3.13.71 \(\int (a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx\) [1271]

Optimal. Leaf size=330 \[ -\frac {i (a-i b)^{3/2} (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {i (a+i b)^{3/2} (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {\left (18 a b c d+3 a^2 d^2+b^2 \left (3 c^2-8 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 \sqrt {b} \sqrt {d} f}+\frac {(3 b c+5 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {b \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f} \]

[Out]

-I*(a-I*b)^(3/2)*(c-I*d)^(3/2)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2)/(c+d*tan(f*x+e))^(1/
2))/f+I*(a+I*b)^(3/2)*(c+I*d)^(3/2)*arctanh((c+I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e)
)^(1/2))/f+1/4*(18*a*b*c*d+3*a^2*d^2+b^2*(3*c^2-8*d^2))*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*ta
n(f*x+e))^(1/2))/f/b^(1/2)/d^(1/2)+1/4*(5*a*d+3*b*c)*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/f+1/2*b*(a+
b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2)/f

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Rubi [A]
time = 3.19, antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3651, 3728, 3736, 6857, 65, 223, 212, 95, 214} \begin {gather*} \frac {\left (3 a^2 d^2+18 a b c d+b^2 \left (3 c^2-8 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 \sqrt {b} \sqrt {d} f}+\frac {b \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {(5 a d+3 b c) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}-\frac {i (a-i b)^{3/2} (c-i d)^{3/2} \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} (c+i d)^{3/2} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(a - I*b)^(3/2)*(c - I*d)^(3/2)*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)*(c + I*d)^(3/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(S
qrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/f + ((18*a*b*c*d + 3*a^2*d^2 + b^2*(3*c^2 - 8*d^2))*ArcTanh[(Sqrt[d]*
Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(4*Sqrt[b]*Sqrt[d]*f) + ((3*b*c + 5*a*d)*Sqrt[a
+ b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*f) + (b*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(
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 95

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 212

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

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 223

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 3651

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 3728

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((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 - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, 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[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3736

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 6857

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]

Rubi steps

\begin {align*} \int (a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx &=\frac {b \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {1}{2} \int \frac {\sqrt {c+d \tan (e+f x)} \left (\frac {1}{2} \left (4 a^2 c-b^2 c-3 a b d\right )+2 \left (2 a b c+a^2 d-b^2 d\right ) \tan (e+f x)+\frac {1}{2} b (3 b c+5 a d) \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx\\ &=\frac {(3 b c+5 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {b \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {\int \frac {-\frac {1}{4} b \left (5 b^2 c^2+14 a b c d-a^2 \left (8 c^2-5 d^2\right )\right )+4 b (b c+a d) (a c-b d) \tan (e+f x)+\frac {1}{4} b \left (18 a b c d+3 a^2 d^2+b^2 \left (3 c^2-8 d^2\right )\right ) \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 b}\\ &=\frac {(3 b c+5 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {b \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {\text {Subst}\left (\int \frac {-\frac {1}{4} b \left (5 b^2 c^2+14 a b c d-a^2 \left (8 c^2-5 d^2\right )\right )+4 b (b c+a d) (a c-b d) x+\frac {1}{4} b \left (18 a b c d+3 a^2 d^2+b^2 \left (3 c^2-8 d^2\right )\right ) x^2}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 b f}\\ &=\frac {(3 b c+5 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {b \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {\text {Subst}\left (\int \left (\frac {b \left (18 a b c d+3 a^2 d^2+b^2 \left (3 c^2-8 d^2\right )\right )}{4 \sqrt {a+b x} \sqrt {c+d x}}+\frac {2 (b (a c-b c-a d-b d) (a c+b c+a d-b d)+2 b (b c+a d) (a c-b d) x)}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{2 b f}\\ &=\frac {(3 b c+5 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {b \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {\text {Subst}\left (\int \frac {b (a c-b c-a d-b d) (a c+b c+a d-b d)+2 b (b c+a d) (a c-b d) x}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{b f}+\frac {\left (18 a b c d+3 a^2 d^2+b^2 \left (3 c^2-8 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {(3 b c+5 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {b \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {\text {Subst}\left (\int \left (\frac {-2 b (b c+a d) (a c-b d)+i b (a c-b c-a d-b d) (a c+b c+a d-b d)}{2 (i-x) \sqrt {a+b x} \sqrt {c+d x}}+\frac {2 b (b c+a d) (a c-b d)+i b (a c-b c-a d-b d) (a c+b c+a d-b d)}{2 (i+x) \sqrt {a+b x} \sqrt {c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{b f}+\frac {\left (18 a b c d+3 a^2 d^2+b^2 \left (3 c^2-8 d^2\right )\right ) \text {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 )}{4 b f}\\ &=\frac {(3 b c+5 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {b \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {\left (i (a-i b)^2 (c-i d)^2\right ) \text {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 (i (a+i b)^2 (c+i d)^2\right ) \text {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 (18 a b c d+3 a^2 d^2+b^2 \left (3 c^2-8 d^2\right )\right ) \text {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 )}{4 b f}\\ &=\frac {\left (18 a b c d+3 a^2 d^2+b^2 \left (3 c^2-8 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 \sqrt {b} \sqrt {d} f}+\frac {(3 b c+5 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {b \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac {\left (i (a-i b)^2 (c-i d)^2\right ) \text {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 (i (a+i b)^2 (c+i d)^2\right ) \text {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} (c-i d)^{3/2} \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} (c+i d)^{3/2} \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 {\left (18 a b c d+3 a^2 d^2+b^2 \left (3 c^2-8 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 \sqrt {b} \sqrt {d} f}+\frac {(3 b c+5 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {b \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2566\) vs. \(2(330)=660\).
time = 6.14, size = 2566, normalized size = 7.78 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*I)*(-a - I*b)*(-((-a - I*b)*(-((-c - I*d)*((-2*(-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]])])/(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*d*Sqrt[a + b*Tan[e + f*x]]*Sqr
t[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)))))))/(b*
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)]))) - (2*(b*c -
a*d)*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))))^(5/2)*((3*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)])])/(8*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))))^(5/2)) + (3/(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))))^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))))^(-1))/4))/(b*(b/((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))^(3/2)*Sqrt[(b*(c + d*Tan[e + f*x]
))/(b*c - a*d)])))/f - ((I/2)*(-a + I*b)*(-((-a + I*b)*(-((-c + I*d)*((-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]])])/Sqrt[-a + I*b] + (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)]*Arc
Sinh[(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*d*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
)))))))/(b*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)]))) +
(2*(b*c - a*d)*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))))^(5/2)*((3*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)])])/(8*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))))^(5/2)) + (3/(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))))^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))))^(-1))/4))/(b*(b/((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))^(3/2)*Sqrt[(b*(c + d*T
an[e + f*x]))/(b*c - a*d)])))/f

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(e + f*x))**(3/2)*(c + d*tan(e + f*x))**(3/2), 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((a+b*tan(f*x+e))^(3/2)*(c+d*tan(f*x+e))^(3/2),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 [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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