\(\int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{3/2}} \, dx\) [1269]

Optimal result

Integrand size = 29, antiderivative size = 206 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{3/2}} \, dx=-\frac {i \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} f}+\frac {i \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} f}-\frac {2 b \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3649, 3697, 3696, 95, 214} \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{3/2}} \, dx=-\frac {2 b \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}-\frac {i \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{3/2}}+\frac {i \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{3/2}} \]

[In]

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

[Out]

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

Rule 95

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 214

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

Rule 3649

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 3696

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 3697

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 b \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}}-\frac {2 \int \frac {\frac {1}{2} (-a c-b d)+\frac {1}{2} (b c-a d) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{a^2+b^2} \\ & = -\frac {2 b \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d) \int \frac {1+i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b)}+\frac {(c+i d) \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b)} \\ & = -\frac {2 b \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a-i b) f}+\frac {(c+i d) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a+i b) f} \\ & = -\frac {2 b \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d) \text {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 )}{(a-i b) f}+\frac {(c+i d) \text {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 )}{(a+i b) f} \\ & = -\frac {i \sqrt {c-i d} \text {arctanh}\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 )}{(a-i b)^{3/2} f}+\frac {i \sqrt {c+i d} \text {arctanh}\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 )}{(a+i b)^{3/2} f}-\frac {2 b \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.81 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{3/2}} \, dx=\frac {\frac {i \sqrt {-c+i d} \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-a+i b)^{3/2}}+\frac {\frac {(i a+b) \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2}}-\frac {2 b \sqrt {c+d \tan (e+f x)}}{(a+i b) \sqrt {a+b \tan (e+f x)}}}{a-i b}}{f} \]

[In]

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

[Out]

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

Maple [F(-1)]

Timed out.

\[\int \frac {\sqrt {c +d \tan \left (f x +e \right )}}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11981 vs. \(2 (158) = 316\).

Time = 7.67 (sec) , antiderivative size = 11981, normalized size of antiderivative = 58.16 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {c + d \tan {\left (e + f x \right )}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((2*b*d+2*a*c)^2>0)', see `ass
ume?` for mo

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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