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

[Out]

-I*(a-I*b)^(7/2)*arctanh((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))/(c-I*d)^(3
/2)/f+I*(a+I*b)^(7/2)*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))/(c+I*
d)^(3/2)/f-b^(5/2)*(-7*a*d+3*b*c)*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan(f*x+e))^(1/2))/d^(5/
2)/f-b*(2*a*d*(-a*d+2*b*c)-b^2*(3*c^2+d^2))*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/d^2/(c^2+d^2)/f-2*(-
a*d+b*c)^2*(a+b*tan(f*x+e))^(3/2)/d/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)

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Rubi [A]  time = 4.03, antiderivative size = 356, 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 = {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 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 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 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 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]

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 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]

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*}

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Mathematica [C]  time = 6.36, size = 1877, normalized size = 5.27 \[ \text {result too large to display} \]

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]

((-1/2*I)*(-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]*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)*Sqrt[c + I*d]) + (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)))))) - (Sq
rt[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*Sq
rt[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)*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)*Sqr
t[c + d*Tan[e + f*x]]) - (-a + I*b)*(-((-a + I*b)*((-2*Sqrt[-a + I*b]*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)*Sqrt[-c + I*d]) + (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*T
an[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*Ta
n[e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))]))))/f

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^(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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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