3.1289 \(\int \frac{(a+b \tan (e+f x))^{7/2}}{(c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=356 \[ -\frac{b \left (2 a d (2 b c-a d)-b^2 \left (3 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d^2 f \left (c^2+d^2\right )}-\frac{b^{5/2} (3 b c-7 a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{d^{5/2} f}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^{3/2}}{d f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{i (a-i b)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c-i d)^{3/2}}+\frac{i (a+i b)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c+i d)^{3/2}} \]

[Out]

((-I)*(a - I*b)^(7/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]]
)])/((c - I*d)^(3/2)*f) + (I*(a + I*b)^(7/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*S
qrt[c + d*Tan[e + f*x]])])/((c + I*d)^(3/2)*f) - (b^(5/2)*(3*b*c - 7*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e +
f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(d^(5/2)*f) - (2*(b*c - a*d)^2*(a + b*Tan[e + f*x])^(3/2))/(d*(c^2
+ d^2)*f*Sqrt[c + d*Tan[e + f*x]]) - (b*(2*a*d*(2*b*c - a*d) - b^2*(3*c^2 + d^2))*Sqrt[a + b*Tan[e + f*x]]*Sq
rt[c + d*Tan[e + f*x]])/(d^2*(c^2 + d^2)*f)

________________________________________________________________________________________

Rubi [A]  time = 4.02561, antiderivative size = 356, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {3565, 3647, 3655, 6725, 63, 217, 206, 93, 208} \[ -\frac{b \left (2 a d (2 b c-a d)-b^2 \left (3 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d^2 f \left (c^2+d^2\right )}-\frac{b^{5/2} (3 b c-7 a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{d^{5/2} f}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^{3/2}}{d f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{i (a-i b)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c-i d)^{3/2}}+\frac{i (a+i b)^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c+i d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^(7/2)/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

((-I)*(a - I*b)^(7/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]]
)])/((c - I*d)^(3/2)*f) + (I*(a + I*b)^(7/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*S
qrt[c + d*Tan[e + f*x]])])/((c + I*d)^(3/2)*f) - (b^(5/2)*(3*b*c - 7*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e +
f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(d^(5/2)*f) - (2*(b*c - a*d)^2*(a + b*Tan[e + f*x])^(3/2))/(d*(c^2
+ d^2)*f*Sqrt[c + d*Tan[e + f*x]]) - (b*(2*a*d*(2*b*c - a*d) - b^2*(3*c^2 + d^2))*Sqrt[a + b*Tan[e + f*x]]*Sq
rt[c + d*Tan[e + f*x]])/(d^2*(c^2 + d^2)*f)

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, 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] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

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 \frac{(a+b \tan (e+f x))^{7/2}}{(c+d \tan (e+f x))^{3/2}} \, dx &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^{3/2}}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 \int \frac{\sqrt{a+b \tan (e+f x)} \left (\frac{1}{2} \left (3 b^3 c^2+a^3 c d-7 a b^2 c d+5 a^2 b d^2\right )+\frac{1}{2} d \left (3 a^2 b c-b^3 c-a^3 d+3 a b^2 d\right ) \tan (e+f x)-\frac{1}{2} b \left (2 a d (2 b c-a d)-b^2 \left (3 c^2+d^2\right )\right ) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{d \left (c^2+d^2\right )}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^{3/2}}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{b \left (2 a d (2 b c-a d)-b^2 \left (3 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d^2 \left (c^2+d^2\right ) f}+\frac{2 \int \frac{\frac{1}{4} \left (2 a^4 c d^2-12 a^2 b^2 c d^2+8 a^3 b d^3+a b^3 d \left (7 c^2-d^2\right )-b^4 c \left (3 c^2+d^2\right )\right )+\frac{1}{2} d^2 \left (4 a^3 b c-4 a b^3 c-a^4 d+6 a^2 b^2 d-b^4 d\right ) \tan (e+f x)-\frac{1}{4} b^3 (3 b c-7 a d) \left (c^2+d^2\right ) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{d^2 \left (c^2+d^2\right )}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^{3/2}}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{b \left (2 a d (2 b c-a d)-b^2 \left (3 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d^2 \left (c^2+d^2\right ) f}+\frac{2 \operatorname{Subst}\left (\int \frac{\frac{1}{4} \left (2 a^4 c d^2-12 a^2 b^2 c d^2+8 a^3 b d^3+a b^3 d \left (7 c^2-d^2\right )-b^4 c \left (3 c^2+d^2\right )\right )+\frac{1}{2} d^2 \left (4 a^3 b c-4 a b^3 c-a^4 d+6 a^2 b^2 d-b^4 d\right ) x-\frac{1}{4} b^3 (3 b c-7 a d) \left (c^2+d^2\right ) x^2}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{d^2 \left (c^2+d^2\right ) f}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^{3/2}}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{b \left (2 a d (2 b c-a d)-b^2 \left (3 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d^2 \left (c^2+d^2\right ) f}+\frac{2 \operatorname{Subst}\left (\int \left (-\frac{b^3 (3 b c-7 a d) \left (c^2+d^2\right )}{4 \sqrt{a+b x} \sqrt{c+d x}}+\frac{d^2 \left (a^4 c-6 a^2 b^2 c+b^4 c+4 a^3 b d-4 a b^3 d\right )+d^2 \left (4 a^3 b c-4 a b^3 c-a^4 d+6 a^2 b^2 d-b^4 d\right ) x}{2 \sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{d^2 \left (c^2+d^2\right ) f}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^{3/2}}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{b \left (2 a d (2 b c-a d)-b^2 \left (3 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d^2 \left (c^2+d^2\right ) f}-\frac{\left (b^3 (3 b c-7 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 d^2 f}+\frac{\operatorname{Subst}\left (\int \frac{d^2 \left (a^4 c-6 a^2 b^2 c+b^4 c+4 a^3 b d-4 a b^3 d\right )+d^2 \left (4 a^3 b c-4 a b^3 c-a^4 d+6 a^2 b^2 d-b^4 d\right ) x}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{d^2 \left (c^2+d^2\right ) f}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^{3/2}}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{b \left (2 a d (2 b c-a d)-b^2 \left (3 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d^2 \left (c^2+d^2\right ) f}-\frac{\left (b^2 (3 b c-7 a d)\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 )}{d^2 f}+\frac{\operatorname{Subst}\left (\int \left (\frac{i d^2 \left (a^4 c-6 a^2 b^2 c+b^4 c+4 a^3 b d-4 a b^3 d\right )-d^2 \left (4 a^3 b c-4 a b^3 c-a^4 d+6 a^2 b^2 d-b^4 d\right )}{2 (i-x) \sqrt{a+b x} \sqrt{c+d x}}+\frac{i d^2 \left (a^4 c-6 a^2 b^2 c+b^4 c+4 a^3 b d-4 a b^3 d\right )+d^2 \left (4 a^3 b c-4 a b^3 c-a^4 d+6 a^2 b^2 d-b^4 d\right )}{2 (i+x) \sqrt{a+b x} \sqrt{c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{d^2 \left (c^2+d^2\right ) f}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^{3/2}}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{b \left (2 a d (2 b c-a d)-b^2 \left (3 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d^2 \left (c^2+d^2\right ) f}-\frac{(a+i b)^4 \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (i c-d) f}-\frac{(a-i b)^4 \operatorname{Subst}\left (\int \frac{1}{(i+x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (i c+d) f}-\frac{\left (b^2 (3 b c-7 a d)\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 )}{d^2 f}\\ &=-\frac{b^{5/2} (3 b c-7 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 )}{d^{5/2} f}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^{3/2}}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{b \left (2 a d (2 b c-a d)-b^2 \left (3 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d^2 \left (c^2+d^2\right ) f}-\frac{(a+i b)^4 \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 )}{(i c-d) f}-\frac{(a-i b)^4 \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 )}{(i c+d) f}\\ &=-\frac{i (a-i b)^{7/2} \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 )}{(c-i d)^{3/2} f}+\frac{i (a+i b)^{7/2} \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 )}{(c+i d)^{3/2} f}-\frac{b^{5/2} (3 b c-7 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 )}{d^{5/2} f}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^{3/2}}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{b \left (2 a d (2 b c-a d)-b^2 \left (3 c^2+d^2\right )\right ) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d^2 \left (c^2+d^2\right ) f}\\ \end{align*}

Mathematica [C]  time = 6.26268, size = 1849, normalized size = 5.19 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Tan[e + f*x])^(7/2)/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

((I/2)*(a + I*b)*((-2*b*(b/((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))^(3/2)*Hypergeometric2F1[3/2, 5/2, 7/2,
-((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))))]*(a + b*Tan[e + f*x])
^(5/2)*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(5*(b*c - a*d)*Sqrt[c + d*Tan[e + f*x]]) + (a + I*b)*((a +
I*b)*((-2*Sqrt[a + I*b]*ArcTan[(Sqrt[-c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x
]])])/(-c - I*d)^(3/2) + (2*Sqrt[a + b*Tan[e + f*x]])/((-c - I*d)*Sqrt[c + d*Tan[e + f*x]])) - (2*(b*c - a*d)*
(b/((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))^(3/2)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))^2*Sqrt[(b*(c
+ d*Tan[e + f*x]))/(b*c - a*d)]*(-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*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 - (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]*Sqrt[d]*ArcSi
nh[(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[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[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*d^2*Sqrt[a + b*Tan[e + f*
x]]*Sqrt[c + d*Tan[e + f*x]]*Sqrt[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)))]))))/f - ((I/2)*(-a + I*b)*((2*b*(b/((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))^(3/2)*Hypergeome
tric2F1[3/2, 5/2, 7/2, -((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))))
]*(a + b*Tan[e + f*x])^(5/2)*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(5*(b*c - a*d)*Sqrt[c + d*Tan[e + f*x
]]) - (-a + I*b)*(-((-a + I*b)*((-2*Sqrt[-a + I*b]*ArcTan[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a +
I*b]*Sqrt[c + d*Tan[e + f*x]])])/(c - I*d)^(3/2) + (2*Sqrt[a + b*Tan[e + f*x]])/((c - I*d)*Sqrt[c + d*Tan[e +
f*x]]))) + (2*(b*c - a*d)*(b/((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))^(3/2)*((b^2*c)/(b*c - a*d) - (a*b*d)
/(b*c - a*d))^2*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)]*(-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*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 - (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]*Sqrt[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[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[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
*d^2*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]*Sqrt[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)))]))))/f

________________________________________________________________________________________

Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^(7/2)/(c+d*tan(f*x+e))^(3/2),x)

[Out]

int((a+b*tan(f*x+e))^(7/2)/(c+d*tan(f*x+e))^(3/2),x)

________________________________________________________________________________________

Maxima [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))^(7/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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))^(7/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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))**(7/2)/(c+d*tan(f*x+e))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^(7/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError