\(\int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\) [463]

Optimal result

Integrand size = 35, antiderivative size = 282 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}+\frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]

[Out]

(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a-b)^(5/2)/d+2*B*arctanh(b^(1/2)*tan(
d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b^(5/2)/d-(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))
^(1/2))/(I*a+b)^(5/2)/d+2*a*(2*A*b^3-a*(a^2+3*b^2)*B)*tan(d*x+c)^(1/2)/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2
)+2/3*a*(A*b-B*a)*tan(d*x+c)^(3/2)/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 3.17 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3686, 3726, 3736, 6857, 65, 223, 212, 95, 211, 214} \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \sqrt {\tan (c+d x)}}{b^2 d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}+\frac {(-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d} \]

[In]

Int[(Tan[c + d*x]^(5/2)*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

((I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a - b)^(5/2)*d) + (2*B*Arc
Tanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(b^(5/2)*d) - ((I*A + B)*ArcTanh[(Sqrt[I*a + b]*S
qrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a + b)^(5/2)*d) + (2*a*(A*b - a*B)*Tan[c + d*x]^(3/2))/(3*b*
(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + (2*a*(2*A*b^3 - a*(a^2 + 3*b^2)*B)*Sqrt[Tan[c + d*x]])/(b^2*(a^2 +
b^2)^2*d*Sqrt[a + b*Tan[c + d*x]])

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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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 3686

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e
+ f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

Rule 3726

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] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Ta
n[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

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*} \text {integral}& = \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \int \frac {\sqrt {\tan (c+d x)} \left (-\frac {3}{2} a (A b-a B)+\frac {3}{2} b (A b-a B) \tan (c+d x)+\frac {3}{2} \left (a^2+b^2\right ) B \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^{3/2}} \, dx}{3 b \left (a^2+b^2\right )} \\ & = \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {4 \int \frac {-\frac {3}{4} a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )-\frac {3}{4} b^2 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)+\frac {3}{4} \left (a^2+b^2\right )^2 B \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 b^2 \left (a^2+b^2\right )^2} \\ & = \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {4 \text {Subst}\left (\int \frac {-\frac {3}{4} a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )-\frac {3}{4} b^2 \left (a^2 A-A b^2+2 a b B\right ) x+\frac {3}{4} \left (a^2+b^2\right )^2 B x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{3 b^2 \left (a^2+b^2\right )^2 d} \\ & = \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {4 \text {Subst}\left (\int \left (\frac {3 \left (a^2+b^2\right )^2 B}{4 \sqrt {x} \sqrt {a+b x}}-\frac {3 \left (b^2 \left (2 a A b-a^2 B+b^2 B\right )+b^2 \left (a^2 A-A b^2+2 a b B\right ) x\right )}{4 \sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{3 b^2 \left (a^2+b^2\right )^2 d} \\ & = \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {b^2 \left (2 a A b-a^2 B+b^2 B\right )+b^2 \left (a^2 A-A b^2+2 a b B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b^2 \left (a^2+b^2\right )^2 d}+\frac {B \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{b^2 d} \\ & = \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \left (\frac {-b^2 \left (a^2 A-A b^2+2 a b B\right )+i b^2 \left (2 a A b-a^2 B+b^2 B\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {b^2 \left (a^2 A-A b^2+2 a b B\right )+i b^2 \left (2 a A b-a^2 B+b^2 B\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b^2 \left (a^2+b^2\right )^2 d}+\frac {(2 B) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{b^2 d} \\ & = \frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(A-i B) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d}+\frac {(A+i B) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}+\frac {(2 B) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^2 d} \\ & = \frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d}+\frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(A-i B) \text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a-i b)^2 d}+\frac {(A+i B) \text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^2 d} \\ & = -\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} (a+i b)^2 d}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}+\frac {2 a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(596\) vs. \(2(282)=564\).

Time = 6.46 (sec) , antiderivative size = 596, normalized size of antiderivative = 2.11 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {(A-i B) \left (\frac {3 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{5/2}}+\frac {\tan ^{\frac {3}{2}}(c+d x)}{(a-i b) (a+b \tan (c+d x))^{3/2}}-\frac {3 i \sqrt {\tan (c+d x)}}{(a-i b)^2 \sqrt {a+b \tan (c+d x)}}\right )}{3 d}-\frac {(A+i B) \left (\frac {3 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{5/2}}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{(a+i b) (a+b \tan (c+d x))^{3/2}}-\frac {3 i \sqrt {\tan (c+d x)}}{(a+i b)^2 \sqrt {a+b \tan (c+d x)}}\right )}{3 d}+\frac {(i A-B) \sqrt {a+b \tan (c+d x)} \left (\frac {b^2 \tan ^2(c+d x)}{(a+b \tan (c+d x))^2}+\frac {3 b \tan (c+d x)}{a+b \tan (c+d x)}-\frac {3 \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {\tan (c+d x)}}{\sqrt {a} \sqrt {1+\frac {b \tan (c+d x)}{a}}}\right )}{3 b^3 d \sqrt {\tan (c+d x)}}-\frac {(i A+B) \sqrt {a+b \tan (c+d x)} \left (\frac {b^2 \tan ^2(c+d x)}{(a+b \tan (c+d x))^2}+\frac {3 b \tan (c+d x)}{a+b \tan (c+d x)}-\frac {3 \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {\tan (c+d x)}}{\sqrt {a} \sqrt {1+\frac {b \tan (c+d x)}{a}}}\right )}{3 b^3 d \sqrt {\tan (c+d x)}} \]

[In]

Integrate[(Tan[c + d*x]^(5/2)*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

((A - I*B)*((3*(-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(-a
+ I*b)^(5/2) + Tan[c + d*x]^(3/2)/((a - I*b)*(a + b*Tan[c + d*x])^(3/2)) - ((3*I)*Sqrt[Tan[c + d*x]])/((a - I
*b)^2*Sqrt[a + b*Tan[c + d*x]])))/(3*d) - ((A + I*B)*((3*(-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[
c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(a + I*b)^(5/2) - Tan[c + d*x]^(3/2)/((a + I*b)*(a + b*Tan[c + d*x])^(3/
2)) - ((3*I)*Sqrt[Tan[c + d*x]])/((a + I*b)^2*Sqrt[a + b*Tan[c + d*x]])))/(3*d) + ((I*A - B)*Sqrt[a + b*Tan[c
+ d*x]]*((b^2*Tan[c + d*x]^2)/(a + b*Tan[c + d*x])^2 + (3*b*Tan[c + d*x])/(a + b*Tan[c + d*x]) - (3*Sqrt[b]*Ar
cSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[Tan[c + d*x]])/(Sqrt[a]*Sqrt[1 + (b*Tan[c + d*x])/a])))/(3*b^
3*d*Sqrt[Tan[c + d*x]]) - ((I*A + B)*Sqrt[a + b*Tan[c + d*x]]*((b^2*Tan[c + d*x]^2)/(a + b*Tan[c + d*x])^2 + (
3*b*Tan[c + d*x])/(a + b*Tan[c + d*x]) - (3*Sqrt[b]*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[Tan[c +
d*x]])/(Sqrt[a]*Sqrt[1 + (b*Tan[c + d*x])/a])))/(3*b^3*d*Sqrt[Tan[c + d*x]])

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 2.18 (sec) , antiderivative size = 2978162, normalized size of antiderivative = 10560.86

\[\text {output too large to display}\]

[In]

int(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x)

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27883 vs. \(2 (236) = 472\).

Time = 22.13 (sec) , antiderivative size = 55768, normalized size of antiderivative = 197.76 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(tan(d*x+c)**(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^(5/2)/(b*tan(d*x + c) + a)^(5/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int((tan(c + d*x)^(5/2)*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^(5/2),x)

[Out]

int((tan(c + d*x)^(5/2)*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^(5/2), x)