3.1277 \(\int (a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=429 \[ \frac{\left (-a^2 d^2+14 a b c d+b^2 \left (11 c^2-8 d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b f}+\frac{\left (15 a^2 b c d^2-a^3 d^3+3 a b^2 d \left (15 c^2-8 d^2\right )+5 b^3 \left (c^3-8 c d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{8 b^{3/2} \sqrt{d} f}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}+\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 b f}-\frac{i (a-i b)^{3/2} (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{i (a+i b)^{3/2} (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f} \]

[Out]

((-I)*(a - I*b)^(3/2)*(c - I*d)^(5/2)*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)^(5/2)*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 + ((15*a^2*b*c*d^2 - a^3*d^3 + 3*a*b^2*d*(15*c^2 - 8*d^2) + 5*b^3*(
c^3 - 8*c*d^2))*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(8*b^(3/2)*Sqr
t[d]*f) + ((14*a*b*c*d - a^2*d^2 + b^2*(11*c^2 - 8*d^2))*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(8
*b*f) + (d*(13*b*c - a*d)*(a + b*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]])/(12*b*f) + (d^2*(a + b*Tan[e +
f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]])/(3*b*f)

________________________________________________________________________________________

Rubi [A]  time = 5.46731, antiderivative size = 429, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {3566, 3647, 3655, 6725, 63, 217, 206, 93, 208} \[ \frac{\left (-a^2 d^2+14 a b c d+b^2 \left (11 c^2-8 d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b f}+\frac{\left (15 a^2 b c d^2-a^3 d^3+3 a b^2 d \left (15 c^2-8 d^2\right )+5 b^3 \left (c^3-8 c d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{8 b^{3/2} \sqrt{d} f}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}+\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 b f}-\frac{i (a-i b)^{3/2} (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{i (a+i b)^{3/2} (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(5/2),x]

[Out]

((-I)*(a - I*b)^(3/2)*(c - I*d)^(5/2)*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)^(5/2)*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 + ((15*a^2*b*c*d^2 - a^3*d^3 + 3*a*b^2*d*(15*c^2 - 8*d^2) + 5*b^3*(
c^3 - 8*c*d^2))*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(8*b^(3/2)*Sqr
t[d]*f) + ((14*a*b*c*d - a^2*d^2 + b^2*(11*c^2 - 8*d^2))*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(8
*b*f) + (d*(13*b*c - a*d)*(a + b*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]])/(12*b*f) + (d^2*(a + b*Tan[e +
f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]])/(3*b*f)

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(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 - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3647

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 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} (c+d \tan (e+f x))^{5/2} \, dx &=\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}+\frac{\int \frac{(a+b \tan (e+f x))^{3/2} \left (\frac{1}{2} \left (6 b c^3-5 b c d^2-a d^3\right )+3 b d \left (3 c^2-d^2\right ) \tan (e+f x)+\frac{1}{2} d^2 (13 b c-a d) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 b}\\ &=\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 b f}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}+\frac{\int \frac{\sqrt{a+b \tan (e+f x)} \left (-\frac{3}{4} d \left (13 b^2 c^2 d+a^2 d^3-a b \left (8 c^3-10 c d^2\right )\right )+6 b d \left (b c^3+3 a c^2 d-3 b c d^2-a d^3\right ) \tan (e+f x)+\frac{3}{4} d^2 \left (14 a b c d-a^2 d^2+b^2 \left (11 c^2-8 d^2\right )\right ) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{6 b d}\\ &=\frac{\left (14 a b c d-a^2 d^2+b^2 \left (11 c^2-8 d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b f}+\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 b f}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}+\frac{\int \frac{-\frac{3}{8} d^2 \left (a^3 d^3+b^3 c \left (11 c^2-8 d^2\right )+a b^2 d \left (51 c^2-8 d^2\right )-a^2 b \left (16 c^3-33 c d^2\right )\right )+6 b d^2 \left (2 a b c \left (c^2-3 d^2\right )-b^2 d \left (3 c^2-d^2\right )+a^2 \left (3 c^2 d-d^3\right )\right ) \tan (e+f x)+\frac{3}{8} d^2 \left (15 a^2 b c d^2-a^3 d^3+3 a b^2 d \left (15 c^2-8 d^2\right )+5 b^3 \left (c^3-8 c d^2\right )\right ) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{6 b d^2}\\ &=\frac{\left (14 a b c d-a^2 d^2+b^2 \left (11 c^2-8 d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b f}+\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 b f}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{3}{8} d^2 \left (a^3 d^3+b^3 c \left (11 c^2-8 d^2\right )+a b^2 d \left (51 c^2-8 d^2\right )-a^2 b \left (16 c^3-33 c d^2\right )\right )+6 b d^2 \left (2 a b c \left (c^2-3 d^2\right )-b^2 d \left (3 c^2-d^2\right )+a^2 \left (3 c^2 d-d^3\right )\right ) x+\frac{3}{8} d^2 \left (15 a^2 b c d^2-a^3 d^3+3 a b^2 d \left (15 c^2-8 d^2\right )+5 b^3 \left (c^3-8 c 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 )}{6 b d^2 f}\\ &=\frac{\left (14 a b c d-a^2 d^2+b^2 \left (11 c^2-8 d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b f}+\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 b f}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}+\frac{\operatorname{Subst}\left (\int \left (\frac{3 d^2 \left (15 a^2 b c d^2-a^3 d^3+5 b^3 \left (c^3-8 c d^2\right )+a b^2 \left (45 c^2 d-24 d^3\right )\right )}{8 \sqrt{a+b x} \sqrt{c+d x}}+\frac{6 \left (-b d^2 \left (b^2 c \left (c^2-3 d^2\right )+2 a b d \left (3 c^2-d^2\right )-a^2 \left (c^3-3 c d^2\right )\right )+b d^2 \left (2 a b c \left (c^2-3 d^2\right )-b^2 d \left (3 c^2-d^2\right )+a^2 \left (3 c^2 d-d^3\right )\right ) x\right )}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{6 b d^2 f}\\ &=\frac{\left (14 a b c d-a^2 d^2+b^2 \left (11 c^2-8 d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b f}+\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 b f}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}+\frac{\operatorname{Subst}\left (\int \frac{-b d^2 \left (b^2 c \left (c^2-3 d^2\right )+2 a b d \left (3 c^2-d^2\right )-a^2 \left (c^3-3 c d^2\right )\right )+b d^2 \left (2 a b c \left (c^2-3 d^2\right )-b^2 d \left (3 c^2-d^2\right )+a^2 \left (3 c^2 d-d^3\right )\right ) x}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{b d^2 f}+\frac{\left (15 a^2 b c d^2-a^3 d^3+3 a b^2 d \left (15 c^2-8 d^2\right )+5 b^3 \left (c^3-8 c d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{16 b f}\\ &=\frac{\left (14 a b c d-a^2 d^2+b^2 \left (11 c^2-8 d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b f}+\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 b f}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}+\frac{\operatorname{Subst}\left (\int \left (\frac{-i b d^2 \left (b^2 c \left (c^2-3 d^2\right )+2 a b d \left (3 c^2-d^2\right )-a^2 \left (c^3-3 c d^2\right )\right )-b d^2 \left (2 a b c \left (c^2-3 d^2\right )-b^2 d \left (3 c^2-d^2\right )+a^2 \left (3 c^2 d-d^3\right )\right )}{2 (i-x) \sqrt{a+b x} \sqrt{c+d x}}+\frac{-i b d^2 \left (b^2 c \left (c^2-3 d^2\right )+2 a b d \left (3 c^2-d^2\right )-a^2 \left (c^3-3 c d^2\right )\right )+b d^2 \left (2 a b c \left (c^2-3 d^2\right )-b^2 d \left (3 c^2-d^2\right )+a^2 \left (3 c^2 d-d^3\right )\right )}{2 (i+x) \sqrt{a+b x} \sqrt{c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{b d^2 f}+\frac{\left (15 a^2 b c d^2-a^3 d^3+3 a b^2 d \left (15 c^2-8 d^2\right )+5 b^3 \left (c^3-8 c d^2\right )\right ) \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 )}{8 b^2 f}\\ &=\frac{\left (14 a b c d-a^2 d^2+b^2 \left (11 c^2-8 d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b f}+\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 b f}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}-\frac{\left ((a+i b)^2 (i c-d)^3\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)^3\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 (15 a^2 b c d^2-a^3 d^3+3 a b^2 d \left (15 c^2-8 d^2\right )+5 b^3 \left (c^3-8 c d^2\right )\right ) \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 )}{8 b^2 f}\\ &=\frac{\left (15 a^2 b c d^2-a^3 d^3+3 a b^2 d \left (15 c^2-8 d^2\right )+5 b^3 \left (c^3-8 c 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 )}{8 b^{3/2} \sqrt{d} f}+\frac{\left (14 a b c d-a^2 d^2+b^2 \left (11 c^2-8 d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b f}+\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 b f}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}-\frac{\left ((a+i b)^2 (i c-d)^3\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)^3\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} (c-i d)^{5/2} \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)^{5/2} \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 (15 a^2 b c d^2-a^3 d^3+3 a b^2 d \left (15 c^2-8 d^2\right )+5 b^3 \left (c^3-8 c 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 )}{8 b^{3/2} \sqrt{d} f}+\frac{\left (14 a b c d-a^2 d^2+b^2 \left (11 c^2-8 d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 b f}+\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{12 b f}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f}\\ \end{align*}

Mathematica [A]  time = 8.10674, size = 773, normalized size = 1.8 \[ \frac{\frac{\frac{3 d \left (-a^2 d^2+14 a b c d+b^2 \left (11 c^2-8 d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{-\frac{6 b d^2 \left (\sqrt{-b^2} \left (a^2 \left (-\left (c^3-3 c d^2\right )\right )+a b \left (6 c^2 d-2 d^3\right )+b^2 c \left (c^2-3 d^2\right )\right )-b \left (a^2 \left (3 c^2 d-d^3\right )+2 a b c \left (c^2-3 d^2\right )-b^2 d \left (3 c^2-d^2\right )\right )\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{b d}{\sqrt{-b^2}}+c} \sqrt{a+b \tan (e+f x)}}{\sqrt{\sqrt{-b^2}-a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{\sqrt{-b^2}-a} \sqrt{\frac{b d}{\sqrt{-b^2}}+c}}-\frac{6 b d^2 \left (\sqrt{-b^2} \left (a^2 \left (-\left (c^3-3 c d^2\right )\right )+a b \left (6 c^2 d-2 d^3\right )+b^2 c \left (c^2-3 d^2\right )\right )+b \left (a^2 \left (3 c^2 d-d^3\right )+2 a b c \left (c^2-3 d^2\right )-b^2 d \left (3 c^2-d^2\right )\right )\right ) \tan ^{-1}\left (\frac{\sqrt{-\frac{\sqrt{-b^2} d+b c}{b}} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+\sqrt{-b^2}} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+\sqrt{-b^2}} \sqrt{-\frac{\sqrt{-b^2} d+b c}{b}}}+\frac{3 \sqrt{b} d^{3/2} \sqrt{c-\frac{a d}{b}} \left (15 a^2 b c d^2-a^3 d^3+3 a b^2 d \left (15 c^2-8 d^2\right )+5 b^3 \left (c^3-8 c d^2\right )\right ) \sqrt{\frac{b c+b d \tan (e+f x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c-\frac{a d}{b}}}\right )}{4 \sqrt{c+d \tan (e+f x)}}}{b d f}}{2 d}+\frac{d (13 b c-a d) (a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}{4 f}}{3 b}+\frac{d^2 (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}{3 b f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(5/2),x]

[Out]

(d^2*(a + b*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]])/(3*b*f) + ((d*(13*b*c - a*d)*(a + b*Tan[e + f*x])^(3
/2)*Sqrt[c + d*Tan[e + f*x]])/(4*f) + ((3*d*(14*a*b*c*d - a^2*d^2 + b^2*(11*c^2 - 8*d^2))*Sqrt[a + b*Tan[e + f
*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*f) + ((-6*b*d^2*(Sqrt[-b^2]*(b^2*c*(c^2 - 3*d^2) - a^2*(c^3 - 3*c*d^2) + a*b
*(6*c^2*d - 2*d^3)) - b*(2*a*b*c*(c^2 - 3*d^2) - b^2*d*(3*c^2 - d^2) + a^2*(3*c^2*d - d^3)))*ArcTan[(Sqrt[c +
(b*d)/Sqrt[-b^2]]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[-a + Sqrt
[-b^2]]*Sqrt[c + (b*d)/Sqrt[-b^2]]) - (6*b*d^2*(Sqrt[-b^2]*(b^2*c*(c^2 - 3*d^2) - a^2*(c^3 - 3*c*d^2) + a*b*(6
*c^2*d - 2*d^3)) + b*(2*a*b*c*(c^2 - 3*d^2) - b^2*d*(3*c^2 - d^2) + a^2*(3*c^2*d - d^3)))*ArcTan[(Sqrt[-((b*c
+ Sqrt[-b^2]*d)/b)]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a + Sqrt
[-b^2]]*Sqrt[-((b*c + Sqrt[-b^2]*d)/b)]) + (3*Sqrt[b]*d^(3/2)*Sqrt[c - (a*d)/b]*(15*a^2*b*c*d^2 - a^3*d^3 + 3*
a*b^2*d*(15*c^2 - 8*d^2) + 5*b^3*(c^3 - 8*c*d^2))*ArcSinh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c -
(a*d)/b])]*Sqrt[(b*c + b*d*Tan[e + f*x])/(b*c - a*d)])/(4*Sqrt[c + d*Tan[e + f*x]]))/(b*d*f))/(2*d))/(3*b)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

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))^(5/2),x)

[Out]

int((a+b*tan(f*x+e))^(3/2)*(c+d*tan(f*x+e))^(5/2),x)

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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}}{\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((a+b*tan(f*x+e))^(3/2)*(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^(3/2)*(d*tan(f*x + e) + c)^(5/2), x)

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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((a+b*tan(f*x+e))^(3/2)*(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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((a+b*tan(f*x+e))**(3/2)*(c+d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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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((a+b*tan(f*x+e))^(3/2)*(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out