 C d^2\right )+2 a^3 b^3 \left (20 c (A-C) d+B \left (4 c^2-13 d^2\right )\right )-3 a b^5 \left (8 c (A-C) d+B \left (8 c^2-d^2\right )\right )-b^6 \left (4 c (2 c C+B d)-A \left (8 c^2+d^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{8 b \left (a^2+b^2\right )^3 (b c-a d) f}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac {\left (3 a^3 b B d+a^4 C d+b^4 (4 B c+A d)+a b^3 (8 A c-8 c C-5 B d)-a^2 b^2 (4 B c+7 A d-9 C d)\right ) \sqrt {c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))}-\frac {((A+i B-C) (c+i d)) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (i a-b)^3 f}+\frac {((A-i B-C) (i c+d)) \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)^3 f}-\frac {\left (3 a^5 b B d^2+a^6 C d^2-3 a^4 b^2 d (4 B c+5 A d-6 C d)-3 a^2 b^4 \left (8 A c^2-8 c^2 C-16 B c d-6 A d^2+5 C d^2\right )+2 a^3 b^3 \left (20 c (A-C) d+B \left (4 c^2-13 d^2\right )\right )-3 a b^5 \left (8 c (A-C) d+B \left (8 c^2-d^2\right )\right )-b^6 \left (4 c (2 c C+B d)-A \left (8 c^2+d^2\right )\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 )}{4 b \left (a^2+b^2\right )^3 d (b c-a d) f}\\ &=\frac {\left (3 a^5 b B d^2+a^6 C d^2-3 a^4 b^2 d (4 B c+5 A d-6 C d)-3 a^2 b^4 \left (8 A c^2-8 c^2 C-16 B c d-6 A d^2+5 C d^2\right )+2 a^3 b^3 \left (20 c (A-C) d+B \left (4 c^2-13 d^2\right )\right )-3 a b^5 \left (8 c (A-C) d+B \left (8 c^2-d^2\right )\right )-b^6 \left (4 c (2 c C+B d)-A \left (8 c^2+d^2\right )\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{4 b^{3/2} \left (a^2+b^2\right )^3 (b c-a d)^{3/2} f}-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac {\left (3 a^3 b B d+a^4 C d+b^4 (4 B c+A d)+a b^3 (8 A c-8 c C-5 B d)-a^2 b^2 (4 B c+7 A d-9 C d)\right ) \sqrt {c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))}-\frac {((A+i B-C) (c+i d)) \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)^3 d f}+\frac {((A-i B-C) (i c+d)) \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 )}{(i a+b)^3 d f}\\ &=-\frac {(A-i B-C) \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(i a+b)^3 f}+\frac {(A+i B-C) \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b)^3 f}+\frac {\left (3 a^5 b B d^2+a^6 C d^2-3 a^4 b^2 d (4 B c+5 A d-6 C d)-3 a^2 b^4 \left (8 A c^2-8 c^2 C-16 B c d-6 A d^2+5 C d^2\right )+2 a^3 b^3 \left (20 c (A-C) d+B \left (4 c^2-13 d^2\right )\right )-3 a b^5 \left (8 c (A-C) d+B \left (8 c^2-d^2\right )\right )-b^6 \left (4 c (2 c C+B d)-A \left (8 c^2+d^2\right )\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{4 b^{3/2} \left (a^2+b^2\right )^3 (b c-a d)^{3/2} f}-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac {\left (3 a^3 b B d+a^4 C d+b^4 (4 B c+A d)+a b^3 (8 A c-8 c C-5 B d)-a^2 b^2 (4 B c+7 A d-9 C d)\right ) \sqrt {c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 (b c-a d) f (a+b \tan (e+f x))}\\ \end {align*}

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Mathematica [B]  time = 6.38, size = 2819, normalized size = 5.19 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*C*Sqrt[c + d*Tan[e + f*x]])/(3*b*f*(a + b*Tan[e + f*x])^2) - (2*(-1/2*(((b^2*(-3*A*b*c + 4*b*c*C - a*C*d))
/2 - a*((-3*b^2*(B*c + (A - C)*d))/2 - (a*(b*c*C - 3*b*B*d - a*C*d))/2))*Sqrt[c + d*Tan[e + f*x]])/((a^2 + b^2
)*(b*c - a*d)*f*(a + b*Tan[e + f*x])^2) - (-((((I*Sqrt[c - I*d]*(b*(b*c - a*d)*((3*b*(3*A*b^2 - 3*a*b*B - a^2*
C - 4*b^2*C)*d*(b*c - a*d))/4 + 3*a*b*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d) + (3*b*(b*c
- a*d)*(a^2*C*d + b^2*(4*B*c + A*d) + a*b*(4*A*c - 4*c*C - B*d)))/4) + a*((3*(b*c - a*d)*((b^2*d)/2 - a*(b*c -
a*d))*(a^2*C*d + b^2*(4*B*c + A*d) + a*b*(4*A*c - 4*c*C - B*d)))/4 + (-(b*c) + (a*d)/2)*((3*a*(3*A*b^2 - 3*a*
b*B - a^2*C - 4*b^2*C)*d*(b*c - a*d))/4 - 3*b^2*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d)) -
(d*((3*b^2*(b*c - a*d)*(a^2*C*d + b^2*(4*B*c + A*d) + a*b*(4*A*c - 4*c*C - B*d)))/4 - a*((3*a*(3*A*b^2 - 3*a*
b*B - a^2*C - 4*b^2*C)*d*(b*c - a*d))/4 - 3*b^2*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d))))
/2) - I*(a*(b*c - a*d)*((3*b*(3*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c - a*d))/4 + 3*a*b*(b*c - a*d)*(A*b*c
- a*B*c - b*c*C - a*A*d - b*B*d + a*C*d) + (3*b*(b*c - a*d)*(a^2*C*d + b^2*(4*B*c + A*d) + a*b*(4*A*c - 4*c*C
- B*d)))/4) - b*((3*(b*c - a*d)*((b^2*d)/2 - a*(b*c - a*d))*(a^2*C*d + b^2*(4*B*c + A*d) + a*b*(4*A*c - 4*c*C
- B*d)))/4 + (-(b*c) + (a*d)/2)*((3*a*(3*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c - a*d))/4 - 3*b^2*(b*c - a
*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d)) - (d*((3*b^2*(b*c - a*d)*(a^2*C*d + b^2*(4*B*c + A*d) + a
*b*(4*A*c - 4*c*C - B*d)))/4 - a*((3*a*(3*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c - a*d))/4 - 3*b^2*(b*c - a
*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d))))/2)))*ArcTanh[Sqrt[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*b*(3*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c - a*d))/4 +
3*a*b*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d) + (3*b*(b*c - a*d)*(a^2*C*d + b^2*(4*B*c +
A*d) + a*b*(4*A*c - 4*c*C - B*d)))/4) + a*((3*(b*c - a*d)*((b^2*d)/2 - a*(b*c - a*d))*(a^2*C*d + b^2*(4*B*c +
A*d) + a*b*(4*A*c - 4*c*C - B*d)))/4 + (-(b*c) + (a*d)/2)*((3*a*(3*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c -
a*d))/4 - 3*b^2*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d)) - (d*((3*b^2*(b*c - a*d)*(a^2*C*
d + b^2*(4*B*c + A*d) + a*b*(4*A*c - 4*c*C - B*d)))/4 - a*((3*a*(3*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c -
a*d))/4 - 3*b^2*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d))))/2) + I*(a*(b*c - a*d)*((3*b*(3
*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c - a*d))/4 + 3*a*b*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*
d + a*C*d) + (3*b*(b*c - a*d)*(a^2*C*d + b^2*(4*B*c + A*d) + a*b*(4*A*c - 4*c*C - B*d)))/4) - b*((3*(b*c - a*d
)*((b^2*d)/2 - a*(b*c - a*d))*(a^2*C*d + b^2*(4*B*c + A*d) + a*b*(4*A*c - 4*c*C - B*d)))/4 + (-(b*c) + (a*d)/2
)*((3*a*(3*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c - a*d))/4 - 3*b^2*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*
A*d - b*B*d + a*C*d)) - (d*((3*b^2*(b*c - a*d)*(a^2*C*d + b^2*(4*B*c + A*d) + a*b*(4*A*c - 4*c*C - B*d)))/4 -
a*((3*a*(3*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c - a*d))/4 - 3*b^2*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*
A*d - b*B*d + a*C*d))))/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]*(-(a*b*(b*c - a*d)*((3*b*(3*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c - a*d))/4 + 3*a*b*(b*c
- a*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d) + (3*b*(b*c - a*d)*(a^2*C*d + b^2*(4*B*c + A*d) + a*b*(
4*A*c - 4*c*C - B*d)))/4)) + (a^2*d*((3*b^2*(b*c - a*d)*(a^2*C*d + b^2*(4*B*c + A*d) + a*b*(4*A*c - 4*c*C - B*
d)))/4 - a*((3*a*(3*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c - a*d))/4 - 3*b^2*(b*c - a*d)*(A*b*c - a*B*c - b
*c*C - a*A*d - b*B*d + a*C*d))))/2 + b^2*((3*(b*c - a*d)*((b^2*d)/2 - a*(b*c - a*d))*(a^2*C*d + b^2*(4*B*c + A
*d) + a*b*(4*A*c - 4*c*C - B*d)))/4 + (-(b*c) + (a*d)/2)*((3*a*(3*A*b^2 - 3*a*b*B - a^2*C - 4*b^2*C)*d*(b*c -
a*d))/4 - 3*b^2*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d))))*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))/((a^2 + b^2)*(b*c - a*d))) - (((3*b^2*(b
*c - a*d)*(a^2*C*d + b^2*(4*B*c + A*d) + a*b*(4*A*c - 4*c*C - B*d)))/4 - a*((3*a*(3*A*b^2 - 3*a*b*B - a^2*C -
4*b^2*C)*d*(b*c - a*d))/4 - 3*b^2*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d)))*Sqrt[c + d*Tan
[e + f*x]])/((a^2 + b^2)*(b*c - a*d)*f*(a + b*Tan[e + f*x])))/(2*(a^2 + b^2)*(b*c - a*d))))/(3*b)

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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))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^3,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))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^3,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.88, size = 9797, normalized size = 18.04 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^3,x)

[Out]

result too large to display

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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((c+d*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^3,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?

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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))^(1/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/(a + b*tan(e + f*x))^3,x)

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(a+b*tan(f*x+e))**3,x)

[Out]

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

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