3.13.38 \(\int \frac {(c+d \tan (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx\) [1238]

Optimal. Leaf size=170 \[ \frac {(c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(i a+b) f}-\frac {(c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b) f}-\frac {2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \left (a^2+b^2\right ) f} \]

[Out]

(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(I*a+b)/f-(c+I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(
1/2)/(c+I*d)^(1/2))/(I*a-b)/f-2*(-a*d+b*c)^(3/2)*arctanh(b^(1/2)*(c+d*tan(f*x+e))^(1/2)/(-a*d+b*c)^(1/2))/(a^2
+b^2)/f/b^(1/2)

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Rubi [A]
time = 0.34, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3654, 3620, 3618, 65, 214, 3715} \begin {gather*} -\frac {2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {b} f \left (a^2+b^2\right )}+\frac {(c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (b+i a)}-\frac {(c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (-b+i a)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((I*a + b)*f) - ((c + I*d)^(3/2)*ArcTanh[Sqr
t[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((I*a - b)*f) - (2*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e +
f*x]])/Sqrt[b*c - a*d]])/(Sqrt[b]*(a^2 + b^2)*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 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 3654

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1
/(c^2 + d^2), Int[Simp[a^2*c - b^2*c + 2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e +
f*x]], x], x] + Dist[(b*c - a*d)^2/(c^2 + d^2), Int[(1 + Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*(c + 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] && NeQ[c^2 + d^2
, 0]

Rule 3715

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

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Mathematica [A]
time = 0.41, size = 168, normalized size = 0.99 \begin {gather*} \frac {\sqrt {b} (-i a+b) (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {b} (i a+b) (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )-2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \left (a^2+b^2\right ) f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(956\) vs. \(2(142)=284\).
time = 0.56, size = 957, normalized size = 5.63

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (\frac {\frac {\frac {\left (-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,c^{2}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,d^{2}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b c d \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 \sqrt {c^{2}+d^{2}}\, a \,d^{2}+2 \sqrt {c^{2}+d^{2}}\, b c d -\frac {\left (-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,c^{2}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,d^{2}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b c d \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 (-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,c^{2}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,d^{2}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b c d \right ) \ln \left (\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d \tan \left (f x +e \right )-c -\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 \sqrt {c^{2}+d^{2}}\, a \,d^{2}-2 \sqrt {c^{2}+d^{2}}\, b c d +\frac {\left (-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,c^{2}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,d^{2}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b c d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 d}}{d^{2} \left (a^{2}+b^{2}\right )}+\frac {\left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {\left (a d -b c \right ) b}}\right )}{d^{2} \left (a^{2}+b^{2}\right ) \sqrt {\left (a d -b c \right ) b}}\right )}{f}\)	\(957\)
default	\(\frac {2 d^{2} \left (\frac {\frac {\frac {\left (-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,c^{2}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,d^{2}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b c d \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 \sqrt {c^{2}+d^{2}}\, a \,d^{2}+2 \sqrt {c^{2}+d^{2}}\, b c d -\frac {\left (-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,c^{2}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,d^{2}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b c d \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 (-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,c^{2}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,d^{2}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b c d \right ) \ln \left (\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d \tan \left (f x +e \right )-c -\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 \sqrt {c^{2}+d^{2}}\, a \,d^{2}-2 \sqrt {c^{2}+d^{2}}\, b c d +\frac {\left (-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,c^{2}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a \,d^{2}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b c d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{4 d}}{d^{2} \left (a^{2}+b^{2}\right )}+\frac {\left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {\left (a d -b c \right ) b}}\right )}{d^{2} \left (a^{2}+b^{2}\right ) \sqrt {\left (a d -b c \right ) b}}\right )}{f}\)	\(957\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*d^2*(1/d^2/(a^2+b^2)*(1/4/d*(1/2*(-(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c-(c^2+d^2)^(1/2)*(2*(c
^2+d^2)^(1/2)+2*c)^(1/2)*b*d+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c^2-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*d^2+2*(2*(c^2
+d^2)^(1/2)+2*c)^(1/2)*b*c*d)*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*(c^2+d^2)^(1/2)*a*d^2+2*(c^2+d^2)^(1/2)*b*c*d-1/2*(-(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2
)*a*c-(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b*d+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c^2-(2*(c^2+d^2)^(1/2)
+2*c)^(1/2)*a*d^2+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b*c*d)*(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)))+1/4/d*
(-1/2*(-(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c-(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b*d+(2
*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c^2-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*d^2+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b*c*d)*l
n((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*(2*(c^2+d^2)^(1/2)*a*
d^2-2*(c^2+d^2)^(1/2)*b*c*d+1/2*(-(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c-(c^2+d^2)^(1/2)*(2*(c^2+d^
2)^(1/2)+2*c)^(1/2)*b*d+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c^2-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*d^2+2*(2*(c^2+d^2)
^(1/2)+2*c)^(1/2)*b*c*d)*(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^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-b*c)^2/d^2/(a^2+b^2)/((a*d-b*c)*
b)^(1/2)*arctan(b*(c+d*tan(f*x+e))^(1/2)/((a*d-b*c)*b)^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*tan(e + f*x))**(3/2)/(a + b*tan(e + f*x)), 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))^(3/2)/(a+b*tan(f*x+e)),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 = 13.01, size = 2500, normalized size = 14.71 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan(-((((((32*(a*b^5*d^15*f^2 + 4*a^5*b*d^15*f^2 - b^6*c*d^14*f^2 - 15*a^3*b^3*d^15*f^2 + 23*b^6*c^3*d^12*f^
2 + 21*b^6*c^5*d^10*f^2 - 3*b^6*c^7*d^8*f^2 - 29*a^2*b^4*c^3*d^12*f^2 - 81*a^2*b^4*c^5*d^10*f^2 + a^2*b^4*c^7*
d^8*f^2 + 59*a^3*b^3*c^2*d^13*f^2 + 75*a^3*b^3*c^4*d^11*f^2 + a^3*b^3*c^6*d^9*f^2 - 32*a^4*b^2*c^3*d^12*f^2 -
2*a^4*b^2*c^5*d^10*f^2 - 61*a*b^5*c^2*d^13*f^2 - 25*a*b^5*c^4*d^11*f^2 + 37*a*b^5*c^6*d^9*f^2 + 53*a^2*b^4*c*d
^14*f^2 - 30*a^4*b^2*c*d^14*f^2 + 4*a^5*b*c^2*d^13*f^2))/f^5 + ((((-b*(a*d - b*c)^3)^(1/2)*((32*(4*a^2*b^6*d^1
2*f^4 + 8*a^4*b^4*d^12*f^4 + 4*a^6*b^2*d^12*f^4 + 12*b^8*c^2*d^10*f^4 + 12*b^8*c^4*d^8*f^4 + 28*a^2*b^6*c^2*d^
10*f^4 + 24*a^2*b^6*c^4*d^8*f^4 - 32*a^3*b^5*c^3*d^9*f^4 + 20*a^4*b^4*c^2*d^10*f^4 + 12*a^4*b^4*c^4*d^8*f^4 -
16*a^5*b^3*c^3*d^9*f^4 + 4*a^6*b^2*c^2*d^10*f^4 - 16*a*b^7*c*d^11*f^4 - 16*a*b^7*c^3*d^9*f^4 - 32*a^3*b^5*c*d^
11*f^4 - 16*a^5*b^3*c*d^11*f^4))/f^5 - (32*(-b*(a*d - b*c)^3)^(1/2)*(c + d*tan(e + f*x))^(1/2)*(16*b^9*d^10*f^
4 + 16*a^2*b^7*d^10*f^4 - 16*a^4*b^5*d^10*f^4 - 16*a^6*b^3*d^10*f^4 + 24*b^9*c^2*d^8*f^4 + 40*a^2*b^7*c^2*d^8*
f^4 + 8*a^4*b^5*c^2*d^8*f^4 - 8*a^6*b^3*c^2*d^8*f^4 + 8*a*b^8*c*d^9*f^4 + 24*a^3*b^6*c*d^9*f^4 + 24*a^5*b^4*c*
d^9*f^4 + 8*a^7*b^2*c*d^9*f^4))/(f^4*(b^3*f + a^2*b*f))))/(b^3*f + a^2*b*f) - (32*(c + d*tan(e + f*x))^(1/2)*(
22*b^7*c*d^12*f^2 - 14*a*b^6*d^13*f^2 + 4*a^3*b^4*d^13*f^2 - 14*a^5*b^2*d^13*f^2 + 28*b^7*c^3*d^10*f^2 - 18*b^
7*c^5*d^8*f^2 + 24*a^2*b^5*c^3*d^10*f^2 + 12*a^2*b^5*c^5*d^8*f^2 - 88*a^3*b^4*c^2*d^11*f^2 - 28*a^3*b^4*c^4*d^
9*f^2 + 60*a^4*b^3*c^3*d^10*f^2 - 2*a^4*b^3*c^5*d^8*f^2 - 44*a^5*b^2*c^2*d^11*f^2 + 2*a^5*b^2*c^4*d^9*f^2 + 8*
a^6*b*c*d^12*f^2 + 20*a*b^6*c^2*d^11*f^2 + 66*a*b^6*c^4*d^9*f^2 - 28*a^2*b^5*c*d^12*f^2 + 54*a^4*b^3*c*d^12*f^
2))/f^4)*(-b*(a*d - b*c)^3)^(1/2))/(b^3*f + a^2*b*f))*(-b*(a*d - b*c)^3)^(1/2))/(b^3*f + a^2*b*f) - (32*(c + d
*tan(e + f*x))^(1/2)*(b^5*d^16 + 2*a^4*b*d^16 + 4*b^5*c^2*d^14 + 8*b^5*c^4*d^12 - 8*b^5*c^6*d^10 + 3*b^5*c^8*d
^8 - 8*a*b^4*c^3*d^13 + 48*a*b^4*c^5*d^11 - 8*a*b^4*c^7*d^9 - 8*a^3*b^2*c*d^15 - 12*a^4*b*c^2*d^14 + 2*a^4*b*c
^4*d^12 + 12*a^2*b^3*c^2*d^14 - 72*a^2*b^3*c^4*d^12 + 12*a^2*b^3*c^6*d^10 + 48*a^3*b^2*c^3*d^13 - 8*a^3*b^2*c^
5*d^11))/f^4)*(-b*(a*d - b*c)^3)^(1/2)*1i)/(b^3*f + a^2*b*f) - (((((32*(a*b^5*d^15*f^2 + 4*a^5*b*d^15*f^2 - b^
6*c*d^14*f^2 - 15*a^3*b^3*d^15*f^2 + 23*b^6*c^3*d^12*f^2 + 21*b^6*c^5*d^10*f^2 - 3*b^6*c^7*d^8*f^2 - 29*a^2*b^
4*c^3*d^12*f^2 - 81*a^2*b^4*c^5*d^10*f^2 + a^2*b^4*c^7*d^8*f^2 + 59*a^3*b^3*c^2*d^13*f^2 + 75*a^3*b^3*c^4*d^11
*f^2 + a^3*b^3*c^6*d^9*f^2 - 32*a^4*b^2*c^3*d^12*f^2 - 2*a^4*b^2*c^5*d^10*f^2 - 61*a*b^5*c^2*d^13*f^2 - 25*a*b
^5*c^4*d^11*f^2 + 37*a*b^5*c^6*d^9*f^2 + 53*a^2*b^4*c*d^14*f^2 - 30*a^4*b^2*c*d^14*f^2 + 4*a^5*b*c^2*d^13*f^2)
)/f^5 + ((((-b*(a*d - b*c)^3)^(1/2)*((32*(4*a^2*b^6*d^12*f^4 + 8*a^4*b^4*d^12*f^4 + 4*a^6*b^2*d^12*f^4 + 12*b^
8*c^2*d^10*f^4 + 12*b^8*c^4*d^8*f^4 + 28*a^2*b^6*c^2*d^10*f^4 + 24*a^2*b^6*c^4*d^8*f^4 - 32*a^3*b^5*c^3*d^9*f^
4 + 20*a^4*b^4*c^2*d^10*f^4 + 12*a^4*b^4*c^4*d^8*f^4 - 16*a^5*b^3*c^3*d^9*f^4 + 4*a^6*b^2*c^2*d^10*f^4 - 16*a*
b^7*c*d^11*f^4 - 16*a*b^7*c^3*d^9*f^4 - 32*a^3*b^5*c*d^11*f^4 - 16*a^5*b^3*c*d^11*f^4))/f^5 + (32*(-b*(a*d - b
*c)^3)^(1/2)*(c + d*tan(e + f*x))^(1/2)*(16*b^9*d^10*f^4 + 16*a^2*b^7*d^10*f^4 - 16*a^4*b^5*d^10*f^4 - 16*a^6*
b^3*d^10*f^4 + 24*b^9*c^2*d^8*f^4 + 40*a^2*b^7*c^2*d^8*f^4 + 8*a^4*b^5*c^2*d^8*f^4 - 8*a^6*b^3*c^2*d^8*f^4 + 8
*a*b^8*c*d^9*f^4 + 24*a^3*b^6*c*d^9*f^4 + 24*a^5*b^4*c*d^9*f^4 + 8*a^7*b^2*c*d^9*f^4))/(f^4*(b^3*f + a^2*b*f))
))/(b^3*f + a^2*b*f) + (32*(c + d*tan(e + f*x))^(1/2)*(22*b^7*c*d^12*f^2 - 14*a*b^6*d^13*f^2 + 4*a^3*b^4*d^13*
f^2 - 14*a^5*b^2*d^13*f^2 + 28*b^7*c^3*d^10*f^2 - 18*b^7*c^5*d^8*f^2 + 24*a^2*b^5*c^3*d^10*f^2 + 12*a^2*b^5*c^
5*d^8*f^2 - 88*a^3*b^4*c^2*d^11*f^2 - 28*a^3*b^4*c^4*d^9*f^2 + 60*a^4*b^3*c^3*d^10*f^2 - 2*a^4*b^3*c^5*d^8*f^2
- 44*a^5*b^2*c^2*d^11*f^2 + 2*a^5*b^2*c^4*d^9*f^2 + 8*a^6*b*c*d^12*f^2 + 20*a*b^6*c^2*d^11*f^2 + 66*a*b^6*c^4
*d^9*f^2 - 28*a^2*b^5*c*d^12*f^2 + 54*a^4*b^3*c*d^12*f^2))/f^4)*(-b*(a*d - b*c)^3)^(1/2))/(b^3*f + a^2*b*f))*(
-b*(a*d - b*c)^3)^(1/2))/(b^3*f + a^2*b*f) + (32*(c + d*tan(e + f*x))^(1/2)*(b^5*d^16 + 2*a^4*b*d^16 + 4*b^5*c
^2*d^14 + 8*b^5*c^4*d^12 - 8*b^5*c^6*d^10 + 3*b^5*c^8*d^8 - 8*a*b^4*c^3*d^13 + 48*a*b^4*c^5*d^11 - 8*a*b^4*c^7
*d^9 - 8*a^3*b^2*c*d^15 - 12*a^4*b*c^2*d^14 + 2*a^4*b*c^4*d^12 + 12*a^2*b^3*c^2*d^14 - 72*a^2*b^3*c^4*d^12 + 1
2*a^2*b^3*c^6*d^10 + 48*a^3*b^2*c^3*d^13 - 8*a^3*b^2*c^5*d^11))/f^4)*(-b*(a*d - b*c)^3)^(1/2)*1i)/(b^3*f + a^2
*b*f))/((64*(a^2*b^2*d^18 + b^4*c^2*d^16 + 5*b^4*c^4*d^14 + 7*b^4*c^6*d^12 + 3*b^4*c^8*d^10 - 12*a*b^3*c^3*d^1
5 - 18*a*b^3*c^5*d^13 - 8*a*b^3*c^7*d^11 - 4*a^3*b*c^3*d^15 - 2*a^3*b*c^5*d^13 + 9*a^2*b^2*c^2*d^16 + 15*a^2*b
^2*c^4*d^14 + 7*a^2*b^2*c^6*d^12 - 2*a*b^3*c*d^17 - 2*a^3*b*c*d^17))/f^5 + (((((32*(a*b^5*d^15*f^2 + 4*a^5*b*d
^15*f^2 - b^6*c*d^14*f^2 - 15*a^3*b^3*d^15*f^2 + 23*b^6*c^3*d^12*f^2 + 21*b^6*c^5*d^10*f^2 - 3*b^6*c^7*d^8*f^2
- 29*a^2*b^4*c^3*d^12*f^2 - 81*a^2*b^4*c^5*d^1...

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