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

Optimal. Leaf size=343 \[ \frac{2 b \sqrt{c+d \tan (e+f x)} \left (6 a d \left (d^2 (A+C)-B c d+2 c^2 C\right )-b \left (c d^2 (3 A+5 C)-6 B c^2 d-3 B d^3+8 c^3 C\right )\right )}{3 d^3 f \left (c^2+d^2\right )}-\frac{2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{(a-i b)^2 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}}-\frac{(a+i b)^2 (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}}+\frac{2 b^2 \tan (e+f x) \left (d^2 (3 A+C)-3 B c d+4 c^2 C\right ) \sqrt{c+d \tan (e+f x)}}{3 d^2 f \left (c^2+d^2\right )} \]

[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)^(3/2)*f)) - ((a + I
*b)^2*(B - I*(A - C))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((c + I*d)^(3/2)*f) - (2*(c^2*C - B*c*d
+ A*d^2)*(a + b*Tan[e + f*x])^2)/(d*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]) + (2*b*(6*a*d*(2*c^2*C - B*c*d +
(A + C)*d^2) - b*(8*c^3*C - 6*B*c^2*d + c*(3*A + 5*C)*d^2 - 3*B*d^3))*Sqrt[c + d*Tan[e + f*x]])/(3*d^3*(c^2 +
d^2)*f) + (2*b^2*(4*c^2*C - 3*B*c*d + (3*A + C)*d^2)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d^2*(c^2 + d^2)
*f)

________________________________________________________________________________________

Rubi [A]  time = 1.35366, antiderivative size = 343, 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, 3637, 3630, 3539, 3537, 63, 208} \[ \frac{2 b \sqrt{c+d \tan (e+f x)} \left (6 a d \left (d^2 (A+C)-B c d+2 c^2 C\right )-b \left (c d^2 (3 A+5 C)-6 B c^2 d-3 B d^3+8 c^3 C\right )\right )}{3 d^3 f \left (c^2+d^2\right )}-\frac{2 \left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{(a-i b)^2 (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}}-\frac{(a+i b)^2 (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}}+\frac{2 b^2 \tan (e+f x) \left (d^2 (3 A+C)-3 B c d+4 c^2 C\right ) \sqrt{c+d \tan (e+f x)}}{3 d^2 f \left (c^2+d^2\right )} \]

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])^(3/2),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)^(3/2)*f)) - ((a + I
*b)^2*(B - I*(A - C))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((c + I*d)^(3/2)*f) - (2*(c^2*C - B*c*d
+ A*d^2)*(a + b*Tan[e + f*x])^2)/(d*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]) + (2*b*(6*a*d*(2*c^2*C - B*c*d +
(A + C)*d^2) - b*(8*c^3*C - 6*B*c^2*d + c*(3*A + 5*C)*d^2 - 3*B*d^3))*Sqrt[c + d*Tan[e + f*x]])/(3*d^3*(c^2 +
d^2)*f) + (2*b^2*(4*c^2*C - 3*B*c*d + (3*A + C)*d^2)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d^2*(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 3637

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*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, 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))^{3/2}} \, dx &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 \int \frac{(a+b \tan (e+f x)) \left (\frac{1}{2} \left (A d (a c+4 b d)+2 \left (2 b c-\frac{a d}{2}\right ) (c C-B d)\right )+\frac{1}{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-3 B c d+(3 A+C) d^2\right ) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{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}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 \left (4 c^2 C-3 B c d+(3 A+C) d^2\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}-\frac{4 \int \frac{\frac{1}{4} \left (2 b^2 c \left (4 c^2 C-3 B c d+(3 A+C) d^2\right )-3 a d (A d (a c+4 b d)+(4 b c-a d) (c C-B d))\right )-\frac{3}{4} d^2 \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} b \left (6 a d \left (2 c^2 C-B c d+(A+C) d^2\right )-b \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )\right ) \tan ^2(e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d^2 \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}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 b \left (6 a d \left (2 c^2 C-B c d+(A+C) d^2\right )-b \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{2 b^2 \left (4 c^2 C-3 B c d+(3 A+C) d^2\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}-\frac{4 \int \frac{-\frac{3}{4} d^2 \left (a^2 (A c-c C+B d)-b^2 (A c-c C+B d)-2 a b (B c-(A-C) d)\right )-\frac{3}{4} d^2 \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)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d^2 \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}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 b \left (6 a d \left (2 c^2 C-B c d+(A+C) d^2\right )-b \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{2 b^2 \left (4 c^2 C-3 B c d+(3 A+C) d^2\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \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)}+\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)}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 b \left (6 a d \left (2 c^2 C-B c d+(A+C) d^2\right )-b \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{2 b^2 \left (4 c^2 C-3 B c d+(3 A+C) d^2\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \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) 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) f}\\ &=-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 b \left (6 a d \left (2 c^2 C-B c d+(A+C) d^2\right )-b \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{2 b^2 \left (4 c^2 C-3 B c d+(3 A+C) d^2\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \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) 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) 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)^{3/2} 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)^{3/2} f}-\frac{2 \left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 b \left (6 a d \left (2 c^2 C-B c d+(A+C) d^2\right )-b \left (8 c^3 C-6 B c^2 d+c (3 A+5 C) d^2-3 B d^3\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{2 b^2 \left (4 c^2 C-3 B c d+(3 A+C) d^2\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}\\ \end{align*}

Mathematica [C]  time = 6.57688, size = 476, normalized size = 1.39 \[ \frac{2 C (a+b \tan (e+f x))^2}{3 d f \sqrt{c+d \tan (e+f x)}}+\frac{2 \left (\frac{(4 a C d+3 b B d-4 b c C) (a+b \tan (e+f x))}{d f \sqrt{c+d \tan (e+f x)}}+\frac{-\frac{2 \left (8 a^2 C d^2+9 a b B d^2-16 a b c C d+3 A b^2 d^2-6 b^2 B c d+8 b^2 c^2 C-3 b^2 C d^2\right )}{d \sqrt{c+d \tan (e+f x)}}+\frac{2 \left (\frac{\left (-\frac{3}{2} c d^3 \left (a^2 B+2 a b (A-C)-b^2 B\right )-\frac{3}{2} d^4 \left (a^2 (-(A-C))+2 a b B+b^2 (A-C)\right )\right ) \left (\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c+i d}\right )}{(-d+i c) \sqrt{c+d \tan (e+f x)}}-\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{c+d \tan (e+f x)}{c-i d}\right )}{(d+i c) \sqrt{c+d \tan (e+f x)}}\right )}{d}+\frac{3}{2} d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \left (\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}\right )\right )}{d}}{2 d f}\right )}{3 d} \]

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])^(3/2),x]

[Out]

(2*C*(a + b*Tan[e + f*x])^2)/(3*d*f*Sqrt[c + d*Tan[e + f*x]]) + (2*(((-4*b*c*C + 3*b*B*d + 4*a*C*d)*(a + b*Tan
[e + f*x]))/(d*f*Sqrt[c + d*Tan[e + f*x]]) + ((-2*(8*b^2*c^2*C - 6*b^2*B*c*d - 16*a*b*c*C*d + 3*A*b^2*d^2 + 9*
a*b*B*d^2 + 8*a^2*C*d^2 - 3*b^2*C*d^2))/(d*Sqrt[c + d*Tan[e + f*x]]) + (2*((3*(a^2*B - b^2*B + 2*a*b*(A - C))*
d^2*(((-I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sqrt[c - I*d] + (I*ArcTanh[Sqrt[c + d*Tan[e + f*x]
]/Sqrt[c + I*d]])/Sqrt[c + I*d]))/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[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)]/((I*c + d)*Sqrt[c
+ d*Tan[e + f*x]])) + Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)]/((I*c - d)*Sqrt[c + d*T
an[e + f*x]])))/d))/d)/(2*d*f)))/(3*d)

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Maple [B]  time = 0.198, size = 36710, normalized size = 107. \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))^(3/2),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))^(3/2),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))^(3/2),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{3}{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))**(3/2),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))**(3/2), 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{3}{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))^(3/2),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)^(3/2), x)