3.1266 \(\int (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=258 \[ \frac{b \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}-\frac{i (a-i b)^{3/2} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{i (a+i b)^{3/2} \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{\sqrt{b} (3 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{d} f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 2.25524, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {3570, 3655, 6725, 63, 217, 206, 93, 208} \[ \frac{b \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}-\frac{i (a-i b)^{3/2} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{i (a+i b)^{3/2} \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{\sqrt{b} (3 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{d} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3570

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[1/(m + n - 1), Int[(a +
b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a^2*c*(m + n - 1) - b*(b*c*(m - 1) + a*d*n) + (2*a*b
*c + a^2*d - b^2*d)*(m + n - 1)*Tan[e + f*x] + b*(b*c*n + a*d*(2*m + n - 2))*Tan[e + f*x]^2, x], x], x] /; Fre
eQ[{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] && GtQ[m, 1] && GtQ[n
, 0] && IntegerQ[2*n]

Rule 3655

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] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[((a + b*ff*x)^m*(c + d*ff*x)^n*(A + B*ff*x + C*ff^2*x^2))/(1 + ff^2*x^2), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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 (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)} \, dx &=\frac{b \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}+\int \frac{\frac{1}{2} \left (2 a^2 c-b (b c+a d)\right )+\left (2 a b c+a^2 d-b^2 d\right ) \tan (e+f x)+\frac{1}{2} b (b c+3 a d) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{b \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} \left (2 a^2 c-b (b c+a d)\right )+\left (2 a b c+a^2 d-b^2 d\right ) x+\frac{1}{2} b (b c+3 a d) x^2}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{b \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}+\frac{\operatorname{Subst}\left (\int \left (\frac{b (b c+3 a d)}{2 \sqrt{a+b x} \sqrt{c+d x}}+\frac{a^2 c-b^2 c-2 a b d+\left (2 a b c+a^2 d-b^2 d\right ) x}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{b \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}+\frac{\operatorname{Subst}\left (\int \frac{a^2 c-b^2 c-2 a b d+\left (2 a b c+a^2 d-b^2 d\right ) x}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}+\frac{(b (b c+3 a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{b \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}+\frac{\operatorname{Subst}\left (\int \left (\frac{-2 a b c-a^2 d+b^2 d+i \left (a^2 c-b^2 c-2 a b d\right )}{2 (i-x) \sqrt{a+b x} \sqrt{c+d x}}+\frac{2 a b c+a^2 d-b^2 d+i \left (a^2 c-b^2 c-2 a b d\right )}{2 (i+x) \sqrt{a+b x} \sqrt{c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{f}+\frac{(b c+3 a d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b \tan (e+f x)}\right )}{f}\\ &=\frac{b \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}+\frac{\left ((a+i b)^2 (i c-d)\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{\left ((a-i b)^2 (i c+d)\right ) \operatorname{Subst}\left (\int \frac{1}{(i+x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{(b c+3 a d) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{f}\\ &=\frac{\sqrt{b} (b c+3 a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{d} f}+\frac{b \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}+\frac{\left ((a+i b)^2 (i c-d)\right ) \operatorname{Subst}\left (\int \frac{1}{a+i b-(c+i d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{\left ((a-i b)^2 (i c+d)\right ) \operatorname{Subst}\left (\int \frac{1}{-a+i b-(-c+i d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{f}\\ &=-\frac{i (a-i b)^{3/2} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{i (a+i b)^{3/2} \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{\sqrt{b} (b c+3 a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{d} f}+\frac{b \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}\\ \end{align*}

Mathematica [B]  time = 6.10236, size = 1520, normalized size = 5.89 \[ \frac{i (a+i b) \left ((a+i b) \left (\frac{2 (c+i d) \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+i b} \sqrt{-c-i d}}-\frac{2 \sqrt{d} \sqrt{b c-a d} \sqrt{\frac{b}{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b c-a d} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}}\right ) \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}}{b^{3/2} \sqrt{c+d \tan (e+f x)}}\right )-\frac{2 \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}+1\right )^{3/2} \left (\frac{\sqrt{b c-a d} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b c-a d} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}}\right )}{2 \sqrt{b} \sqrt{d} \sqrt{a+b \tan (e+f x)} \left (\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}+1\right )^{3/2}}+\frac{1}{2 \left (\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}+1\right )}\right )}{\sqrt{\frac{b}{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}} \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}}\right )}{2 f}-\frac{i (i b-a) \left (\frac{2 \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}+1\right )^{3/2} \left (\frac{\sqrt{b c-a d} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b c-a d} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}}\right )}{2 \sqrt{b} \sqrt{d} \sqrt{a+b \tan (e+f x)} \left (\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}+1\right )^{3/2}}+\frac{1}{2 \left (\frac{b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}\right )}+1\right )}\right )}{\sqrt{\frac{b}{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}} \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}}-(i b-a) \left (\frac{2 \sqrt{d} \sqrt{b c-a d} \sqrt{\frac{b}{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b c-a d} \sqrt{\frac{b^2 c}{b c-a d}-\frac{a b d}{b c-a d}}}\right ) \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}}{b^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{2 (i d-c) \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{i b-a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{i b-a} \sqrt{c-i d}}\right )\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(a + I*b)*((a + I*b)*((2*(c + I*d)*ArcTan[(Sqrt[-c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt
[c + d*Tan[e + f*x]])])/(Sqrt[a + I*b]*Sqrt[-c - I*d]) - (2*Sqrt[d]*Sqrt[b*c - a*d]*Sqrt[b/((b^2*c)/(b*c - a*d
) - (a*b*d)/(b*c - a*d))]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)]*ArcSinh[(Sqrt[b]*Sqrt[d]*Sqrt[a + b*
Tan[e + f*x]])/(Sqrt[b*c - a*d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)])]*Sqrt[(b*(c + d*Tan[e + f*x])
)/(b*c - a*d)])/(b^(3/2)*Sqrt[c + d*Tan[e + f*x]])) - (2*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]*(1
+ (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))))^(3/2)*((Sqrt[b*c - a*d
]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)]*ArcSinh[(Sqrt[b]*Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b*c
- a*d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)])])/(2*Sqrt[b]*Sqrt[d]*Sqrt[a + b*Tan[e + f*x]]*(1 + (b
*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))))^(3/2)) + 1/(2*(1 + (b*d*(a
+ b*Tan[e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))))))/(Sqrt[b/((b^2*c)/(b*c - a*d)
- (a*b*d)/(b*c - a*d))]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])))/f - ((I/2)*(-a + I*b)*(-((-a + I*b)*((-
2*(-c + I*d)*ArcTan[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt
[-a + I*b]*Sqrt[c - I*d]) + (2*Sqrt[d]*Sqrt[b*c - a*d]*Sqrt[b/((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))]*Sqr
t[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)]*ArcSinh[(Sqrt[b]*Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b*c - a*
d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)])]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(b^(3/2)*Sqrt
[c + d*Tan[e + f*x]]))) + (2*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]*(1 + (b*d*(a + b*Tan[e + f*x]))
/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))))^(3/2)*((Sqrt[b*c - a*d]*Sqrt[(b^2*c)/(b*c - a*d) -
(a*b*d)/(b*c - a*d)]*ArcSinh[(Sqrt[b]*Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b*c - a*d]*Sqrt[(b^2*c)/(b*c -
a*d) - (a*b*d)/(b*c - a*d)])])/(2*Sqrt[b]*Sqrt[d]*Sqrt[a + b*Tan[e + f*x]]*(1 + (b*d*(a + b*Tan[e + f*x]))/((b
*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))))^(3/2)) + 1/(2*(1 + (b*d*(a + b*Tan[e + f*x]))/((b*c -
a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))))))/(Sqrt[b/((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))]*Sqr
t[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])))/f

________________________________________________________________________________________

Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c+d\tan \left ( fx+e \right ) } \left ( a+b\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

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))^(3/2),x)

[Out]

int((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^(3/2),x)

________________________________________________________________________________________

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

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))^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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))^(1/2)*(a+b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

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))**(3/2),x)

[Out]

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

________________________________________________________________________________________

Giac [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))^(1/2)*(a+b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out