3.106 \(\int (c+d \tan (e+f x))^{5/2} (A+B \tan (e+f x)+C \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=229 \[ \frac{2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{(c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}-\frac{(c+i d)^{5/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{2 C (c+d \tan (e+f x))^{7/2}}{7 d f} \]

[Out]

-(((I*A + B - I*C)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f) - ((B - I*(A - C))*(c +
I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*(2*c*(A - C)*d + B*(c^2 - d^2))*Sqrt[c + d
*Tan[e + f*x]])/f + (2*(B*c + (A - C)*d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (2*B*(c + d*Tan[e + f*x])^(5/2))/
(5*f) + (2*C*(c + d*Tan[e + f*x])^(7/2))/(7*d*f)

________________________________________________________________________________________

Rubi [A]  time = 0.628563, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3630, 3528, 3539, 3537, 63, 208} \[ \frac{2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{(c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}-\frac{(c+i d)^{5/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{2 C (c+d \tan (e+f x))^{7/2}}{7 d f} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]

[Out]

-(((I*A + B - I*C)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f) - ((B - I*(A - C))*(c +
I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*(2*c*(A - C)*d + B*(c^2 - d^2))*Sqrt[c + d
*Tan[e + f*x]])/f + (2*(B*c + (A - C)*d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (2*B*(c + d*Tan[e + f*x])^(5/2))/
(5*f) + (2*C*(c + d*Tan[e + f*x])^(7/2))/(7*d*f)

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 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 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]

Rubi steps

\begin{align*} \int (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac{2 C (c+d \tan (e+f x))^{7/2}}{7 d f}+\int (A-C+B \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx\\ &=\frac{2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{2 C (c+d \tan (e+f x))^{7/2}}{7 d f}+\int (c+d \tan (e+f x))^{3/2} (A c-c C-B d+(B c+(A-C) d) \tan (e+f x)) \, dx\\ &=\frac{2 (B c+(A-C) d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{2 C (c+d \tan (e+f x))^{7/2}}{7 d f}+\int \sqrt{c+d \tan (e+f x)} \left (-c^2 C-2 B c d+C d^2+A \left (c^2-d^2\right )+\left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \tan (e+f x)\right ) \, dx\\ &=\frac{2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 (B c+(A-C) d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{2 C (c+d \tan (e+f x))^{7/2}}{7 d f}+\int \frac{-c^3 C-3 B c^2 d+3 c C d^2+B d^3+A \left (c^3-3 c d^2\right )+\left ((A-C) 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\\ &=\frac{2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 (B c+(A-C) d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{2 C (c+d \tan (e+f x))^{7/2}}{7 d f}+\frac{1}{2} \left ((A-i B-C) (c-i d)^3\right ) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{1}{2} \left ((A+i B-C) (c+i d)^3\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 (B c+(A-C) d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{2 C (c+d \tan (e+f x))^{7/2}}{7 d f}+\frac{\left ((i A+B-i C) (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 f}-\frac{\left (i (A+i B-C) (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 f}\\ &=\frac{2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 (B c+(A-C) d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{2 C (c+d \tan (e+f x))^{7/2}}{7 d f}-\frac{\left ((A-i B-C) (c-i d)^3\right ) \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 )}{d f}-\frac{\left ((A+i B-C) (c+i d)^3\right ) \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 )}{d f}\\ &=-\frac{(i A+B-i C) (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}-\frac{(B-i (A-C)) (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 (B c+(A-C) d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\\ \end{align*}

Mathematica [A]  time = 2.03062, size = 262, normalized size = 1.14 \[ \frac{7 i (A-i B-C) \left (\frac{2}{5} (c+d \tan (e+f x))^{5/2}+\frac{2}{3} (c-i d) \left (\sqrt{c+d \tan (e+f x)} (4 c+d \tan (e+f x)-3 i d)-3 (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )\right )\right )-7 i (A+i B-C) \left (\frac{2}{5} (c+d \tan (e+f x))^{5/2}+\frac{2}{3} (c+i d) \left (\sqrt{c+d \tan (e+f x)} (4 c+d \tan (e+f x)+3 i d)-3 (c+i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )\right )\right )+\frac{4 C (c+d \tan (e+f x))^{7/2}}{d}}{14 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]

[Out]

((4*C*(c + d*Tan[e + f*x])^(7/2))/d + (7*I)*(A - I*B - C)*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (2*(c - I*d)*(-3
*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c - (3*I)*d + d
*Tan[e + f*x])))/3) - (7*I)*(A + I*B - C)*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (2*(c + I*d)*(-3*(c + I*d)^(3/2)
*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c + (3*I)*d + d*Tan[e + f*x])))
/3))/(14*f)

________________________________________________________________________________________

Maple [B]  time = 0.123, size = 3614, normalized size = 15.8 \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)+C*tan(f*x+e)^2),x)

[Out]

2/5*B*(c+d*tan(f*x+e))^(5/2)/f+1/f/(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^2+d^2)^(1/2)*c^2+1/4*d/f*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))*A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)-3/4*
d/f*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))*A*(2*(c^2+d^2)^(1/
2)+2*c)^(1/2)*c-1/4*d/f*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)
)*C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)+1/4/d/f*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))*C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c^2+2/f*B*c^2*(c+d*tan(f*x
+e))^(1/2)+2/3/f*B*(c+d*tan(f*x+e))^(3/2)*c+2/3*d/f*A*(c+d*tan(f*x+e))^(3/2)-2/3*d/f*C*(c+d*tan(f*x+e))^(3/2)-
2*d^2/f*B*(c+d*tan(f*x+e))^(1/2)+2/7*C*(c+d*tan(f*x+e))^(7/2)/f/d+2*d/f/(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*(c^2+d^2)^(1/2)*c+2*d
/f/(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+d^2)^(1/2)*c-2*d/f/(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*(c^2+d^2)^(1/2)*c+1/4/d/f*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))*A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/
2)*c^2-1/4/d/f*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))*C*(2*(c
^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c^2-1/4/d/f*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))*A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c^2-2*d/f/(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
+d^2)^(1/2)*c+1/4/d/f*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))*
C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3+d^3/f/(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-d^3/f/(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-3/4/f*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))*B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+1/f/(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+3/4/f*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))
*B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-1/f/(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+1/4*d^2/f*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))*B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-4*d/f*c*C*(c+d*tan(f*x+e))^(
1/2)-1/4*d^2/f*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))*B*(2*(c
^2+d^2)^(1/2)+2*c)^(1/2)+4*d/f*A*c*(c+d*tan(f*x+e))^(1/2)+d^3/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*t
an(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-d^3/f/(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-1/4/d/
f*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))*A*(2*(c^2+d^2)^(1/2)
+2*c)^(1/2)*c^3-1/2/f*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))*
B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c-1/f/(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^2+d^2)^(1/2)*c^2+1/2/f*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))*B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^
2+d^2)^(1/2)*c-3*d/f/(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*c^2-3*d^2/f/(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*d/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arcta
n((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-d^2/f/(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^2+d^2)^(1/2)+3/4*d/f*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))*A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c+1/4*d/f*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))*C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)-3/4*d/f*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))*C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c+d^2/f/(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^2+d^2)^(1/2)+3/4*d/f*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))*C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c+1/4/d/f*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))*A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3-1/4*d/f*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))*A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)-1/4/d/
f*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))*C*(2*(c^2+d^2)^(1/2)
+2*c)^(1/2)*c^3+3*d/f/(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*c^2-3*d/f/(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+3*d^2/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*ar
ctan(((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

________________________________________________________________________________________

Maxima [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)+C*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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)+C*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}} \left (A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \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)+C*tan(f*x+e)**2),x)

[Out]

Integral((c + d*tan(e + f*x))**(5/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)**2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )}{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{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)+C*tan(f*x+e)^2),x, algorithm="giac")

[Out]

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(d*tan(f*x + e) + c)^(5/2), x)