∫ 1_____________ (a+b tan(e+fx))2(c+d tan(e+fx))5∕2 dx

3.1264 \(\int \frac {1}{(a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=425 \[ -\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {b^{7/2} \left (-9 a^2 d+4 a b c-5 b^2 d\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{f \left (a^2+b^2\right )^2 (b c-a d)^{7/2}}+\frac {d \left (4 a^3 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )+4 a b^2 c d^3-\left (b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right )^2 (b c-a d)^3 \sqrt {c+d \tan (e+f x)}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (a-i b)^2 (c-i d)^{5/2}}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (a+i b)^2 (c+i d)^{5/2}} \]

[Out]

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

+I*d)^(1/2))/(a+I*b)^2/(c+I*d)^(5/2)/f-b^(7/2)*(-9*a^2*d+4*a*b*c-5*b^2*d)*arctanh(b^(1/2)*(c+d*tan(f*x+e))^(1/

2)/(-a*d+b*c)^(1/2))/(a^2+b^2)^2/(-a*d+b*c)^(7/2)/f+d*(4*a^3*c*d^3+4*a*b^2*c*d^3-4*a^2*b*d^2*(2*c^2+d^2)-b^3*(

c^4+10*c^2*d^2+5*d^4))/(a^2+b^2)/(-a*d+b*c)^3/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)-1/3*d*(2*a^2*d^2+b^2*(3*c^2

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

/(c+d*tan(f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A] time = 2.51, antiderivative size = 425, normalized size of antiderivative = 1.00, number of steps used = 14, 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} \[ \frac {d \left (-4 a^2 b d^2 \left (2 c^2+d^2\right )+4 a^3 c d^3+4 a b^2 c d^3+b^3 \left (-\left (10 c^2 d^2+c^4+5 d^4\right )\right )\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right )^2 (b c-a d)^3 \sqrt {c+d \tan (e+f x)}}-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {b^{7/2} \left (-9 a^2 d+4 a b c-5 b^2 d\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{f \left (a^2+b^2\right )^2 (b c-a d)^{7/2}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (a-i b)^2 (c-i d)^{5/2}}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (a+i b)^2 (c+i d)^{5/2}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)^2*(c + I*d)^(5/2)*f) - (b^(7/2)*(4*a*b*c - 9*a^2*d - 5*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)^(7/2)*f) - (d*(2*a^2*d^2 + b

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

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

2*(2*c^2 + d^2) - b^3*(c^4 + 10*c^2*d^2 + 5*d^4)))/((a^2 + b^2)*(b*c - a*d)^3*(c^2 + d^2)^2*f*Sqrt[c + d*Tan[e

+ f*x]])

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 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 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 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 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]

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, -

Rubi steps

\begin {align*} \int \frac {1}{(a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {\int \frac {\frac {1}{2} \left (-2 a b c+2 a^2 d+5 b^2 d\right )+b (b c-a d) \tan (e+f x)+\frac {5}{2} b^2 d \tan ^2(e+f x)}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}} \, dx}{\left (a^2+b^2\right ) (b c-a d)}\\ &=-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {2 \int \frac {-\frac {3}{4} \left (2 a^3 c d^2-4 a^2 b d \left (c^2+d^2\right )-5 b^3 d \left (c^2+d^2\right )+2 a b^2 c \left (c^2+2 d^2\right )\right )+\frac {3}{2} (b c-a d)^2 (b c+a d) \tan (e+f x)+\frac {3}{4} b \left (2 a^2 d^3+b^2 \left (3 c^2 d+5 d^3\right )\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx}{3 \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 (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}-\frac {4 \int \frac {\frac {3}{8} \left (2 a^4 d^3 \left (c^2-d^2\right )+5 b^4 d \left (c^2+d^2\right )^2-2 a^3 b c d^2 \left (3 c^2+d^2\right )-2 a b^3 c \left (c^4+5 c^2 d^2+2 d^4\right )+2 a^2 b^2 d \left (3 c^4+7 c^2 d^2+2 d^4\right )\right )+\frac {3}{4} (b c-a d)^3 \left (2 a c d+b \left (c^2-d^2\right )\right ) \tan (e+f x)-\frac {3}{8} b d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2}\\ &=-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\left (b^4 \left (4 a b c-9 a^2 d-5 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)^3}-\frac {4 \int \frac {-\frac {3}{4} (b c-a d)^3 (a c-b c-a d-b d) (a c+b c+a d-b d)+\frac {3}{2} (b c-a d)^3 (b c+a d) (a c-b d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right )^2}\\ &=-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \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 {c+d \tan (e+f x)}} \, dx}{2 (a-i b)^2 (c-i d)^2}+\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)^2}+\frac {\left (b^4 \left (4 a b c-9 a^2 d-5 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)^3 f}\\ &=-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {i \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 (c-i d)^2 f}-\frac {i \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 (c+i d)^2 f}+\frac {\left (b^4 \left (4 a b c-9 a^2 d-5 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)^3 f}\\ &=-\frac {b^{7/2} \left (4 a b c-9 a^2 d-5 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)^{7/2} f}-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \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)^2 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)^2 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)^{5/2} 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)^{5/2} f}-\frac {b^{7/2} \left (4 a b c-9 a^2 d-5 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)^{7/2} f}-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] time = 6.27, size = 2536, normalized size = 5.97 \[ \text {Result too large to show} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

a^2*d + 5*b^2*d))/2 - c*((-5*b^2*c*d)/2 + b*d*(b*c - a*d))))/(3*(-(b*c) + a*d)*(c^2 + d^2)*f*(c + d*Tan[e + f*

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

*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2 + d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - (3*b*d*(2*a^2*d^3 + b^2*(3*

c^2*d + 5*d^3)))/4))/2 + a*((-3*((b*d^2)/2 - (c*(-(b*c) + a*d))/2)*(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^

3*d*(c^2 + d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - (a*d*((3*d*(b*c - a*d)^2*(b*c + a*d))/2 - (3*b*c*(2*a^2*d^3 +

b^2*(3*c^2*d + 5*d^3)))/4))/2 - (b*((-3*d^2*(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2 + d^2) + 2*a*b

^2*c*(c^2 + 2*d^2)))/4 - c*((3*d*(b*c - a*d)^2*(b*c + a*d))/2 - (3*b*c*(2*a^2*d^3 + b^2*(3*c^2*d + 5*d^3)))/4)

))/2) - I*((a*(-(b*c) + a*d)*((-3*c*(b*c - a*d)^2*(b*c + a*d))/2 - (3*d*(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) -

5*b^3*d*(c^2 + d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - (3*b*d*(2*a^2*d^3 + b^2*(3*c^2*d + 5*d^3)))/4))/2 - b*((-

3*((b*d^2)/2 - (c*(-(b*c) + a*d))/2)*(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2 + d^2) + 2*a*b^2*c*(c

^2 + 2*d^2)))/4 - (a*d*((3*d*(b*c - a*d)^2*(b*c + a*d))/2 - (3*b*c*(2*a^2*d^3 + b^2*(3*c^2*d + 5*d^3)))/4))/2

- (b*((-3*d^2*(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2 + d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - c*((3

*d*(b*c - a*d)^2*(b*c + a*d))/2 - (3*b*c*(2*a^2*d^3 + b^2*(3*c^2*d + 5*d^3)))/4)))/2)))*ArcTanh[Sqrt[c + d*Tan

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

d))/2 - (3*d*(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2 + d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - (3*b*d

*(2*a^2*d^3 + b^2*(3*c^2*d + 5*d^3)))/4))/2 + a*((-3*((b*d^2)/2 - (c*(-(b*c) + a*d))/2)*(2*a^3*c*d^2 - 4*a^2*b

*d*(c^2 + d^2) - 5*b^3*d*(c^2 + d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - (a*d*((3*d*(b*c - a*d)^2*(b*c + a*d))/2 -

(3*b*c*(2*a^2*d^3 + b^2*(3*c^2*d + 5*d^3)))/4))/2 - (b*((-3*d^2*(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*

d*(c^2 + d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - c*((3*d*(b*c - a*d)^2*(b*c + a*d))/2 - (3*b*c*(2*a^2*d^3 + b^2*(

3*c^2*d + 5*d^3)))/4)))/2) + I*((a*(-(b*c) + a*d)*((-3*c*(b*c - a*d)^2*(b*c + a*d))/2 - (3*d*(2*a^3*c*d^2 - 4*

a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2 + d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - (3*b*d*(2*a^2*d^3 + b^2*(3*c^2*d + 5

*d^3)))/4))/2 - b*((-3*((b*d^2)/2 - (c*(-(b*c) + a*d))/2)*(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2

+ d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - (a*d*((3*d*(b*c - a*d)^2*(b*c + a*d))/2 - (3*b*c*(2*a^2*d^3 + b^2*(3*c^

2*d + 5*d^3)))/4))/2 - (b*((-3*d^2*(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2 + d^2) + 2*a*b^2*c*(c^2

+ 2*d^2)))/4 - c*((3*d*(b*c - a*d)^2*(b*c + a*d))/2 - (3*b*c*(2*a^2*d^3 + b^2*(3*c^2*d + 5*d^3)))/4)))/2)))*A

rcTanh[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]*(-1/2*(a*b*(-

(b*c) + a*d)*((-3*c*(b*c - a*d)^2*(b*c + a*d))/2 - (3*d*(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2 +

d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - (3*b*d*(2*a^2*d^3 + b^2*(3*c^2*d + 5*d^3)))/4)) + (a^2*b*((-3*d^2*(2*a^3*

c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2 + d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - c*((3*d*(b*c - a*d)^2*(b*c

+ a*d))/2 - (3*b*c*(2*a^2*d^3 + b^2*(3*c^2*d + 5*d^3)))/4)))/2 + b^2*((-3*((b*d^2)/2 - (c*(-(b*c) + a*d))/2)*

(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2 + d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - (a*d*((3*d*(b*c - a

*d)^2*(b*c + a*d))/2 - (3*b*c*(2*a^2*d^3 + b^2*(3*c^2*d + 5*d^3)))/4))/2))*ArcTanh[(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*((-3*d^2*

(2*a^3*c*d^2 - 4*a^2*b*d*(c^2 + d^2) - 5*b^3*d*(c^2 + d^2) + 2*a*b^2*c*(c^2 + 2*d^2)))/4 - c*((3*d*(b*c - a*d)

^2*(b*c + a*d))/2 - (3*b*c*(2*a^2*d^3 + b^2*(3*c^2*d + 5*d^3)))/4)))/((-(b*c) + a*d)*(c^2 + d^2)*f*Sqrt[c + d*

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

________________________________________________________________________________________

fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))^(5/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab