3.1.91 \(\int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x)) \, dx\) [91]

Optimal. Leaf size=325 \[ -\frac {(a-i b)^2 (B+i (A-C)) \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(a+i b)^2 (B-i (A-C)) \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (8 c^2 C-14 B c d+35 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{3/2}}{105 d^3 f}-\frac {2 b (4 b c C-7 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{35 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f} \]

[Out]

-(a-I*b)^2*(B+I*(A-C))*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c-I*d)^(1/2)/f-(a+I*b)^2*(B-I*(A-C))*arc
tanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))*(c+I*d)^(1/2)/f+2*(a^2*B-b^2*B+2*a*b*(A-C))*(c+d*tan(f*x+e))^(1/2)/
f+2/105*(20*a^2*C*d^2-14*a*b*d*(-5*B*d+2*C*c)+b^2*(8*c^2*C-14*B*c*d+35*(A-C)*d^2))*(c+d*tan(f*x+e))^(3/2)/d^3/
f-2/35*b*(-7*B*b*d-4*C*a*d+4*C*b*c)*tan(f*x+e)*(c+d*tan(f*x+e))^(3/2)/d^2/f+2/7*C*(a+b*tan(f*x+e))^2*(c+d*tan(
f*x+e))^(3/2)/d/f

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Rubi [A]
time = 0.86, antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {3728, 3718, 3711, 3609, 3620, 3618, 65, 214} \begin {gather*} \frac {2 (c+d \tan (e+f x))^{3/2} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (35 d^2 (A-C)-14 B c d+8 c^2 C\right )\right )}{105 d^3 f}+\frac {2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {(a-i b)^2 \sqrt {c-i d} (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 \sqrt {c+i d} (B-i (A-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-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{35 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - I*b)^2*(B + I*(A - C))*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f) - ((a + I*b)^
2*(B - I*(A - C))*Sqrt[c + I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*(a^2*B - b^2*B + 2*a*b
*(A - C))*Sqrt[c + d*Tan[e + f*x]])/f + (2*(20*a^2*C*d^2 - 14*a*b*d*(2*c*C - 5*B*d) + b^2*(8*c^2*C - 14*B*c*d
+ 35*(A - C)*d^2))*(c + d*Tan[e + f*x])^(3/2))/(105*d^3*f) - (2*b*(4*b*c*C - 7*b*B*d - 4*a*C*d)*Tan[e + f*x]*(
c + d*Tan[e + f*x])^(3/2))/(35*d^2*f) + (2*C*(a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x])^(3/2))/(7*d*f)

Rule 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3609

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 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^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 3620

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 3711

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 3718

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 3728

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 \sqrt {c+d \tan (e+f x)} \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))^{3/2}}{7 d f}+\frac {2 \int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left (\frac {1}{2} (-4 b c C+a (7 A-3 C) d)+\frac {7}{2} (A b+a B-b C) d \tan (e+f x)-\frac {1}{2} (4 b c C-7 b B d-4 a C d) \tan ^2(e+f x)\right ) \, dx}{7 d}\\ &=-\frac {2 b (4 b c C-7 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{35 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {4 \int \sqrt {c+d \tan (e+f x)} \left (\frac {1}{4} \left (28 a b c C d-5 a^2 (7 A-3 C) d^2-4 b^2 \left (2 c^2 C-\frac {7 B c d}{2}\right )\right )-\frac {35}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)-\frac {1}{4} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (8 c^2 C-14 B c d+35 (A-C) d^2\right )\right ) \tan ^2(e+f x)\right ) \, dx}{35 d^2}\\ &=\frac {2 \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (8 c^2 C-14 B c d+35 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{3/2}}{105 d^3 f}-\frac {2 b (4 b c C-7 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{35 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {4 \int \sqrt {c+d \tan (e+f x)} \left (\frac {35}{4} \left (2 a b B-a^2 (A-C)+b^2 (A-C)\right ) d^2-\frac {35}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)\right ) \, dx}{35 d^2}\\ &=\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (8 c^2 C-14 B c d+35 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{3/2}}{105 d^3 f}-\frac {2 b (4 b c C-7 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{35 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {4 \int \frac {-\frac {35}{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 {35}{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}{35 d^2}\\ &=\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (8 c^2 C-14 B c d+35 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{3/2}}{105 d^3 f}-\frac {2 b (4 b c C-7 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{35 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}+\frac {1}{2} \left ((a-i b)^2 (A-i B-C) (c-i d)\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)\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (8 c^2 C-14 B c d+35 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{3/2}}{105 d^3 f}-\frac {2 b (4 b c C-7 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{35 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\left (i (a+i b)^2 (A+i B-C) (c+i d)\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 f}+\frac {\left ((a-i b)^2 (A-i B-C) (i c+d)\right ) \text {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 (a^2 B-b^2 B+2 a b (A-C)\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (8 c^2 C-14 B c d+35 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{3/2}}{105 d^3 f}-\frac {2 b (4 b c C-7 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{35 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\left ((a+i b)^2 (A+i B-C) (c+i d)\right ) \text {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 (i A+B-i C) (i c+d)\right ) \text {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 (B+i (A-C)) \sqrt {c-i d} \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)) \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (8 c^2 C-14 B c d+35 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{3/2}}{105 d^3 f}-\frac {2 b (4 b c C-7 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{35 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}\\ \end {align*}

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Mathematica [A]
time = 3.18, size = 314, normalized size = 0.97 \begin {gather*} \frac {2 \left (\left (20 a^2 C d^2+14 a b d (-2 c C+5 B d)+b^2 \left (8 c^2 C-14 B c d+35 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{3/2}+3 b d (-4 b c C+7 b B d+4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{3/2}+15 C d^2 (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}+\frac {105}{2} (a-i b)^2 (i A+B-i C) d^3 \left (-\sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)}\right )+\frac {105}{2} (a+i b)^2 (-i A+B+i C) d^3 \left (-\sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)}\right )\right )}{105 d^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((20*a^2*C*d^2 + 14*a*b*d*(-2*c*C + 5*B*d) + b^2*(8*c^2*C - 14*B*c*d + 35*(A - C)*d^2))*(c + d*Tan[e + f*x]
)^(3/2) + 3*b*d*(-4*b*c*C + 7*b*B*d + 4*a*C*d)*Tan[e + f*x]*(c + d*Tan[e + f*x])^(3/2) + 15*C*d^2*(a + b*Tan[e
+ f*x])^2*(c + d*Tan[e + f*x])^(3/2) + (105*(a - I*b)^2*(I*A + B - I*C)*d^3*(-(Sqrt[c - I*d]*ArcTanh[Sqrt[c +
d*Tan[e + f*x]]/Sqrt[c - I*d]]) + Sqrt[c + d*Tan[e + f*x]]))/2 + (105*(a + I*b)^2*((-I)*A + B + I*C)*d^3*(-(S
qrt[c + I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]]) + Sqrt[c + d*Tan[e + f*x]]))/2))/(105*d^3*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2207\) vs. \(2(291)=582\).
time = 0.49, size = 2208, normalized size = 6.79

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2208\)
default	\(\text {Expression too large to display}\)	\(2208\)

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))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

2/f/d^3*(1/7*b^2*C*(c+d*tan(f*x+e))^(7/2)+1/5*B*b^2*d*(c+d*tan(f*x+e))^(5/2)+2/5*C*a*b*d*(c+d*tan(f*x+e))^(5/2
)-2/5*C*b^2*c*(c+d*tan(f*x+e))^(5/2)+1/3*A*b^2*d^2*(c+d*tan(f*x+e))^(3/2)+2/3*B*a*b*d^2*(c+d*tan(f*x+e))^(3/2)
-1/3*B*b^2*c*d*(c+d*tan(f*x+e))^(3/2)+1/3*C*a^2*d^2*(c+d*tan(f*x+e))^(3/2)-2/3*C*a*b*c*d*(c+d*tan(f*x+e))^(3/2
)+1/3*C*b^2*c^2*(c+d*tan(f*x+e))^(3/2)-1/3*C*b^2*d^2*(c+d*tan(f*x+e))^(3/2)+2*A*a*b*d^3*(c+d*tan(f*x+e))^(1/2)
+B*a^2*d^3*(c+d*tan(f*x+e))^(1/2)-B*b^2*d^3*(c+d*tan(f*x+e))^(1/2)-2*C*a*b*d^3*(c+d*tan(f*x+e))^(1/2)+d^3*(1/4
/d*(1/2*(-A*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2+A*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*
b^2+A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*c-2*A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b*d-A*(2*(c^2+d^2)^(1/2)+2*c)^(1
/2)*b^2*c+2*B*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b-B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*d-2*B*(2*(
c^2+d^2)^(1/2)+2*c)^(1/2)*a*b*c+B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2*d+C*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c
)^(1/2)*a^2-C*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2-C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*c+2*C*(2*(
c^2+d^2)^(1/2)+2*c)^(1/2)*a*b*d+C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2*c)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2
)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))+2*(-4*A*(c^2+d^2)^(1/2)*a*b*d-2*B*(c^2+d^2)^(1/2)*a^2*d+2*B*(
c^2+d^2)^(1/2)*b^2*d+4*C*(c^2+d^2)^(1/2)*a*b*d-1/2*(-A*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2+A*(c^
2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2+A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*c-2*A*(2*(c^2+d^2)^(1/2)+2*
c)^(1/2)*a*b*d-A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2*c+2*B*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b-B*(
2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*d-2*B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b*c+B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2
*d+C*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2-C*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2-C*(
2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*c+2*C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b*d+C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2
*c)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)
^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)))+1/4/d*(1/2*(A*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)
*a^2-A*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2-A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*c+2*A*(2*(c^2+d^2
)^(1/2)+2*c)^(1/2)*a*b*d+A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2*c-2*B*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/
2)*a*b+B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*d+2*B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b*c-B*(2*(c^2+d^2)^(1/2)+2*c)
^(1/2)*b^2*d-C*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2+C*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/
2)*b^2+C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*c-2*C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b*d-C*(2*(c^2+d^2)^(1/2)+2*c)
^(1/2)*b^2*c)*ln(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))+2*(-4*A*
(c^2+d^2)^(1/2)*a*b*d-2*B*(c^2+d^2)^(1/2)*a^2*d+2*B*(c^2+d^2)^(1/2)*b^2*d+4*C*(c^2+d^2)^(1/2)*a*b*d+1/2*(A*(c^
2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2-A*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2-A*(2*(c^2+d
^2)^(1/2)+2*c)^(1/2)*a^2*c+2*A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b*d+A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2*c-2*B*(
c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b+B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*d+2*B*(2*(c^2+d^2)^(1/2)+
2*c)^(1/2)*a*b*c-B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2*d-C*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2+C*(
c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2+C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*c-2*C*(2*(c^2+d^2)^(1/2)+
2*c)^(1/2)*a*b*d-C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2*c)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)
^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (e + f x \right )}\right )^{2} \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 )\, dx \end {gather*}

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))**2*(A+B*tan(f*x+e)+C*tan(f*x+e)**2),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

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

[Out]

\text{Hanged}

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