3.1295 \(\int \frac {1}{(a+b \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=417 \[ -\frac {4 b^2 \left (-5 a^2 d+3 a b c-2 b^2 d\right )}{3 f \left (a^2+b^2\right )^2 (b c-a d)^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 b^2}{3 f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {2 d \left (3 a^4 d^3+a^2 b^2 d \left (11 c^2+17 d^2\right )-6 a b^3 c \left (c^2+d^2\right )+b^4 d \left (5 c^2+8 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 f \left (a^2+b^2\right )^2 \left (c^2+d^2\right ) (b c-a d)^3 \sqrt {c+d \tan (e+f x)}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2} (c-i d)^{3/2}}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2} (c+i d)^{3/2}} \]

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

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Rubi [A]  time = 1.98, antiderivative size = 417, 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 = {3569, 3649, 3616, 3615, 93, 208} \[ \frac {2 d \left (a^2 b^2 d \left (11 c^2+17 d^2\right )+3 a^4 d^3-6 a b^3 c \left (c^2+d^2\right )+b^4 d \left (5 c^2+8 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 f \left (a^2+b^2\right )^2 \left (c^2+d^2\right ) (b c-a d)^3 \sqrt {c+d \tan (e+f x)}}-\frac {4 b^2 \left (-5 a^2 d+3 a b c-2 b^2 d\right )}{3 f \left (a^2+b^2\right )^2 (b c-a d)^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 b^2}{3 f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2} (c-i d)^{3/2}}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2} (c+i d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

Rule 93

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 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 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 3615

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 3616

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 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])))

Rubi steps

\begin {align*} \int \frac {1}{(a+b \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx &=-\frac {2 b^2}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 \int \frac {\frac {1}{2} \left (4 b^2 d-3 a (b c-a d)\right )+\frac {3}{2} b (b c-a d) \tan (e+f x)+2 b^2 d \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx}{3 \left (a^2+b^2\right ) (b c-a d)}\\ &=-\frac {2 b^2}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {4 b^2 \left (3 a b c-5 a^2 d-2 b^2 d\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {4 \int \frac {\frac {1}{4} \left (-6 a^3 b c d-6 a b^3 c d+3 a^4 d^2-b^4 \left (3 c^2-8 d^2\right )+a^2 b^2 \left (3 c^2+17 d^2\right )\right )-\frac {3}{2} a b (b c-a d)^2 \tan (e+f x)-b^2 d \left (3 a b c-5 a^2 d-2 b^2 d\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 )^2 (b c-a d)^2}\\ &=-\frac {2 b^2}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {4 b^2 \left (3 a b c-5 a^2 d-2 b^2 d\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {2 d \left (3 a^4 d^3-6 a b^3 c \left (c^2+d^2\right )+b^4 d \left (5 c^2+8 d^2\right )+a^2 b^2 d \left (11 c^2+17 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {8 \int \frac {\frac {3}{8} (b c-a d)^3 \left (a^2 c-b^2 c-2 a b d\right )-\frac {3}{8} (b c-a d)^3 \left (2 a b c+a^2 d-b^2 d\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 )^2 (b c-a d)^3 \left (c^2+d^2\right )}\\ &=-\frac {2 b^2}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {4 b^2 \left (3 a b c-5 a^2 d-2 b^2 d\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {2 d \left (3 a^4 d^3-6 a b^3 c \left (c^2+d^2\right )+b^4 d \left (5 c^2+8 d^2\right )+a^2 b^2 d \left (11 c^2+17 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) 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)^2 (c-i d)}+\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)^2 (c+i d)}\\ &=-\frac {2 b^2}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {4 b^2 \left (3 a b c-5 a^2 d-2 b^2 d\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {2 d \left (3 a^4 d^3-6 a b^3 c \left (c^2+d^2\right )+b^4 d \left (5 c^2+8 d^2\right )+a^2 b^2 d \left (11 c^2+17 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {\operatorname {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)^2 (c-i d) f}+\frac {\operatorname {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)^2 (c+i d) f}\\ &=-\frac {2 b^2}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {4 b^2 \left (3 a b c-5 a^2 d-2 b^2 d\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {2 d \left (3 a^4 d^3-6 a b^3 c \left (c^2+d^2\right )+b^4 d \left (5 c^2+8 d^2\right )+a^2 b^2 d \left (11 c^2+17 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {\operatorname {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)^2 (c-i d) f}+\frac {\operatorname {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)^2 (c+i d) 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)^{5/2} (c-i d)^{3/2} 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)^{5/2} (c+i d)^{3/2} f}-\frac {2 b^2}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {4 b^2 \left (3 a b c-5 a^2 d-2 b^2 d\right )}{3 \left (a^2+b^2\right )^2 (b c-a d)^2 f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {2 d \left (3 a^4 d^3-6 a b^3 c \left (c^2+d^2\right )+b^4 d \left (5 c^2+8 d^2\right )+a^2 b^2 d \left (11 c^2+17 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 6.49, size = 601, normalized size = 1.44 \[ -\frac {2 b^2}{3 f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (-\frac {2 \left (\frac {1}{2} b^2 \left (4 b^2 d-3 a (b c-a d)\right )-a \left (\frac {3}{2} b^2 (b c-a d)-2 a b^2 d\right )\right )}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 \left (-\frac {2 \left (\frac {1}{4} d^2 \left (3 a^4 d^2-6 a^3 b c d+a^2 b^2 \left (3 c^2+17 d^2\right )-6 a b^3 c d-b^4 \left (3 c^2-8 d^2\right )\right )-c \left (b^2 c d \left (-5 a^2 d+3 a b c-2 b^2 d\right )-\frac {3}{2} a b d (b c-a d)^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (a d-b c) \sqrt {c+d \tan (e+f x)}}-\frac {3 (b c-a d)^3 \left (\frac {(a-i b)^2 (d+i c) \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} \sqrt {c+i d}}+\frac {(a+i b)^2 (-d+i c) \tanh ^{-1}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+i b} \sqrt {-c+i d}}\right )}{4 f \left (c^2+d^2\right ) (a d-b c)}\right )}{\left (a^2+b^2\right ) (b c-a d)}\right )}{3 \left (a^2+b^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2)/(3*(a^2 + b^2)*(b*c - a*d)*f*(a + b*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]) - (2*((-2*((b^2*(4*
b^2*d - 3*a*(b*c - a*d)))/2 - a*(-2*a*b^2*d + (3*b^2*(b*c - a*d))/2)))/((a^2 + b^2)*(b*c - a*d)*f*Sqrt[a + b*T
an[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) - (2*((-3*(b*c - a*d)^3*(((a + I*b)^2*(I*c - d)*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]) + ((a -
I*b)^2*(I*c + d)*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*(-(c*((-3*a*b*d*(b*c - a*d)^2)/2 + b^2*c
*d*(3*a*b*c - 5*a^2*d - 2*b^2*d))) + (d^2*(-6*a^3*b*c*d - 6*a*b^3*c*d + 3*a^4*d^2 - b^4*(3*c^2 - 8*d^2) + a^2*
b^2*(3*c^2 + 17*d^2)))/4)*Sqrt[a + b*Tan[e + f*x]])/((-(b*c) + a*d)*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]])))/
((a^2 + b^2)*(b*c - a*d))))/(3*(a^2 + b^2)*(b*c - a*d))

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

[Out]

Timed out

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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