3.1233 \(\int \frac{\sqrt{c+d \tan (e+f x)}}{(a+b \tan (e+f x))^2} \, dx\)
Optimal. Leaf size=231 \[ -\frac{b \sqrt{c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (a+b \tan (e+f x))}-\frac{\sqrt{b} \left (-3 a^2 d+4 a b c+b^2 d\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right )^2 \sqrt{b c-a d}}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2}+\frac{i \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2} \]
[Out]
((-I)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)^2*f) + (I*Sqrt[c + I*d]*ArcTan
h[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)^2*f) - (Sqrt[b]*(4*a*b*c - 3*a^2*d + b^2*d)*ArcTanh[(Sqr
t[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)^2*Sqrt[b*c - a*d]*f) - (b*Sqrt[c + d*Tan[e + f*x
]])/((a^2 + b^2)*f*(a + b*Tan[e + f*x]))
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Rubi [A] time = 0.77513, antiderivative size = 231, normalized size of antiderivative = 1.,
number of steps used = 12, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used
= {3568, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{b \sqrt{c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (a+b \tan (e+f x))}-\frac{\sqrt{b} \left (-3 a^2 d+4 a b c+b^2 d\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right )^2 \sqrt{b c-a d}}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2}+\frac{i \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[c + d*Tan[e + f*x]]/(a + b*Tan[e + f*x])^2,x]
[Out]
((-I)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)^2*f) + (I*Sqrt[c + I*d]*ArcTan
h[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)^2*f) - (Sqrt[b]*(4*a*b*c - 3*a^2*d + b^2*d)*ArcTanh[(Sqr
t[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)^2*Sqrt[b*c - a*d]*f) - (b*Sqrt[c + d*Tan[e + f*x
]])/((a^2 + b^2)*f*(a + b*Tan[e + f*x]))
Rule 3568
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n)/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(a^2
+ b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*(m + 1) - b*d*n - (b*c - a*d)*
(m + 1)*Tan[e + f*x] - b*d*(m + n + 1)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[2*m]
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{\sqrt{c+d \tan (e+f x)}}{(a+b \tan (e+f x))^2} \, dx &=-\frac{b \sqrt{c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac{\int \frac{\frac{1}{2} (-2 a c-b d)+(b c-a d) \tan (e+f x)+\frac{1}{2} b d \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{a^2+b^2}\\ &=-\frac{b \sqrt{c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac{\int \frac{-a^2 c+b^2 c-2 a b d+\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}+\frac{\left (b \left (4 a b c-3 a^2 d+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}\\ &=-\frac{b \sqrt{c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{(c-i d) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b)^2}+\frac{(c+i d) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b)^2}+\frac{\left (b \left (4 a b c-3 a^2 d+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 f}\\ &=-\frac{b \sqrt{c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac{(i c-d) \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 f}+\frac{(i c+d) \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 f}+\frac{\left (b \left (4 a b c-3 a^2 d+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 f}\\ &=-\frac{\sqrt{b} \left (4 a b c-3 a^2 d+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 \sqrt{b c-a d} f}-\frac{b \sqrt{c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac{(c-i d) \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 d f}-\frac{(c+i d) \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 d f}\\ &=-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(a-i b)^2 f}+\frac{i \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(a+i b)^2 f}-\frac{\sqrt{b} \left (4 a b c-3 a^2 d+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 \sqrt{b c-a d} f}-\frac{b \sqrt{c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}\\ \end{align*}
Mathematica [A] time = 1.95359, size = 276, normalized size = 1.19 \[ -\frac{\frac{\sqrt{b} \sqrt{b c-a d} \left (-3 a^2 d+4 a b c+b^2 d\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{a^2+b^2}+\frac{i \left ((a-i b)^2 \sqrt{c+i d} (a d-b c) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )+(a+i b)^2 \sqrt{c-i d} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )\right )}{a^2+b^2}+\frac{b^2 (c+d \tan (e+f x))^{3/2}}{a+b \tan (e+f x)}-b d \sqrt{c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d)} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[c + d*Tan[e + f*x]]/(a + b*Tan[e + f*x])^2,x]
[Out]
-(((I*((a + I*b)^2*Sqrt[c - I*d]*(b*c - a*d)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + (a - I*b)^2*Sqr
t[c + I*d]*(-(b*c) + a*d)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]]))/(a^2 + b^2) + (Sqrt[b]*Sqrt[b*c -
a*d]*(4*a*b*c - 3*a^2*d + b^2*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(a^2 + b^2) - b*
d*Sqrt[c + d*Tan[e + f*x]] + (b^2*(c + d*Tan[e + f*x])^(3/2))/(a + b*Tan[e + f*x]))/((a^2 + b^2)*(b*c - a*d)*f
))
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Maple [B] time = 0.095, size = 1890, normalized size = 8.2 \begin{align*} \text{result too large to display} \end{align*}
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,x)
[Out]
1/4/f/d/(a^2+b^2)^2*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2
*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2*c+1/4/f/d/(a^2+b^2)^2*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-
d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^2-1/4/f/d/(a^2+b^2)^2*ln((c+d*
tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*
(c^2+d^2)^(1/2)*b^2-1/4/f/d/(a^2+b^2)^2*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c
-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*c-1/2/f/(a^2+b^2)^2*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)
^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b-1/4/f/d/(a^2+b^2)^2*ln(d*t
an(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*(c^2+d^2)^(1/2)+2*c)^(1/2
)*(c^2+d^2)^(1/2)*a^2+1/4/f/d/(a^2+b^2)^2*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*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b^2-2/f/(a^2+b^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^
(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(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)*a*b+2/f/(a^2+b^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+
e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a*b*c+1/4/f/d/(a^2+b^2)^2*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*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*c+1/2/f/(a^2+b^2)^2*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*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*
b-1/4/f/d/(a^2+b^2)^2*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*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2*c+2/f/(a^2+b^2)^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))*(c^2+d^2)^(1/2)*a*b-2/f/(a^2+b^2)^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))*a*b*c-1/f*d/(a^2+b^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan
(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a^2+1/f*d/(a^2+b^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*
(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b^2+1/f*d/(a^2+b^2)^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))*a^2-1/f*d/(a^2+b^2)^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))*b^2-1/f*d*b/(a^2+b^2)^2*(c+d*tan(f*x+e))^(1/2)/(tan(f*x+e)*b*d
+a*d)*a^2-1/f*d*b^3/(a^2+b^2)^2*(c+d*tan(f*x+e))^(1/2)/(tan(f*x+e)*b*d+a*d)-3/f*d*b/(a^2+b^2)^2/((a*d-b*c)*b)^
(1/2)*arctan((c+d*tan(f*x+e))^(1/2)*b/((a*d-b*c)*b)^(1/2))*a^2+4/f*b^2/(a^2+b^2)^2/((a*d-b*c)*b)^(1/2)*arctan(
(c+d*tan(f*x+e))^(1/2)*b/((a*d-b*c)*b)^(1/2))*a*c+1/f*d*b^3/(a^2+b^2)^2/((a*d-b*c)*b)^(1/2)*arctan((c+d*tan(f*
x+e))^(1/2)*b/((a*d-b*c)*b)^(1/2))
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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((c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^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((c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^2,x, algorithm="fricas")
[Out]
Timed out
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c + d \tan{\left (e + f x \right )}}}{\left (a + b \tan{\left (e + f x \right )}\right )^{2}}\, dx \end{align*}
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,x)
[Out]
Integral(sqrt(c + d*tan(e + f*x))/(a + b*tan(e + f*x))**2, x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d \tan \left (f x + e\right ) + c}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
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,x, algorithm="giac")
[Out]
integrate(sqrt(d*tan(f*x + e) + c)/(b*tan(f*x + e) + a)^2, x)