3.161 \(\int \frac{\sqrt{a+b \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=373 \[ -\frac{2 \left (A d^2-B c d+c^2 C\right ) \sqrt{a+b \tan (e+f x)}}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac{2 \sqrt{a+b \tan (e+f x)} \left (3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (-c^2 d^2 (5 A-7 C)+A d^4+2 B c^3 d-4 B c d^3+c^4 C\right )\right )}{3 d f \left (c^2+d^2\right )^2 (b c-a d) \sqrt{c+d \tan (e+f x)}}-\frac{\sqrt{a-i b} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}-\frac{\sqrt{a+i b} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c+i d)^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.92182, antiderivative size = 373, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 49, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3645, 3649, 3616, 3615, 93, 208} \[ -\frac{2 \left (A d^2-B c d+c^2 C\right ) \sqrt{a+b \tan (e+f x)}}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac{2 \sqrt{a+b \tan (e+f x)} \left (3 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )+b \left (-c^2 d^2 (5 A-7 C)+A d^4+2 B c^3 d-4 B c d^3+c^4 C\right )\right )}{3 d f \left (c^2+d^2\right )^2 (b c-a d) \sqrt{c+d \tan (e+f x)}}-\frac{\sqrt{a-i b} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}-\frac{\sqrt{a+i b} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c+i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*Tan[e + f*x]]*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

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

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 3616

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A + I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 -
I*Tan[e + f*x]), x], x] + Dist[(A - I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 + I*Tan[e +
f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2
+ B^2, 0]

Rule 3615

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[A^2/f, Subst[Int[((a + b*x)^m*(c + d*x)^n)/(A - B*x), x], x, Tan[e
+ f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 +
B^2, 0]

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

Mathematica [A]  time = 6.91028, size = 609, normalized size = 1.63 \[ -\frac{C \sqrt{a+b \tan (e+f x)}}{d f (c+d \tan (e+f x))^{3/2}}-\frac{-\frac{2 \sqrt{a+b \tan (e+f x)} \left (\frac{1}{2} d^2 (-a d (2 A-3 C)-b c C)-c \left (d^2 (-(a B+A b-b C))-\frac{1}{2} c (a C d-2 b B d-b c C)\right )\right )}{3 f \left (c^2+d^2\right ) (a d-b c) (c+d \tan (e+f x))^{3/2}}-\frac{2 \left (-\frac{2 \sqrt{a+b \tan (e+f x)} \left (-\frac{1}{2} d^2 (b c-a d) \left (3 a d (A c+B d-c C)+b \left (A d^2-B c d+c^2 C\right )\right )-c \left (\frac{1}{2} b c (b c-a d) \left (-d^2 (2 A-3 C)+2 B c d+c^2 C\right )-\frac{3}{2} d^2 (b c-a d) (-a A d+a B c+a C d+A b c+b B d-b c C)\right )\right )}{f \left (c^2+d^2\right ) (a d-b c) \sqrt{c+d \tan (e+f x)}}-\frac{3 d (b c-a d)^2 \left (\frac{\sqrt{a+i b} (c-i d)^2 (B-i (A-C)) \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{-c-i d}}+\frac{\sqrt{-a+i b} (c+i d)^2 (i A+B-i C) \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{-a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{c-i d}}\right )}{2 f \left (c^2+d^2\right ) (a d-b c)}\right )}{3 \left (c^2+d^2\right ) (a d-b c)}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*Tan[e + f*x]]*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

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

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{(A+B\tan \left ( fx+e \right ) +C \left ( \tan \left ( fx+e \right ) \right ) ^{2})\sqrt{a+b\tan \left ( fx+e \right ) } \left ( c+d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(5/2),x)

[Out]

int((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(5/2),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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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))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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))**(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(c+d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError