\(\int \frac {1}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx\) [1301]

Optimal result

Integrand size = 29, antiderivative size = 295 \[ \int \frac {1}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a-i b} (c-i d)^{5/2} f}+\frac {i \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} (c+i d)^{5/2} f}+\frac {2 d^2 \sqrt {a+b \tan (e+f x)}}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {4 d^2 \left (3 a c d-b \left (4 c^2+d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 (b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \]

[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)^(5/2)/f/(a-I*b)^
(1/2)+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)^(5/2)/f/(a+
I*b)^(1/2)-4/3*d^2*(3*a*c*d-b*(4*c^2+d^2))*(a+b*tan(f*x+e))^(1/2)/(-a*d+b*c)^2/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^
(1/2)+2/3*d^2*(a+b*tan(f*x+e))^(1/2)/(-a*d+b*c)/(c^2+d^2)/f/(c+d*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 1.49 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3650, 3730, 3697, 3696, 95, 214} \[ \int \frac {1}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b} (c-i d)^{5/2}}+\frac {i \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a+i b} (c+i d)^{5/2}}-\frac {4 d^2 \left (3 a c d-b \left (4 c^2+d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}+\frac {2 d^2 \sqrt {a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}} \]

[In]

Int[1/(Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

((-I)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a - I*
b]*(c - I*d)^(5/2)*f) + (I*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e +
f*x]])])/(Sqrt[a + I*b]*(c + I*d)^(5/2)*f) + (2*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)) - (4*d^2*(3*a*c*d - b*(4*c^2 + d^2))*Sqrt[a + b*Tan[e + f*x]])/(3*(b*c - a*d)^2*(c^2
+ d^2)^2*f*Sqrt[c + d*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 3650

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 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]

Rule 3730

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*Ta
n[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])))

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

Mathematica [A] (verified)

Time = 2.17 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx=\frac {3 i (b c-a d)^2 \left (\frac {(c+i d)^2 \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+i b} \sqrt {-c+i d}}+\frac {(c-i d)^2 \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} \sqrt {c+i d}}\right )+\frac {2 d^2 (b c-a d) \left (c^2+d^2\right ) \sqrt {a+b \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}}+\frac {4 d^2 \left (-3 a c d+b \left (4 c^2+d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{3 (b c-a d)^2 \left (c^2+d^2\right )^2 f} \]

[In]

Integrate[1/(Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

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

Maple [F(-1)]

Timed out.

\[\int \frac {1}{\sqrt {a +b \tan \left (f x +e \right )}\, \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27028 vs. \(2 (239) = 478\).

Time = 56.28 (sec) , antiderivative size = 27028, normalized size of antiderivative = 91.62 \[ \int \frac {1}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

\[ \int \frac {1}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {1}{\sqrt {a + b \tan {\left (e + f x \right )}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/(a+b*tan(f*x+e))**(1/2)/(c+d*tan(f*x+e))**(5/2),x)

[Out]

Integral(1/(sqrt(a + b*tan(e + f*x))*(c + d*tan(e + f*x))**(5/2)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {b \tan \left (f x + e\right ) + a} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integrate(1/(sqrt(b*tan(f*x + e) + a)*(d*tan(f*x + e) + c)^(5/2)), x)

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx=\text {Hanged} \]

[In]

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

[Out]

\text{Hanged}