3.7.47 \(\int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) [647]

Optimal. Leaf size=317 \[ -\frac {i \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}-\frac {5 a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{7/2} d}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^3 \left (a^2+b^2\right )^2 d} \]

[Out]

-I*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-5*a*arctanh(b^(1/2)*tan(d*x+c
)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b^(7/2)/d+I*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+(5*a^4+10*a^2*b^2+b^4)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/b^3/(a^2+b^2)^2/d-2/3*a^2*(5*a^2+1
1*b^2)*tan(d*x+c)^(3/2)/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)-2/3*a^2*tan(d*x+c)^(5/2)/b/(a^2+b^2)/d/(a+b*t
an(d*x+c))^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 1.59, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3646, 3726, 3728, 3736, 6857, 65, 223, 212, 95, 211, 214} \begin {gather*} -\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^3 d \left (a^2+b^2\right )^2}-\frac {i \text {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 {5 a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{7/2} d}+\frac {i \tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(9/2)/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

((-I)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a - b)^(5/2)*d) - (5*a*ArcTanh[
(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(b^(7/2)*d) + (I*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d
*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a + b)^(5/2)*d) - (2*a^2*Tan[c + d*x]^(5/2))/(3*b*(a^2 + b^2)*d*(a + b*Ta
n[c + d*x])^(3/2)) - (2*a^2*(5*a^2 + 11*b^2)*Tan[c + d*x]^(3/2))/(3*b^2*(a^2 + b^2)^2*d*Sqrt[a + b*Tan[c + d*x
]]) + ((5*a^4 + 10*a^2*b^2 + b^4)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(b^3*(a^2 + b^2)^2*d)

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 3646

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 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 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 \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx &=-\frac {2 a^2 \tan ^{\frac {5}{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 {\tan ^{\frac {3}{2}}(c+d x) \left (\frac {5 a^2}{2}-\frac {3}{2} a b \tan (c+d x)+\frac {1}{2} \left (5 a^2+3 b^2\right ) \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^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {4 \int \frac {\sqrt {\tan (c+d x)} \left (\frac {3}{4} a^2 \left (5 a^2+11 b^2\right )-\frac {3}{2} a b^3 \tan (c+d x)+\frac {3}{4} \left (5 a^4+10 a^2 b^2+b^4\right ) \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx}{3 b^2 \left (a^2+b^2\right )^2}\\ &=-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^3 \left (a^2+b^2\right )^2 d}+\frac {4 \int \frac {-\frac {3}{8} a \left (5 a^4+10 a^2 b^2+b^4\right )+\frac {3}{4} b^3 \left (a^2-b^2\right ) \tan (c+d x)-\frac {15}{8} a \left (a^2+b^2\right )^2 \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 b^3 \left (a^2+b^2\right )^2}\\ &=-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^3 \left (a^2+b^2\right )^2 d}+\frac {4 \text {Subst}\left (\int \frac {-\frac {3}{8} a \left (5 a^4+10 a^2 b^2+b^4\right )+\frac {3}{4} b^3 \left (a^2-b^2\right ) x-\frac {15}{8} a \left (a^2+b^2\right )^2 x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{3 b^3 \left (a^2+b^2\right )^2 d}\\ &=-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^3 \left (a^2+b^2\right )^2 d}+\frac {4 \text {Subst}\left (\int \left (-\frac {15 a \left (a^2+b^2\right )^2}{8 \sqrt {x} \sqrt {a+b x}}+\frac {3 \left (2 a b^4+b^3 \left (a^2-b^2\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^3 \left (a^2+b^2\right )^2 d}\\ &=-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^3 \left (a^2+b^2\right )^2 d}-\frac {(5 a) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 b^3 d}+\frac {\text {Subst}\left (\int \frac {2 a b^4+b^3 \left (a^2-b^2\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b^3 \left (a^2+b^2\right )^2 d}\\ &=-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^3 \left (a^2+b^2\right )^2 d}-\frac {(5 a) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{b^3 d}+\frac {\text {Subst}\left (\int \left (\frac {2 i a b^4-b^3 \left (a^2-b^2\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {2 i a b^4+b^3 \left (a^2-b^2\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b^3 \left (a^2+b^2\right )^2 d}\\ &=-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^3 \left (a^2+b^2\right )^2 d}+\frac {\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 {\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 {(5 a) \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^3 d}\\ &=-\frac {5 a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{7/2} d}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^3 \left (a^2+b^2\right )^2 d}+\frac {\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 {\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 \tan ^{-1}\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 {5 a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{7/2} d}+\frac {i \tanh ^{-1}\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^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^3 \left (a^2+b^2\right )^2 d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 6.17, size = 407, normalized size = 1.28 \begin {gather*} \frac {\sqrt [4]{-1} \text {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) (-a+i b)^{3/2} d}-\frac {\sqrt [4]{-1} \text {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) (a+i b)^{3/2} d}+\frac {\tan ^{\frac {3}{2}}(c+d x)}{3 (-a+i b) d (a+b \tan (c+d x))^{3/2}}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{3 (a+i b) d (a+b \tan (c+d x))^{3/2}}-\frac {i \sqrt {\tan (c+d x)}}{(a-i b) (-a+i b) d \sqrt {a+b \tan (c+d x)}}+\frac {i \sqrt {\tan (c+d x)}}{(-a-i b) (a+i b) d \sqrt {a+b \tan (c+d x)}}+\frac {2 \, _2F_1\left (\frac {5}{2},\frac {7}{2};\frac {9}{2};-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {7}{2}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{7 a^2 d \sqrt {a+b \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(9/2)/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

((-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)*(-a +
I*b)^(3/2)*d) - ((-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)*(a + I*b)^(3/2)*d) + Tan[c + d*x]^(3/2)/(3*(-a + I*b)*d*(a + b*Tan[c + d*x])^(3/2)) - Tan[c + d*x]^(
3/2)/(3*(a + I*b)*d*(a + b*Tan[c + d*x])^(3/2)) - (I*Sqrt[Tan[c + d*x]])/((a - I*b)*(-a + I*b)*d*Sqrt[a + b*Ta
n[c + d*x]]) + (I*Sqrt[Tan[c + d*x]])/((-a - I*b)*(a + I*b)*d*Sqrt[a + b*Tan[c + d*x]]) + (2*Hypergeometric2F1
[5/2, 7/2, 9/2, -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(7/2)*Sqrt[1 + (b*Tan[c + d*x])/a])/(7*a^2*d*Sqrt[a + b*Ta
n[c + d*x]])

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 1.32, size = 1493684, normalized size = 4711.94 \[\text {output too large to display}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^(5/2),x)

[Out]

result too large to display

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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(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(tan(d*x + c)^(9/2)/(b*tan(d*x + c) + a)^(5/2), 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(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(9/2)/(a+b*tan(d*x+c))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 7317 deep

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Giac [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(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{9/2}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(9/2)/(a + b*tan(c + d*x))^(5/2),x)

[Out]

int(tan(c + d*x)^(9/2)/(a + b*tan(c + d*x))^(5/2), x)

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