3.1244 \(\int \frac{(c+d \tan (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx\)

Optimal. Leaf size=195 \[ -\frac{2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{3/2} f \left (a^2+b^2\right )}+\frac{(c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a)}-\frac{(c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a)}+\frac{2 d^2 \sqrt{c+d \tan (e+f x)}}{b f} \]

[Out]

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

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Rubi [A]  time = 0.915193, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {3566, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{3/2} f \left (a^2+b^2\right )}+\frac{(c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a)}-\frac{(c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-b+i a)}+\frac{2 d^2 \sqrt{c+d \tan (e+f x)}}{b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3566

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 - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[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)} \, dx &=\frac{2 d^2 \sqrt{c+d \tan (e+f x)}}{b f}+\frac{2 \int \frac{\frac{1}{2} \left (b c^3-a d^3\right )+\frac{1}{2} b d \left (3 c^2-d^2\right ) \tan (e+f x)+\frac{1}{2} d^2 (3 b c-a d) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{b}\\ &=\frac{2 d^2 \sqrt{c+d \tan (e+f x)}}{b f}+\frac{2 \int \frac{\frac{1}{2} b \left (a c^3+3 b c^2 d-3 a c d^2-b d^3\right )+\frac{1}{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)}{\sqrt{c+d \tan (e+f x)}} \, dx}{b \left (a^2+b^2\right )}+\frac{(b c-a d)^3 \int \frac{1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac{2 d^2 \sqrt{c+d \tan (e+f x)}}{b f}+\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)}+\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)}+\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{b \left (a^2+b^2\right ) f}\\ &=\frac{2 d^2 \sqrt{c+d \tan (e+f x)}}{b f}-\frac{\left (i (c+i d)^3\right ) \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) f}-\frac{(i c+d)^3 \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) f}+\frac{\left (2 (b c-a d)^3\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 )}{b \left (a^2+b^2\right ) d f}\\ &=-\frac{2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right ) f}+\frac{2 d^2 \sqrt{c+d \tan (e+f x)}}{b 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) 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) 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) 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) f}-\frac{2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right ) f}+\frac{2 d^2 \sqrt{c+d \tan (e+f x)}}{b f}\\ \end{align*}

Mathematica [A]  time = 0.571066, size = 199, normalized size = 1.02 \[ \frac{2 \sqrt{b} d^2 \left (a^2+b^2\right ) \sqrt{c+d \tan (e+f x)}+b^{3/2} (b-i a) (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )+b^{3/2} (b+i a) (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )-2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{3/2} f \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.076, size = 2802, normalized size = 14.4 \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)),x)

[Out]

1/f/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2
*(c^2+d^2)^(1/2)-2*c)^(1/2))*b*c^3-1/f/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2
)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b*c^3+2*d^2*(c+d*tan(f*x+e))^(1/2)/b/f+3/4/f/(
a^2+b^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^
(1/2)+2*c)^(1/2)*b*c^2+1/4/f*d^2/(a^2+b^2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e
)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b-1/4/f*d^2/(a^2+b^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1
/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b+1/f*d^3/(a^2+b^2)/(2*(c^2+d
^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(
1/2))*a-1/f*d^3/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*
c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a+2/f*b^2/(a^2+b^2)/((a*d-b*c)*b)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)
*b/((a*d-b*c)*b)^(1/2))*c^3-3/4/f/(a^2+b^2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+
e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b*c^2-1/4/f/d/(a^2+b^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))
^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*c^2-2/f*
d/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(
c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*a*c+1/4/f/d/(a^2+b^2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+
2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*c^2+2/f*d/(a^2+b^2)
/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1
/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*a*c-6/f*d*b/(a^2+b^2)/((a*d-b*c)*b)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)*b/((a*
d-b*c)*b)^(1/2))*a*c^2+1/2/f/(a^2+b^2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-
(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b*c-1/f/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)
*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2
)*b*c^2+3/f*d/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)
^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a*c^2-3/f*d^2/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d
^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b*c+1/4/f*d/(a^2+b^2)*ln(d*tan(f
*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c
^2+d^2)^(1/2)*a+3/f*d^2/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^
(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b*c-3/f*d/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*
(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a*c^2+1/f*d^2/(a^2+b^2)/(2
*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)
-2*c)^(1/2))*(c^2+d^2)^(1/2)*b+6/f*d^2/(a^2+b^2)/((a*d-b*c)*b)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)*b/((a*d-b*c
)*b)^(1/2))*a^2*c-2/f*d^3/b/(a^2+b^2)/((a*d-b*c)*b)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)*b/((a*d-b*c)*b)^(1/2))
*a^3-1/f*d^2/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^
(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*b+1/4/f/d/(a^2+b^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(
1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c^3-3/4/f*d/(a^2+b^2)*ln(d
*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1
/2)*a*c-1/4/f*d/(a^2+b^2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/
2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a-1/4/f/d/(a^2+b^2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(
1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c^3+3/4/f*d/(a^2+b^2)*ln((c+d*
tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*
a*c+1/f/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)
)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*b*c^2-1/2/f/(a^2+b^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2
)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b*c

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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)),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)),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)),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}}}{b \tan \left (f x + e\right ) + a}\,{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)),x, algorithm="giac")

[Out]

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