\(\int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) [648]

Optimal result

Integrand size = 25, antiderivative size = 251 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (a^2+3 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 2.28 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3646, 3726, 3736, 6857, 65, 223, 212, 95, 211, 214} \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (a^2+3 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}-\frac {\arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]

[In]

Int[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

-(ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/((I*a - b)^(5/2)*d)) + (2*ArcTanh[(Sqrt[
b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(b^(5/2)*d) - ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqr
t[a + b*Tan[c + d*x]]]/((I*a + b)^(5/2)*d) - (2*a^2*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^2 + 3*b^2)*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 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 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^2 \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 a^2}{2}-\frac {3}{2} a b \tan (c+d x)+\frac {3}{2} \left (a^2+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 {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^2 \left (a^2+3 b^2\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^2 \left (a^2+3 b^2\right )-\frac {3}{2} a b^3 \tan (c+d x)+\frac {3}{4} \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^2 \left (a^2+b^2\right )^2} \\ & = -\frac {2 a^2 \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^2 \left (a^2+3 b^2\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^2 \left (a^2+3 b^2\right )-\frac {3}{2} a b^3 x+\frac {3}{4} \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^2 \left (a^2+b^2\right )^2 d} \\ & = -\frac {2 a^2 \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^2 \left (a^2+3 b^2\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}{4 \sqrt {x} \sqrt {a+b x}}+\frac {3 \left (b^2 \left (a^2-b^2\right )-2 a b^3 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^2 \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^2 \left (a^2+3 b^2\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 {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{b^2 d}+\frac {\text {Subst}\left (\int \frac {b^2 \left (a^2-b^2\right )-2 a b^3 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 {2 a^2 \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^2 \left (a^2+3 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{b^2 d}+\frac {\text {Subst}\left (\int \left (\frac {2 a b^3+i b^2 \left (a^2-b^2\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {-2 a b^3+i b^2 \left (a^2-b^2\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 a^2 \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^2 \left (a^2+3 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {i \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 {i \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 \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 \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^2 \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^2 \left (a^2+3 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {i \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 \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 {\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 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d}-\frac {\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^2 \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^2 \left (a^2+3 b^2\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 [A] (verified)

Time = 6.25 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.69 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {(-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{5/2} d}-\frac {(-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{5/2} d}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{3 (i a-b) d (a+b \tan (c+d x))^{3/2}}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d (a+b \tan (c+d x))^{3/2}}+\frac {\tan ^{\frac {3}{2}}(c+d x)}{3 (i a+b) d (a+b \tan (c+d x))^{3/2}}-\frac {\sqrt {\tan (c+d x)}}{(a-i b)^2 d \sqrt {a+b \tan (c+d x)}}-\frac {\sqrt {\tan (c+d x)}}{(a+i b)^2 d \sqrt {a+b \tan (c+d x)}}-\frac {2 \sqrt {\tan (c+d x)}}{b^2 d \sqrt {a+b \tan (c+d x)}}+\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {a+b \tan (c+d x)}}{\sqrt {a} b^{5/2} d \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]

[In]

Integrate[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

-(((-1)^(3/4)*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)*d)) - ((-1)^(3/4)*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)*d) - Tan[c + d*x]^(3/2)/(3*(I*a - b)*d*(a + b*Tan[c + d*x])^(3/2)) - (2*Tan[c + d*x]^(3/2))/(3*b*d*(a
+ b*Tan[c + d*x])^(3/2)) + Tan[c + d*x]^(3/2)/(3*(I*a + b)*d*(a + b*Tan[c + d*x])^(3/2)) - Sqrt[Tan[c + d*x]]/
((a - I*b)^2*d*Sqrt[a + b*Tan[c + d*x]]) - Sqrt[Tan[c + d*x]]/((a + I*b)^2*d*Sqrt[a + b*Tan[c + d*x]]) - (2*Sq
rt[Tan[c + d*x]])/(b^2*d*Sqrt[a + b*Tan[c + d*x]]) + (2*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[a +
b*Tan[c + d*x]])/(Sqrt[a]*b^(5/2)*d*Sqrt[1 + (b*Tan[c + d*x])/a])

Maple [B] (warning: unable to verify)

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

Time = 1.20 (sec) , antiderivative size = 1490358, normalized size of antiderivative = 5937.68

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

[In]

int(tan(d*x+c)^(7/2)/(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. 11532 vs. \(2 (209) = 418\).

Time = 4.33 (sec) , antiderivative size = 23066, normalized size of antiderivative = 91.90 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(tan(d*x+c)**(7/2)/(a+b*tan(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{\frac {7}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(tan(d*x + c)^(7/2)/(b*tan(d*x + c) + a)^(5/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(tan(c + d*x)^(7/2)/(a + b*tan(c + d*x))^(5/2),x)

[Out]

int(tan(c + d*x)^(7/2)/(a + b*tan(c + d*x))^(5/2), x)