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

Optimal. Leaf size=396 \[ \frac {2 (c+d \tan (e+f x))^{5/2} \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (63 d^2 (A-C)-18 B c d+8 c^2 C\right )\right )}{315 d^3 f}+\frac {2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \sqrt {c+d \tan (e+f x)} \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{f}-\frac {(a-i b)^2 (c-i d)^{3/2} (B+i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(a+i b)^2 (c+i d)^{3/2} (i A-B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}-\frac {2 b \tan (e+f x) (-4 a C d-9 b B d+4 b c C) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f} \]

[Out]

-(a-I*b)^2*(B+I*(A-C))*(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f+(a+I*b)^2*(I*A-B-I*C)*(c+
I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f+2*(2*a*b*(A*c-B*d-C*c)+a^2*(B*c+(A-C)*d)-b^2*(B*c+(
A-C)*d))*(c+d*tan(f*x+e))^(1/2)/f+2/3*(a^2*B-b^2*B+2*a*b*(A-C))*(c+d*tan(f*x+e))^(3/2)/f+2/315*(28*a^2*C*d^2-1
8*a*b*d*(-7*B*d+2*C*c)+b^2*(8*c^2*C-18*B*c*d+63*(A-C)*d^2))*(c+d*tan(f*x+e))^(5/2)/d^3/f-2/63*b*(-9*B*b*d-4*C*
a*d+4*C*b*c)*tan(f*x+e)*(c+d*tan(f*x+e))^(5/2)/d^2/f+2/9*C*(a+b*tan(f*x+e))^2*(c+d*tan(f*x+e))^(5/2)/d/f

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Rubi [A]  time = 1.73, antiderivative size = 396, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {3647, 3637, 3630, 3528, 3539, 3537, 63, 208} \[ \frac {2 (c+d \tan (e+f x))^{5/2} \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (63 d^2 (A-C)-18 B c d+8 c^2 C\right )\right )}{315 d^3 f}+\frac {2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \sqrt {c+d \tan (e+f x)} \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{f}-\frac {(a-i b)^2 (c-i d)^{3/2} (B+i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(a+i b)^2 (c+i d)^{3/2} (i A-B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}-\frac {2 b \tan (e+f x) (-4 a C d-9 b B d+4 b c C) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - I*b)^2*(B + I*(A - C))*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f) + ((a + I*b
)^2*(I*A - B - I*C)*(c + I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*(2*a*b*(A*c - c*C
- B*d) + a^2*(B*c + (A - C)*d) - b^2*(B*c + (A - C)*d))*Sqrt[c + d*Tan[e + f*x]])/f + (2*(a^2*B - b^2*B + 2*a*
b*(A - C))*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (2*(28*a^2*C*d^2 - 18*a*b*d*(2*c*C - 7*B*d) + b^2*(8*c^2*C - 18
*B*c*d + 63*(A - C)*d^2))*(c + d*Tan[e + f*x])^(5/2))/(315*d^3*f) - (2*b*(4*b*c*C - 9*b*B*d - 4*a*C*d)*Tan[e +
f*x]*(c + d*Tan[e + f*x])^(5/2))/(63*d^2*f) + (2*C*(a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x])^(5/2))/(9*d*f)

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 3528

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

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 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 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 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}+\frac {2 \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \left (\frac {1}{2} (-4 b c C+a (9 A-5 C) d)+\frac {9}{2} (A b+a B-b C) d \tan (e+f x)-\frac {1}{2} (4 b c C-9 b B d-4 a C d) \tan ^2(e+f x)\right ) \, dx}{9 d}\\ &=-\frac {2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}-\frac {4 \int (c+d \tan (e+f x))^{3/2} \left (\frac {1}{4} \left (36 a b c C d-7 a^2 (9 A-5 C) d^2-4 b^2 \left (2 c^2 C-\frac {9 B c d}{2}\right )\right )-\frac {63}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)-\frac {1}{4} \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) \tan ^2(e+f x)\right ) \, dx}{63 d^2}\\ &=\frac {2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac {2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}-\frac {4 \int (c+d \tan (e+f x))^{3/2} \left (\frac {63}{4} \left (2 a b B-a^2 (A-C)+b^2 (A-C)\right ) d^2-\frac {63}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)\right ) \, dx}{63 d^2}\\ &=\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac {2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}-\frac {4 \int \sqrt {c+d \tan (e+f x)} \left (-\frac {63}{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 {63}{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)\right ) \, dx}{63 d^2}\\ &=\frac {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 ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac {2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}-\frac {4 \int \frac {\frac {63}{4} 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 {63}{4} 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}{63 d^2}\\ &=\frac {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 ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac {2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}+\frac {1}{2} \left ((a-i b)^2 (A-i B-C) (c-i d)^2\right ) \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx+\frac {1}{2} \left ((a+i b)^2 (A+i B-C) (c+i d)^2\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {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 ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac {2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}+\frac {\left ((a-i b)^2 (i A+B-i C) (c-i d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 f}-\frac {\left (i (a+i b)^2 (A+i B-C) (c+i d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 f}\\ &=\frac {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 ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac {2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}-\frac {\left ((a-i b)^2 (A-i B-C) (c-i d)^2\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 )}{d f}-\frac {\left ((a+i b)^2 (A+i B-C) (c+i d)^2\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 )}{d f}\\ &=-\frac {(a-i b)^2 (i A+B-i C) (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(a+i b)^2 (B-i (A-C)) (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {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 ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \left (28 a^2 C d^2-18 a b d (2 c C-7 B d)+b^2 \left (8 c^2 C-18 B c d+63 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{5/2}}{315 d^3 f}-\frac {2 b (4 b c C-9 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{63 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}}{9 d f}\\ \end {align*}

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Mathematica [A]  time = 6.20, size = 350, normalized size = 0.88 \[ \frac {2 \left ((c+d \tan (e+f x))^{5/2} \left (28 a^2 C d^2+18 a b d (7 B d-2 c C)+b^2 \left (63 d^2 (A-C)-18 B c d+8 c^2 C\right )\right )+\frac {105}{2} d^3 (a-i b)^2 (i A+B-i C) \left (\sqrt {c+d \tan (e+f x)} (4 c+d \tan (e+f x)-3 i d)-3 (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )\right )+\frac {105}{2} d^3 (a+i b)^2 (-i A+B+i C) \left (\sqrt {c+d \tan (e+f x)} (4 c+d \tan (e+f x)+3 i d)-3 (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )\right )+5 b d \tan (e+f x) (4 a C d+9 b B d-4 b c C) (c+d \tan (e+f x))^{5/2}+35 C d^2 (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}\right )}{315 d^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((28*a^2*C*d^2 + 18*a*b*d*(-2*c*C + 7*B*d) + b^2*(8*c^2*C - 18*B*c*d + 63*(A - C)*d^2))*(c + d*Tan[e + f*x]
)^(5/2) + 5*b*d*(-4*b*c*C + 9*b*B*d + 4*a*C*d)*Tan[e + f*x]*(c + d*Tan[e + f*x])^(5/2) + 35*C*d^2*(a + b*Tan[e
+ f*x])^2*(c + d*Tan[e + f*x])^(5/2) + (105*(a - I*b)^2*(I*A + B - I*C)*d^3*(-3*(c - I*d)^(3/2)*ArcTanh[Sqrt[
c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c - (3*I)*d + d*Tan[e + f*x])))/2 + (105*(a +
I*b)^2*((-I)*A + B + I*C)*d^3*(-3*(c + I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + Sqrt[c +
d*Tan[e + f*x]]*(4*c + (3*I)*d + d*Tan[e + f*x])))/2))/(315*d^3*f)

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

[Out]

Timed out

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maple [B]  time = 0.60, size = 8031, normalized size = 20.28 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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