3.1246 \(\int \frac{(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=355 \[ \frac{\sqrt{b c-a d} \left (-6 a^2 b^2 \left (4 c^2-3 d^2\right )+8 a^3 b c d+a^4 d^2-56 a b^3 c d+b^4 \left (8 c^2-15 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{3/2} f \left (a^2+b^2\right )^3}-\frac{(b c-a d) \left (a^2 d+8 a b c+9 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{4 b f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))}-\frac{(b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{2 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}-\frac{(c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a)^3}+\frac{(c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a)^3} \]

[Out]

-(((c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((I*a + b)^3*f)) + ((c + I*d)^(5/2)*ArcTan
h[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((I*a - b)^3*f) + (Sqrt[b*c - a*d]*(8*a^3*b*c*d - 56*a*b^3*c*d + a^
4*d^2 + b^4*(8*c^2 - 15*d^2) - 6*a^2*b^2*(4*c^2 - 3*d^2))*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c
- a*d]])/(4*b^(3/2)*(a^2 + b^2)^3*f) - ((b*c - a*d)^2*Sqrt[c + d*Tan[e + f*x]])/(2*b*(a^2 + b^2)*f*(a + b*Tan[
e + f*x])^2) - ((b*c - a*d)*(8*a*b*c + a^2*d + 9*b^2*d)*Sqrt[c + d*Tan[e + f*x]])/(4*b*(a^2 + b^2)^2*f*(a + b*
Tan[e + f*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.85775, antiderivative size = 355, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {3565, 3649, 3653, 3539, 3537, 63, 208, 3634} \[ \frac{\sqrt{b c-a d} \left (-6 a^2 b^2 \left (4 c^2-3 d^2\right )+8 a^3 b c d+a^4 d^2-56 a b^3 c d+b^4 \left (8 c^2-15 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{3/2} f \left (a^2+b^2\right )^3}-\frac{(b c-a d) \left (a^2 d+8 a b c+9 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{4 b f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))}-\frac{(b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{2 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}-\frac{(c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a)^3}+\frac{(c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((I*a + b)^3*f)) + ((c + I*d)^(5/2)*ArcTan
h[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((I*a - b)^3*f) + (Sqrt[b*c - a*d]*(8*a^3*b*c*d - 56*a*b^3*c*d + a^
4*d^2 + b^4*(8*c^2 - 15*d^2) - 6*a^2*b^2*(4*c^2 - 3*d^2))*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c
- a*d]])/(4*b^(3/2)*(a^2 + b^2)^3*f) - ((b*c - a*d)^2*Sqrt[c + d*Tan[e + f*x]])/(2*b*(a^2 + b^2)*f*(a + b*Tan[
e + f*x])^2) - ((b*c - a*d)*(8*a*b*c + a^2*d + 9*b^2*d)*Sqrt[c + d*Tan[e + f*x]])/(4*b*(a^2 + b^2)^2*f*(a + b*
Tan[e + f*x]))

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(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] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

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, -
1]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^3} \, dx &=-\frac{(b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\int \frac{\frac{1}{2} \left (9 b^2 c^2 d+a^2 d^3+a b \left (4 c^3-6 c d^2\right )\right )+2 b \left (a d \left (3 c^2-d^2\right )-b \left (c^3-3 c d^2\right )\right ) \tan (e+f x)+\frac{1}{2} d \left (\left (a^2+4 b^2\right ) d^2-3 b c (b c-2 a d)\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}} \, dx}{2 b \left (a^2+b^2\right )}\\ &=-\frac{(b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac{(b c-a d) \left (8 a b c+a^2 d+9 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\int \frac{-\frac{1}{4} (b c-a d) \left (8 a^2 b c^3-8 b^3 c^3+40 a b^2 c^2 d-17 a^2 b c d^2+15 b^3 c d^2+a^3 d^3-7 a b^2 d^3\right )+2 b (b c-a d) \left (2 a b c \left (c^2-3 d^2\right )+b^2 d \left (3 c^2-d^2\right )-a^2 \left (3 c^2 d-d^3\right )\right ) \tan (e+f x)+\frac{1}{4} d (b c-a d)^2 \left (8 a b c+a^2 d+9 b^2 d\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{2 b \left (a^2+b^2\right )^2 (b c-a d)}\\ &=-\frac{(b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac{(b c-a d) \left (8 a b c+a^2 d+9 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\int \frac{-2 b (b c-a d) (a c+b d) \left (a^2 c^2-3 b^2 c^2+8 a b c d-3 a^2 d^2+b^2 d^2\right )+2 b (b c-a d)^2 \left (8 a b c d-b^2 \left (c^2-3 d^2\right )+a^2 \left (3 c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 b \left (a^2+b^2\right )^3 (b c-a d)}-\frac{\left ((b c-a d) \left (8 a^3 b c d-56 a b^3 c d+a^4 d^2+b^4 \left (8 c^2-15 d^2\right )-6 a^2 b^2 \left (4 c^2-3 d^2\right )\right )\right ) \int \frac{1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{8 b \left (a^2+b^2\right )^3}\\ &=-\frac{(b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac{(b c-a d) \left (8 a b c+a^2 d+9 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}+\frac{(c-i d)^3 \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b)^3}+\frac{(c+i d)^3 \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b)^3}-\frac{\left ((b c-a d) \left (8 a^3 b c d-56 a b^3 c d+a^4 d^2+b^4 \left (8 c^2-15 d^2\right )-6 a^2 b^2 \left (4 c^2-3 d^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{8 b \left (a^2+b^2\right )^3 f}\\ &=-\frac{(b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac{(b c-a d) \left (8 a b c+a^2 d+9 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}+\frac{(c-i d)^3 \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (i a+b)^3 f}-\frac{(c+i d)^3 \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (i a-b)^3 f}-\frac{\left ((b c-a d) \left (8 a^3 b c d-56 a b^3 c d+a^4 d^2+b^4 \left (8 c^2-15 d^2\right )-6 a^2 b^2 \left (4 c^2-3 d^2\right )\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 )}{4 b \left (a^2+b^2\right )^3 d f}\\ &=\frac{\sqrt{b c-a d} \left (8 a^3 b c d-56 a b^3 c d+a^4 d^2+b^4 \left (8 c^2-15 d^2\right )-6 a^2 b^2 \left (4 c^2-3 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{3/2} \left (a^2+b^2\right )^3 f}-\frac{(b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac{(b c-a d) \left (8 a b c+a^2 d+9 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{(c-i d)^3 \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)^3 d f}-\frac{(c+i d)^3 \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)^3 d f}\\ &=-\frac{(c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(i a+b)^3 f}+\frac{(c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(i a-b)^3 f}+\frac{\sqrt{b c-a d} \left (8 a^3 b c d-56 a b^3 c d+a^4 d^2+b^4 \left (8 c^2-15 d^2\right )-6 a^2 b^2 \left (4 c^2-3 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{4 b^{3/2} \left (a^2+b^2\right )^3 f}-\frac{(b c-a d)^2 \sqrt{c+d \tan (e+f x)}}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac{(b c-a d) \left (8 a b c+a^2 d+9 b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{4 b \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}\\ \end{align*}

Mathematica [B]  time = 6.45764, size = 2946, normalized size = 8.3 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*(c + d*Tan[e + f*x])^(7/2))/(2*(a^2 + b^2)*(b*c - a*d)*f*(a + b*Tan[e + f*x])^2) - (-((b*d*(c + d*Tan[e
+ f*x])^(5/2))/(f*(a + b*Tan[e + f*x]))) + (2*((-3*b*d*(b*c - a*d)*(c + d*Tan[e + f*x])^(3/2))/(2*f*(a + b*Tan
[e + f*x])) + (2*((-3*b*d*(b*c - a*d)^2*Sqrt[c + d*Tan[e + f*x]])/(4*f*(a + b*Tan[e + f*x])) - (2*(-((((I*Sqrt
[c - I*d]*(b*(b*c - a*d)*((3*a*b^3*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*b^3*d*(b*c - a*
d)*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^2 - 4*d^2)))/16 + (3*b^3*(b*c - a*d)*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 -
6*c*d^2)))/16) + a*((3*b^2*(b*c - a*d)*((b^2*d)/2 - a*(b*c - a*d))*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d
^2)))/16 + (-(b*c) + (a*d)/2)*((-3*b^4*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*a*b^2*d*(b*
c - a*d)*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^2 - 4*d^2)))/16) - (d*((3*b^4*(b*c - a*d)*(9*b^2*c^2*d + a^2*d^3 + a*
b*(4*c^3 - 6*c*d^2)))/16 - a*((-3*b^4*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*a*b^2*d*(b*c
- a*d)*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^2 - 4*d^2)))/16)))/2) - I*(a*(b*c - a*d)*((3*a*b^3*(b*c - a*d)*(b*c^3
- 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*b^3*d*(b*c - a*d)*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^2 - 4*d^2)))/16 + (
3*b^3*(b*c - a*d)*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16) - b*((3*b^2*(b*c - a*d)*((b^2*d)/2 - a*
(b*c - a*d))*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16 + (-(b*c) + (a*d)/2)*((-3*b^4*(b*c - a*d)*(b*
c^3 - 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*a*b^2*d*(b*c - a*d)*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^2 - 4*d^2)))/
16) - (d*((3*b^4*(b*c - a*d)*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16 - a*((-3*b^4*(b*c - a*d)*(b*c
^3 - 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*a*b^2*d*(b*c - a*d)*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^2 - 4*d^2)))/1
6)))/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*a*b^3*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*b^3*d*(b*c - a*d)*(6*a*b*c*d + a^2*d^2 - b
^2*(3*c^2 - 4*d^2)))/16 + (3*b^3*(b*c - a*d)*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16) + a*((3*b^2*
(b*c - a*d)*((b^2*d)/2 - a*(b*c - a*d))*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16 + (-(b*c) + (a*d)/
2)*((-3*b^4*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*a*b^2*d*(b*c - a*d)*(6*a*b*c*d + a^2*d
^2 - b^2*(3*c^2 - 4*d^2)))/16) - (d*((3*b^4*(b*c - a*d)*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16 -
a*((-3*b^4*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*a*b^2*d*(b*c - a*d)*(6*a*b*c*d + a^2*d^
2 - b^2*(3*c^2 - 4*d^2)))/16)))/2) + I*(a*(b*c - a*d)*((3*a*b^3*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2 + a
*d^3))/4 - (3*b^3*d*(b*c - a*d)*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^2 - 4*d^2)))/16 + (3*b^3*(b*c - a*d)*(9*b^2*c^
2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16) - b*((3*b^2*(b*c - a*d)*((b^2*d)/2 - a*(b*c - a*d))*(9*b^2*c^2*d +
a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16 + (-(b*c) + (a*d)/2)*((-3*b^4*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2
+ a*d^3))/4 - (3*a*b^2*d*(b*c - a*d)*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^2 - 4*d^2)))/16) - (d*((3*b^4*(b*c - a*d
)*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16 - a*((-3*b^4*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2
+ a*d^3))/4 - (3*a*b^2*d*(b*c - a*d)*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^2 - 4*d^2)))/16)))/2)))*ArcTanh[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]*(-(a*b*(b*c - a*d)*((3*a*b^3*
(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*b^3*d*(b*c - a*d)*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^
2 - 4*d^2)))/16 + (3*b^3*(b*c - a*d)*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16)) + (a^2*d*((3*b^4*(b
*c - a*d)*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16 - a*((-3*b^4*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*
b*c*d^2 + a*d^3))/4 - (3*a*b^2*d*(b*c - a*d)*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^2 - 4*d^2)))/16)))/2 + b^2*((3*b^
2*(b*c - a*d)*((b^2*d)/2 - a*(b*c - a*d))*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^3 - 6*c*d^2)))/16 + (-(b*c) + (a*d
)/2)*((-3*b^4*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*a*b^2*d*(b*c - a*d)*(6*a*b*c*d + a^2
*d^2 - b^2*(3*c^2 - 4*d^2)))/16)))*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))/((a^2 + b^2)*(b*c - a*d))) - (((3*b^4*(b*c - a*d)*(9*b^2*c^2*d + a^2*d^3 + a*b*(4*c^
3 - 6*c*d^2)))/16 - a*((-3*b^4*(b*c - a*d)*(b*c^3 - 3*a*c^2*d - 3*b*c*d^2 + a*d^3))/4 - (3*a*b^2*d*(b*c - a*d)
*(6*a*b*c*d + a^2*d^2 - b^2*(3*c^2 - 4*d^2)))/16))*Sqrt[c + d*Tan[e + f*x]])/((a^2 + b^2)*(b*c - a*d)*f*(a + b
*Tan[e + f*x]))))/b))/b))/(3*b))/(2*(a^2 + b^2)*(b*c - a*d))

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Maple [B]  time = 0.102, size = 6771, normalized size = 19.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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