3.1.93 \(\int \sqrt {c+d \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x)) \, dx\) [93]

Optimal. Leaf size=155 \[ -\frac {(i A+B-i C) \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(B-i (A-C)) \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 B \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 C (c+d \tan (e+f x))^{3/2}}{3 d f} \]

[Out]

-(I*A+B-I*C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c-I*d)^(1/2)/f-(B-I*(A-C))*arctanh((c+d*tan(f*x+e)
)^(1/2)/(c+I*d)^(1/2))*(c+I*d)^(1/2)/f+2*B*(c+d*tan(f*x+e))^(1/2)/f+2/3*C*(c+d*tan(f*x+e))^(3/2)/d/f

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Rubi [A]
time = 0.21, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3711, 3609, 3620, 3618, 65, 214} \begin {gather*} -\frac {\sqrt {c-i d} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {\sqrt {c+i d} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 B \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 C (c+d \tan (e+f x))^{3/2}}{3 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 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 3609

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 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^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 3620

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 3711

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]

Rubi steps

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

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Mathematica [A]
time = 0.39, size = 150, normalized size = 0.97 \begin {gather*} \frac {-3 i (A-i B-C) \sqrt {c-i d} d \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+3 i (A+i B-C) \sqrt {c+i d} d \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+2 \sqrt {c+d \tan (e+f x)} (c C+3 B d+C d \tan (e+f x))}{3 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(845\) vs. \(2(130)=260\).
time = 0.42, size = 846, normalized size = 5.46

method	result	size
derivativedivides	\(\frac {\frac {2 C \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 B d \sqrt {c +d \tan \left (f x +e \right )}+2 d \left (\frac {\frac {\left (A \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-A \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +B \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -C \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+C \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-2 B \sqrt {c^{2}+d^{2}}\, d +\frac {\left (A \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-A \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +B \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -C \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+C \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 d}+\frac {\frac {\left (-A \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+A \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -B \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d +C \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-C \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-2 B \sqrt {c^{2}+d^{2}}\, d -\frac {\left (-A \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+A \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -B \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d +C \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-C \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 d}\right )}{d f}\)	\(846\)
default	\(\frac {\frac {2 C \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 B d \sqrt {c +d \tan \left (f x +e \right )}+2 d \left (\frac {\frac {\left (A \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-A \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +B \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -C \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+C \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-2 B \sqrt {c^{2}+d^{2}}\, d +\frac {\left (A \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-A \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c +B \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -C \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+C \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 d}+\frac {\frac {\left (-A \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+A \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -B \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d +C \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-C \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-2 B \sqrt {c^{2}+d^{2}}\, d -\frac {\left (-A \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+A \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -B \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d +C \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-C \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 d}\right )}{d f}\)	\(846\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

2/f/d*(1/3*C*(c+d*tan(f*x+e))^(3/2)+B*d*(c+d*tan(f*x+e))^(1/2)+d*(1/4/d*(1/2*(A*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(
1/2)+2*c)^(1/2)-A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c+B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d-C*(c^2+d^2)^(1/2)*(2*(c^2+
d^2)^(1/2)+2*c)^(1/2)+C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c)*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*(-2*B*(c^2+d^2)^(1/2)*d+1/2*(A*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1
/2)-A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c+B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d-C*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2
*c)^(1/2)+C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arct
an((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)))+1/4/d*(1/2*(-A*(c^
2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d
+C*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c)*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*(-2*B*(c^2+d^2)^(1/2)*d-1/2*(-A*(c^2+d^2)^(1/2
)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d+C*(c^2+d^2
)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(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)))))

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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((c+d*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*sqrt(d*tan(f*x + e) + c), 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((c+d*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^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 \sqrt {c + d \tan {\left (e + f x \right )}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [B]
time = 17.40, size = 1199, normalized size = 7.74 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {32\,B^2\,d^4\,\sqrt {\frac {B^2\,c}{4\,f^2}-\frac {\sqrt {-B^4\,d^2\,f^4}}{4\,f^4}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{\frac {16\,B\,d^4\,\sqrt {-B^4\,d^2\,f^4}}{f^3}+\frac {16\,B\,c^2\,d^2\,\sqrt {-B^4\,d^2\,f^4}}{f^3}}-\frac {32\,c\,d^2\,\sqrt {\frac {B^2\,c}{4\,f^2}-\frac {\sqrt {-B^4\,d^2\,f^4}}{4\,f^4}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-B^4\,d^2\,f^4}}{\frac {16\,B\,d^4\,\sqrt {-B^4\,d^2\,f^4}}{f}+\frac {16\,B\,c^2\,d^2\,\sqrt {-B^4\,d^2\,f^4}}{f}}\right )\,\sqrt {-\frac {\sqrt {-B^4\,d^2\,f^4}-B^2\,c\,f^2}{4\,f^4}}-2\,\mathrm {atanh}\left (\frac {32\,B^2\,d^4\,\sqrt {\frac {\sqrt {-B^4\,d^2\,f^4}}{4\,f^4}+\frac {B^2\,c}{4\,f^2}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{\frac {16\,B\,d^4\,\sqrt {-B^4\,d^2\,f^4}}{f^3}+\frac {16\,B\,c^2\,d^2\,\sqrt {-B^4\,d^2\,f^4}}{f^3}}+\frac {32\,c\,d^2\,\sqrt {\frac {\sqrt {-B^4\,d^2\,f^4}}{4\,f^4}+\frac {B^2\,c}{4\,f^2}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-B^4\,d^2\,f^4}}{\frac {16\,B\,d^4\,\sqrt {-B^4\,d^2\,f^4}}{f}+\frac {16\,B\,c^2\,d^2\,\sqrt {-B^4\,d^2\,f^4}}{f}}\right )\,\sqrt {\frac {\sqrt {-B^4\,d^2\,f^4}+B^2\,c\,f^2}{4\,f^4}}-\mathrm {atanh}\left (\frac {f^3\,\sqrt {-\frac {\sqrt {-A^4\,d^2\,f^4}+A^2\,c\,f^2}{f^4}}\,\left (\frac {16\,\left (A^2\,d^4-A^2\,c^2\,d^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^2}+\frac {16\,c\,d^2\,\left (\sqrt {-A^4\,d^2\,f^4}+A^2\,c\,f^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^4}\right )}{16\,\left (A^3\,c^2\,d^3+A^3\,d^5\right )}\right )\,\sqrt {-\frac {\sqrt {-A^4\,d^2\,f^4}+A^2\,c\,f^2}{f^4}}-\mathrm {atanh}\left (\frac {f^3\,\sqrt {\frac {\sqrt {-A^4\,d^2\,f^4}-A^2\,c\,f^2}{f^4}}\,\left (\frac {16\,\left (A^2\,d^4-A^2\,c^2\,d^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^2}-\frac {16\,c\,d^2\,\left (\sqrt {-A^4\,d^2\,f^4}-A^2\,c\,f^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^4}\right )}{16\,\left (A^3\,c^2\,d^3+A^3\,d^5\right )}\right )\,\sqrt {\frac {\sqrt {-A^4\,d^2\,f^4}-A^2\,c\,f^2}{f^4}}+\mathrm {atanh}\left (\frac {f^3\,\sqrt {-\frac {\sqrt {-C^4\,d^2\,f^4}+C^2\,c\,f^2}{f^4}}\,\left (\frac {16\,\left (C^2\,d^4-C^2\,c^2\,d^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^2}+\frac {16\,c\,d^2\,\left (\sqrt {-C^4\,d^2\,f^4}+C^2\,c\,f^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^4}\right )}{16\,\left (C^3\,c^2\,d^3+C^3\,d^5\right )}\right )\,\sqrt {-\frac {\sqrt {-C^4\,d^2\,f^4}+C^2\,c\,f^2}{f^4}}+\mathrm {atanh}\left (\frac {f^3\,\sqrt {\frac {\sqrt {-C^4\,d^2\,f^4}-C^2\,c\,f^2}{f^4}}\,\left (\frac {16\,\left (C^2\,d^4-C^2\,c^2\,d^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^2}-\frac {16\,c\,d^2\,\left (\sqrt {-C^4\,d^2\,f^4}-C^2\,c\,f^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^4}\right )}{16\,\left (C^3\,c^2\,d^3+C^3\,d^5\right )}\right )\,\sqrt {\frac {\sqrt {-C^4\,d^2\,f^4}-C^2\,c\,f^2}{f^4}}+\frac {2\,B\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f}+\frac {2\,C\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,d\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*tan(e + f*x))^(1/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2),x)

[Out]

2*atanh((32*B^2*d^4*((B^2*c)/(4*f^2) - (-B^4*d^2*f^4)^(1/2)/(4*f^4))^(1/2)*(c + d*tan(e + f*x))^(1/2))/((16*B*
d^4*(-B^4*d^2*f^4)^(1/2))/f^3 + (16*B*c^2*d^2*(-B^4*d^2*f^4)^(1/2))/f^3) - (32*c*d^2*((B^2*c)/(4*f^2) - (-B^4*
d^2*f^4)^(1/2)/(4*f^4))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(-B^4*d^2*f^4)^(1/2))/((16*B*d^4*(-B^4*d^2*f^4)^(1/2)
)/f + (16*B*c^2*d^2*(-B^4*d^2*f^4)^(1/2))/f))*(-((-B^4*d^2*f^4)^(1/2) - B^2*c*f^2)/(4*f^4))^(1/2) - 2*atanh((3
2*B^2*d^4*((-B^4*d^2*f^4)^(1/2)/(4*f^4) + (B^2*c)/(4*f^2))^(1/2)*(c + d*tan(e + f*x))^(1/2))/((16*B*d^4*(-B^4*
d^2*f^4)^(1/2))/f^3 + (16*B*c^2*d^2*(-B^4*d^2*f^4)^(1/2))/f^3) + (32*c*d^2*((-B^4*d^2*f^4)^(1/2)/(4*f^4) + (B^
2*c)/(4*f^2))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(-B^4*d^2*f^4)^(1/2))/((16*B*d^4*(-B^4*d^2*f^4)^(1/2))/f + (16*
B*c^2*d^2*(-B^4*d^2*f^4)^(1/2))/f))*(((-B^4*d^2*f^4)^(1/2) + B^2*c*f^2)/(4*f^4))^(1/2) - atanh((f^3*(-((-A^4*d
^2*f^4)^(1/2) + A^2*c*f^2)/f^4)^(1/2)*((16*(A^2*d^4 - A^2*c^2*d^2)*(c + d*tan(e + f*x))^(1/2))/f^2 + (16*c*d^2
*((-A^4*d^2*f^4)^(1/2) + A^2*c*f^2)*(c + d*tan(e + f*x))^(1/2))/f^4))/(16*(A^3*d^5 + A^3*c^2*d^3)))*(-((-A^4*d
^2*f^4)^(1/2) + A^2*c*f^2)/f^4)^(1/2) - atanh((f^3*(((-A^4*d^2*f^4)^(1/2) - A^2*c*f^2)/f^4)^(1/2)*((16*(A^2*d^
4 - A^2*c^2*d^2)*(c + d*tan(e + f*x))^(1/2))/f^2 - (16*c*d^2*((-A^4*d^2*f^4)^(1/2) - A^2*c*f^2)*(c + d*tan(e +
f*x))^(1/2))/f^4))/(16*(A^3*d^5 + A^3*c^2*d^3)))*(((-A^4*d^2*f^4)^(1/2) - A^2*c*f^2)/f^4)^(1/2) + atanh((f^3*
(-((-C^4*d^2*f^4)^(1/2) + C^2*c*f^2)/f^4)^(1/2)*((16*(C^2*d^4 - C^2*c^2*d^2)*(c + d*tan(e + f*x))^(1/2))/f^2 +
(16*c*d^2*((-C^4*d^2*f^4)^(1/2) + C^2*c*f^2)*(c + d*tan(e + f*x))^(1/2))/f^4))/(16*(C^3*d^5 + C^3*c^2*d^3)))*
(-((-C^4*d^2*f^4)^(1/2) + C^2*c*f^2)/f^4)^(1/2) + atanh((f^3*(((-C^4*d^2*f^4)^(1/2) - C^2*c*f^2)/f^4)^(1/2)*((
16*(C^2*d^4 - C^2*c^2*d^2)*(c + d*tan(e + f*x))^(1/2))/f^2 - (16*c*d^2*((-C^4*d^2*f^4)^(1/2) - C^2*c*f^2)*(c +
d*tan(e + f*x))^(1/2))/f^4))/(16*(C^3*d^5 + C^3*c^2*d^3)))*(((-C^4*d^2*f^4)^(1/2) - C^2*c*f^2)/f^4)^(1/2) + (
2*B*(c + d*tan(e + f*x))^(1/2))/f + (2*C*(c + d*tan(e + f*x))^(3/2))/(3*d*f)

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