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

Optimal. Leaf size=381 \[ -\frac {\left (a^2 C d^2-2 a b d (2 B d+c C)+b^2 \left (-8 d^2 (A-C)-4 B c d+c^2 C\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 b^{3/2} d^{3/2} f}-\frac {\sqrt {a-i b} \sqrt {c-i d} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}-\frac {\sqrt {a+i b} \sqrt {c+i d} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}-\frac {(-a C d-4 b B d+b c C) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 b d f}+\frac {C \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f} \]

[Out]

-1/4*(a^2*C*d^2-2*a*b*d*(2*B*d+C*c)+b^2*(c^2*C-4*B*c*d-8*(A-C)*d^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^(3/2)/d^(3/2)/f-(I*A+B-I*C)*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)^(1/2)*(c-I*d)^(1/2)/f-(B-I*(A-C))*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)^(1/2)*(c+I*d)^(1/2)/f-1/4*(-4*B*b*d-C*a*d+C*b*c)*(a+b*
tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/b/d/f+1/2*C*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2)/d/f

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Rubi [A]  time = 4.97, antiderivative size = 383, normalized size of antiderivative = 1.01, 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 = {3647, 3655, 6725, 63, 217, 206, 93, 208} \[ -\frac {\left (a^2 C d^2-2 a b d (2 B d+c C)+b^2 \left (-8 d^2 (A-C)-4 B c d+c^2 C\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 b^{3/2} d^{3/2} f}-\frac {\sqrt {a-i b} \sqrt {c-i d} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {\sqrt {a+i b} \sqrt {c+i d} (i A-B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}-\frac {(-a C d-4 b B d+b c C) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 b d f}+\frac {C \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 7.79, size = 619, normalized size = 1.62 \[ \frac {\frac {-\frac {\sqrt {b} \sqrt {c-\frac {a d}{b}} \left (a^2 C d^2-2 a b d (2 B d+c C)+b^2 \left (-8 d^2 (A-C)-4 B c d+c^2 C\right )\right ) \sqrt {\frac {b c+b d \tan (e+f x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right )}{2 \sqrt {d} \sqrt {c+d \tan (e+f x)}}+\frac {2 b d \left (b (a A d+a B c-a C d+A b c-b B d-b c C)-\sqrt {-b^2} (-a (A c-B d-c C)+b d (A-C)+b B c)\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}-c} \sqrt {a+b \tan (e+f x)}}{\sqrt {\sqrt {-b^2}-a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {\sqrt {-b^2}-a} \sqrt {\frac {\sqrt {-b^2} d}{b}-c}}-\frac {2 b d \left (\sqrt {-b^2} (-a (A c-B d-c C)+b d (A-C)+b B c)+b (a A d+a B c-a C d+A b c-b B d-b c C)\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}+c} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {\frac {\sqrt {-b^2} d}{b}+c}}}{b^2 f}+\frac {(a C d+4 b B d-b c C) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{2 b f}}{2 d}+\frac {C \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(2*d*f) + (((-(b*c*C) + 4*b*B*d + a*C*d)*Sqrt[a + b*Ta
n[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(2*b*f) + ((2*b*d*(b*(A*b*c + a*B*c - b*c*C + a*A*d - b*B*d - a*C*d) - S
qrt[-b^2]*(b*B*c + b*(A - C)*d - a*(A*c - c*C - B*d)))*ArcTanh[(Sqrt[-c + (Sqrt[-b^2]*d)/b]*Sqrt[a + b*Tan[e +
f*x]])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[-c + (Sqrt[-b^2]*d)/b])
- (2*b*d*(b*(A*b*c + a*B*c - b*c*C + a*A*d - b*B*d - a*C*d) + Sqrt[-b^2]*(b*B*c + b*(A - C)*d - a*(A*c - c*C
- B*d)))*ArcTanh[(Sqrt[c + (Sqrt[-b^2]*d)/b]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e
+ f*x]])])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[c + (Sqrt[-b^2]*d)/b]) - (Sqrt[b]*Sqrt[c - (a*d)/b]*(a^2*C*d^2 - 2*a*b*d
*(c*C + 2*B*d) + b^2*(c^2*C - 4*B*c*d - 8*(A - C)*d^2))*ArcSinh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sq
rt[c - (a*d)/b])]*Sqrt[(b*c + b*d*Tan[e + f*x])/(b*c - a*d)])/(2*Sqrt[d]*Sqrt[c + d*Tan[e + f*x]]))/(b^2*f))/(
2*d)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)*(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)*sqrt(b*tan(f*x + e) + a)*sqrt(d*tan(f*x + e) + c), x)

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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))^(1/2)*(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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \tan {\left (e + f x \right )}} \sqrt {c + d \tan {\left (e + f x \right )}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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