3.14.2 \(\int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx\) [1302]

Optimal. Leaf size=433 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} (c-i d)^{5/2} f}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} (c+i d)^{5/2} f}-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 d^2+b^2 \left (3 c^2+4 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (6 a^3 c d^4+6 a b^2 c d^4-a^2 b d^3 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4 d+17 c^2 d^3+8 d^5\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \]

[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))/(a-I*b)^(3/2)/(c-I*d)^(5
/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))/(a+I*b)^(3/2)/(c+I*
d)^(5/2)/f+2/3*(6*a^3*c*d^4+6*a*b^2*c*d^4-a^2*b*d^3*(11*c^2+5*d^2)-b^3*(3*c^4*d+17*c^2*d^3+8*d^5))*(a+b*tan(f*
x+e))^(1/2)/(a^2+b^2)/(-a*d+b*c)^3/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)-2*b^2/(a^2+b^2)/(-a*d+b*c)/f/(a+b*tan(
f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2)-2/3*d*(a^2*d^2+b^2*(3*c^2+4*d^2))*(a+b*tan(f*x+e))^(1/2)/(a^2+b^2)/(-a*d+
b*c)^2/(c^2+d^2)/f/(c+d*tan(f*x+e))^(3/2)

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Rubi [A]
time = 1.33, antiderivative size = 433, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3650, 3730, 3697, 3696, 95, 214} \begin {gather*} -\frac {2 d \left (a^2 d^2+b^2 \left (3 c^2+4 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{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 {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac {2 \left (6 a^3 c d^4-a^2 b d^3 \left (11 c^2+5 d^2\right )+6 a b^2 c d^4-b^3 \left (3 c^4 d+17 c^2 d^3+8 d^5\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 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-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} (c-i d)^{5/2}}+\frac {i \tanh ^{-1}\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} (c+i d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Tan[e + f*x])^(3/2)*(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]])])/((a - I*b)^(
3/2)*(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]])])/((a + I*b)^(3/2)*(c + I*d)^(5/2)*f) - (2*b^2)/((a^2 + b^2)*(b*c - a*d)*f*Sqrt[a + b*Tan[e + f*x]]*(
c + d*Tan[e + f*x])^(3/2)) - (2*d*(a^2*d^2 + b^2*(3*c^2 + 4*d^2))*Sqrt[a + b*Tan[e + f*x]])/(3*(a^2 + b^2)*(b*
c - a*d)^2*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + (2*(6*a^3*c*d^4 + 6*a*b^2*c*d^4 - a^2*b*d^3*(11*c^2 + 5
*d^2) - b^3*(3*c^4*d + 17*c^2*d^3 + 8*d^5))*Sqrt[a + b*Tan[e + f*x]])/(3*(a^2 + b^2)*(b*c - a*d)^3*(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*} \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 \int \frac {\frac {1}{2} \left (-a b c+a^2 d+4 b^2 d\right )+\frac {1}{2} b (b c-a d) \tan (e+f x)+2 b^2 d \tan ^2(e+f x)}{\sqrt {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 {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 d^2+b^2 \left (3 c^2+4 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{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 {4 \int \frac {\frac {1}{4} \left (-3 a^3 c d^2-3 a b^2 c \left (c^2+2 d^2\right )+a^2 b d \left (6 c^2+5 d^2\right )+b^3 d \left (9 c^2+8 d^2\right )\right )+\frac {3}{4} (b c-a d)^2 (b c+a d) \tan (e+f x)+\frac {1}{2} b \left (a^2 d^3+b^2 \left (3 c^2 d+4 d^3\right )\right ) \tan ^2(e+f x)}{\sqrt {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 {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 d^2+b^2 \left (3 c^2+4 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{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 {2 \left (6 a^3 c d^4+6 a b^2 c d^4-a^2 b d^3 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4 d+17 c^2 d^3+8 d^5\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \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 {8 \int \frac {\frac {3}{8} (b c-a d)^3 \left (2 b c d-a \left (c^2-d^2\right )\right )+\frac {3}{8} (b c-a d)^3 \left (2 a c d+b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {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 {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 d^2+b^2 \left (3 c^2+4 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{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 {2 \left (6 a^3 c d^4+6 a b^2 c d^4-a^2 b d^3 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4 d+17 c^2 d^3+8 d^5\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \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 {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b) (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 (a+i b) (c+i d)^2}\\ &=-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 d^2+b^2 \left (3 c^2+4 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{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 {2 \left (6 a^3 c d^4+6 a b^2 c d^4-a^2 b d^3 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4 d+17 c^2 d^3+8 d^5\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \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 {\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) (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 (a+i b) (c+i d)^2 f}\\ &=-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 d^2+b^2 \left (3 c^2+4 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{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 {2 \left (6 a^3 c d^4+6 a b^2 c d^4-a^2 b d^3 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4 d+17 c^2 d^3+8 d^5\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \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 {\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) (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 )}{(a+i b) (c+i d)^2 f}\\ &=-\frac {i \tanh ^{-1}\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} (c-i d)^{5/2} f}+\frac {i \tanh ^{-1}\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} (c+i d)^{5/2} f}-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (a^2 d^2+b^2 \left (3 c^2+4 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{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 {2 \left (6 a^3 c d^4+6 a b^2 c d^4-a^2 b d^3 \left (11 c^2+5 d^2\right )-b^3 \left (3 c^4 d+17 c^2 d^3+8 d^5\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \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*}

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Mathematica [A]
time = 6.48, size = 610, normalized size = 1.41 \begin {gather*} -\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (-\frac {2 \left (\frac {1}{2} d^2 \left (-a b c+a^2 d+4 b^2 d\right )-c \left (-2 b^2 c d+\frac {1}{2} b d (b c-a d)\right )\right ) \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 \left (\frac {3 (b c-a d)^3 \left (\frac {(i a-b) (c+i d)^2 \tanh ^{-1}\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 {(i a+b) (c-i d)^2 \tanh ^{-1}\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 )}{4 (-b c+a d) \left (c^2+d^2\right ) f}-\frac {2 \left (\frac {1}{4} d^2 \left (-3 a^3 c d^2-3 a b^2 c \left (c^2+2 d^2\right )+a^2 b d \left (6 c^2+5 d^2\right )+b^3 d \left (9 c^2+8 d^2\right )\right )-c \left (\frac {3}{4} d (b c-a d)^2 (b c+a d)-\frac {1}{2} b c \left (a^2 d^3+b^2 \left (3 c^2 d+4 d^3\right )\right )\right )\right ) \sqrt {a+b \tan (e+f x)}}{(-b c+a d) \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}\right )}{3 (-b c+a d) \left (c^2+d^2\right )}\right )}{\left (a^2+b^2\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

(-2*b^2)/((a^2 + b^2)*(b*c - a*d)*f*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2)) - (2*((-2*((d^2*(-(a*
b*c) + a^2*d + 4*b^2*d))/2 - c*(-2*b^2*c*d + (b*d*(b*c - a*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)) - (2*((3*(b*c - a*d)^3*(((I*a - b)*(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]) + ((
I*a + b)*(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])))/(4*(-(b*c) + a*d)*(c^2 + d^2)*f) - (2*((d^2*(-3*a^3*c*d^2 - 3*a*b^2*c*(c^2
+ 2*d^2) + a^2*b*d*(6*c^2 + 5*d^2) + b^3*d*(9*c^2 + 8*d^2)))/4 - c*((3*d*(b*c - a*d)^2*(b*c + a*d))/4 - (b*c*(
a^2*d^3 + b^2*(3*c^2*d + 4*d^3)))/2))*Sqrt[a + b*Tan[e + f*x]])/((-(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))

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}} \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x)

[Out]

int(1/(a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*tan(f*x + e) + a)^(3/2)*(d*tan(f*x + e) + c)^(5/2)), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e))**(3/2)/(c+d*tan(f*x+e))**(5/2),x)

[Out]

Integral(1/((a + b*tan(e + f*x))**(3/2)*(c + d*tan(e + f*x))**(5/2)), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*tan(e + f*x))^(3/2)*(c + d*tan(e + f*x))^(5/2)),x)

[Out]

int(1/((a + b*tan(e + f*x))^(3/2)*(c + d*tan(e + f*x))^(5/2)), x)

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