3.1258 \(\int \frac{1}{(a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=314 \[ -\frac{d \left (2 a^2 d^2+b^2 \left (c^2+3 d^2\right )\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{b^2}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{b^{5/2} \left (-7 a^2 d+4 a b c-3 b^2 d\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right )^2 (b c-a d)^{5/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2 (c-i d)^{3/2}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2 (c+i d)^{3/2}} \]

[Out]

((-I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)^2*(c - I*d)^(3/2)*f) + (I*ArcTanh[Sqrt[c + d
*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)^2*(c + I*d)^(3/2)*f) - (b^(5/2)*(4*a*b*c - 7*a^2*d - 3*b^2*d)*ArcTan
h[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)^2*(b*c - a*d)^(5/2)*f) - (d*(2*a^2*d^2 + b
^2*(c^2 + 3*d^2)))/((a^2 + b^2)*(b*c - a*d)^2*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]) - b^2/((a^2 + b^2)*(b*c
- a*d)*f*(a + b*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.47136, antiderivative size = 314, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {3569, 3649, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{d \left (2 a^2 d^2+b^2 \left (c^2+3 d^2\right )\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{b^2}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{b^{5/2} \left (-7 a^2 d+4 a b c-3 b^2 d\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right )^2 (b c-a d)^{5/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2 (c-i d)^{3/2}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2 (c+i d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x])^(3/2)),x]

[Out]

((-I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)^2*(c - I*d)^(3/2)*f) + (I*ArcTanh[Sqrt[c + d
*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)^2*(c + I*d)^(3/2)*f) - (b^(5/2)*(4*a*b*c - 7*a^2*d - 3*b^2*d)*ArcTan
h[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)^2*(b*c - a*d)^(5/2)*f) - (d*(2*a^2*d^2 + b
^2*(c^2 + 3*d^2)))/((a^2 + b^2)*(b*c - a*d)^2*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]) - b^2/((a^2 + b^2)*(b*c
- a*d)*f*(a + b*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(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 3653

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

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 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 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 3634

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

Rubi steps

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

Mathematica [A]  time = 6.22077, size = 628, normalized size = 2. \[ -\frac{b^2}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{-\frac{2 \left (\frac{1}{2} d^2 \left (2 a^2 d-2 a b c+3 b^2 d\right )-c \left (b d (b c-a d)-\frac{3}{2} b^2 c d\right )\right )}{f \left (c^2+d^2\right ) (a d-b c) \sqrt{c+d \tan (e+f x)}}-\frac{2 \left (\frac{2 \sqrt{b c-a d} \left (\frac{1}{4} a^2 b d \left (2 a^2 d^2+b^2 \left (c^2+3 d^2\right )\right )+\frac{1}{4} b^2 \left (4 a^2 b d \left (c^2+d^2\right )-2 a^3 c d^2-2 a b^2 c \left (c^2+2 d^2\right )+3 b^3 d \left (c^2+d^2\right )\right )-\frac{1}{2} a b (b c-a d)^2 (a d+b c)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{\sqrt{b} f \left (a^2+b^2\right ) (a d-b c)}+\frac{\frac{i \sqrt{c-i d} \left (-\frac{1}{2} \left (a^2 c-2 a b d-b^2 c\right ) (b c-a d)^2-\frac{1}{2} i \left (a^2 d+2 a b c-b^2 d\right ) (b c-a d)^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (-c+i d)}-\frac{i \sqrt{c+i d} \left (-\frac{1}{2} (b c-a d)^2 \left (a^2 c-2 a b d-b^2 c\right )+\frac{1}{2} i (b c-a d)^2 \left (a^2 d+2 a b c-b^2 d\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-c-i d)}}{a^2+b^2}\right )}{\left (c^2+d^2\right ) (a d-b c)}}{\left (a^2+b^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x])^(3/2)),x]

[Out]

-(b^2/((a^2 + b^2)*(b*c - a*d)*f*(a + b*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])) - ((-2*(((I*Sqrt[c - I*d]*(-(
(b*c - a*d)^2*(a^2*c - b^2*c - 2*a*b*d))/2 - (I/2)*(b*c - a*d)^2*(2*a*b*c + a^2*d - b^2*d))*ArcTanh[Sqrt[c + d
*Tan[e + f*x]]/Sqrt[c - I*d]])/((-c + I*d)*f) - (I*Sqrt[c + I*d]*(-((b*c - a*d)^2*(a^2*c - b^2*c - 2*a*b*d))/2
+ (I/2)*(b*c - a*d)^2*(2*a*b*c + a^2*d - b^2*d))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((-c - I*d)
*f))/(a^2 + b^2) + (2*Sqrt[b*c - a*d]*(-(a*b*(b*c - a*d)^2*(b*c + a*d))/2 + (b^2*(-2*a^3*c*d^2 + 4*a^2*b*d*(c^
2 + d^2) + 3*b^3*d*(c^2 + d^2) - 2*a*b^2*c*(c^2 + 2*d^2)))/4 + (a^2*b*d*(2*a^2*d^2 + b^2*(c^2 + 3*d^2)))/4)*Ar
cTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(Sqrt[b]*(a^2 + b^2)*(-(b*c) + a*d)*f)))/((-(b*c) +
a*d)*(c^2 + d^2)) - (2*((d^2*(-2*a*b*c + 2*a^2*d + 3*b^2*d))/2 - c*((-3*b^2*c*d)/2 + b*d*(b*c - a*d))))/((-(b
*c) + a*d)*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]))/((a^2 + b^2)*(b*c - a*d))

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Maple [B]  time = 0.103, size = 12889, normalized size = 41.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(f*x+e))^2/(c+d*tan(f*x+e))^(3/2),x)

[Out]

result too large to display

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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(1/(a+b*tan(f*x+e))^2/(c+d*tan(f*x+e))^(3/2),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(1/(a+b*tan(f*x+e))^2/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \tan{\left (e + f x \right )}\right )^{2} \left (c + d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e))**2/(c+d*tan(f*x+e))**(3/2),x)

[Out]

Integral(1/((a + b*tan(e + f*x))**2*(c + d*tan(e + f*x))**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e))^2/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*tan(f*x + e) + a)^2*(d*tan(f*x + e) + c)^(3/2)), x)