3.468 \(\int \frac{A+B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=359 \[ \frac{2 b \left (30 a^2 A b^3+8 a^4 A b-17 a^3 b^2 B-3 a^5 B-8 a b^4 B+16 A b^5\right ) \sqrt{\tan (c+d x)}}{3 a^4 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}+\frac{2 b \left (7 a^2 A b-3 a^3 B-4 a b^2 B+8 A b^3\right ) \sqrt{\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 (2 A b-a B)}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{(A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}}-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \]

[Out]

((A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a - b)^(5/2)*d) + ((A - I*
B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a + b)^(5/2)*d) - (2*A)/(3*a*d*Ta
n[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2)) + (2*(2*A*b - a*B))/(a^2*d*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x]
)^(3/2)) + (2*b*(7*a^2*A*b + 8*A*b^3 - 3*a^3*B - 4*a*b^2*B)*Sqrt[Tan[c + d*x]])/(3*a^3*(a^2 + b^2)*d*(a + b*Ta
n[c + d*x])^(3/2)) + (2*b*(8*a^4*A*b + 30*a^2*A*b^3 + 16*A*b^5 - 3*a^5*B - 17*a^3*b^2*B - 8*a*b^4*B)*Sqrt[Tan[
c + d*x]])/(3*a^4*(a^2 + b^2)^2*d*Sqrt[a + b*Tan[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.64235, antiderivative size = 359, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3609, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 b \left (30 a^2 A b^3+8 a^4 A b-17 a^3 b^2 B-3 a^5 B-8 a b^4 B+16 A b^5\right ) \sqrt{\tan (c+d x)}}{3 a^4 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}+\frac{2 b \left (7 a^2 A b-3 a^3 B-4 a b^2 B+8 A b^3\right ) \sqrt{\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 (2 A b-a B)}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{(A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}}-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a - b)^(5/2)*d) + ((A - I*
B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a + b)^(5/2)*d) - (2*A)/(3*a*d*Ta
n[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2)) + (2*(2*A*b - a*B))/(a^2*d*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x]
)^(3/2)) + (2*b*(7*a^2*A*b + 8*A*b^3 - 3*a^3*B - 4*a*b^2*B)*Sqrt[Tan[c + d*x]])/(3*a^3*(a^2 + b^2)*d*(a + b*Ta
n[c + d*x])^(3/2)) + (2*b*(8*a^4*A*b + 30*a^2*A*b^3 + 16*A*b^5 - 3*a^5*B - 17*a^3*b^2*B - 8*a*b^4*B)*Sqrt[Tan[
c + d*x]])/(3*a^4*(a^2 + b^2)^2*d*Sqrt[a + b*Tan[c + d*x]])

Rule 3609

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

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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])

Rubi steps

\begin{align*} \int \frac{A+B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}-\frac{2 \int \frac{\frac{3}{2} (2 A b-a B)+\frac{3}{2} a A \tan (c+d x)+3 A b \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx}{3 a}\\ &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{2 (2 A b-a B)}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{4 \int \frac{-\frac{3}{4} \left (a^2 A-8 A b^2+4 a b B\right )-\frac{3}{4} a^2 B \tan (c+d x)+3 b (2 A b-a B) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}} \, dx}{3 a^2}\\ &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{2 (2 A b-a B)}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{8 \int \frac{-\frac{3}{8} \left (3 a^4 A-14 a^2 A b^2-16 A b^4+9 a^3 b B+8 a b^3 B\right )+\frac{9}{8} a^3 (A b-a B) \tan (c+d x)+\frac{3}{4} b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{9 a^3 \left (a^2+b^2\right )}\\ &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{2 (2 A b-a B)}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt{\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{16 \int \frac{-\frac{9}{16} a^4 \left (a^2 A-A b^2+2 a b B\right )+\frac{9}{16} a^4 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{9 a^4 \left (a^2+b^2\right )^2}\\ &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{2 (2 A b-a B)}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt{\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{(A-i B) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}-\frac{(A+i B) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}\\ &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{2 (2 A b-a B)}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt{\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d}-\frac{(A+i B) \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}\\ &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{2 (2 A b-a B)}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt{\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(a-i b)^2 d}-\frac{(A+i B) \operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(a+i b)^2 d}\\ &=\frac{(A+i B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{2 (2 A b-a B)}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt{\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.42227, size = 383, normalized size = 1.07 \[ \frac{\frac{6 b \left (7 a^2 A b-3 a^3 B-4 a b^2 B+8 A b^3\right ) \sqrt{\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{\frac{6 b \left (30 a^2 A b^3+8 a^4 A b-17 a^3 b^2 B-3 a^5 B-8 a b^4 B+16 A b^5\right ) \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}+9 (-1)^{3/4} a^4 \left (\frac{(a-i b)^2 (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}+\frac{(a+i b)^2 (A-i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}\right )}{a^3 \left (a^2+b^2\right )^2}+\frac{6 (6 A b-3 a B)}{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac{6 A}{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}}{9 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*A)/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2)) + (6*(6*A*b - 3*a*B))/(a*Sqrt[Tan[c + d*x]]*(a + b*Tan
[c + d*x])^(3/2)) + (6*b*(7*a^2*A*b + 8*A*b^3 - 3*a^3*B - 4*a*b^2*B)*Sqrt[Tan[c + d*x]])/(a^2*(a^2 + b^2)*(a +
b*Tan[c + d*x])^(3/2)) + (9*(-1)^(3/4)*a^4*(((a - I*b)^2*(A + I*B)*ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Ta
n[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a - I*b] + ((a + I*b)^2*(A - I*B)*ArcTanh[((-1)^(1/4)*Sqrt[a - I
*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a - I*b]) + (6*b*(8*a^4*A*b + 30*a^2*A*b^3 + 16*A*b^5
- 3*a^5*B - 17*a^3*b^2*B - 8*a*b^4*B)*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]])/(a^3*(a^2 + b^2)^2))/(9*a*
d)

________________________________________________________________________________________

Maple [B]  time = 2.284, size = 2979563, normalized size = 8299.6 \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(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(5/2),x)

[Out]

result too large to display

________________________________________________________________________________________

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(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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(d*x+c))/tan(d*x+c)**(5/2)/(a+b*tan(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError