3.7.54 \(\int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx\) [654]

Optimal. Leaf size=298 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {4 b}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b^2 \left (7 a^2+8 b^2\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]

[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+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+4/3*b^2*(4*a^4+15*a^2*b^2+8*b^4)*tan(d*x+c)^(1/2)/a^4/(a^2+b^2)
^2/d/(a+b*tan(d*x+c))^(1/2)+4*b/a^2/d/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2)+2/3*b^2*(7*a^2+8*b^2)*tan(d*x+c)
^(1/2)/a^3/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)-2/3/a/d/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.80, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3650, 3730, 3731, 3697, 3696, 95, 209, 212} \begin {gather*} \frac {4 b}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{3 a^4 d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}+\frac {2 b^2 \left (7 a^2+8 b^2\right ) \sqrt {\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\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 {2}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {\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[1/(Tan[c + d*x]^(5/2)*(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) + ArcTanh[(Sqrt[I*a +
b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/((I*a + b)^(5/2)*d) - 2/(3*a*d*Tan[c + d*x]^(3/2)*(a + b*Tan[
c + d*x])^(3/2)) + (4*b)/(a^2*d*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)) + (2*b^2*(7*a^2 + 8*b^2)*Sqrt[T
an[c + d*x]])/(3*a^3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + (4*b^2*(4*a^4 + 15*a^2*b^2 + 8*b^4)*Sqrt[Tan[
c + d*x]])/(3*a^4*(a^2 + b^2)^2*d*Sqrt[a + b*Tan[c + d*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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 3650

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3696

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

Rule 3697

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

Rule 3730

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*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3731

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*t
an[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 + a^2*C)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x]
)^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[
e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2)) - a*C*(b*c*(m + 1)
+ a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - b*C)*Tan[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 2)*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^
2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx &=-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}-\frac {2 \int \frac {3 b+\frac {3}{2} a \tan (c+d x)+3 b \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx}{3 a}\\ &=-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {4 b}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 \int \frac {-\frac {3}{4} \left (a^2-8 b^2\right )+6 b^2 \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}} \, dx}{3 a^2}\\ &=-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {4 b}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b^2 \left (7 a^2+8 b^2\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {8 \int \frac {-\frac {3}{8} \left (3 a^4-14 a^2 b^2-16 b^4\right )+\frac {9}{8} a^3 b \tan (c+d x)+\frac {3}{4} b^2 \left (7 a^2+8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{9 a^3 \left (a^2+b^2\right )}\\ &=-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {4 b}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b^2 \left (7 a^2+8 b^2\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {16 \int \frac {-\frac {9}{16} a^4 \left (a^2-b^2\right )+\frac {9}{8} a^5 b \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{9 a^4 \left (a^2+b^2\right )^2}\\ &=-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {4 b}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b^2 \left (7 a^2+8 b^2\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {\int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}-\frac {\int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}\\ &=-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {4 b}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b^2 \left (7 a^2+8 b^2\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {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}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}\\ &=-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {4 b}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b^2 \left (7 a^2+8 b^2\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {1}{1-(i a+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}{1-(-i a+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 {\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 {\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}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {4 b}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b^2 \left (7 a^2+8 b^2\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 6.29, size = 483, normalized size = 1.62 \begin {gather*} -\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}-\frac {2 \left (-\frac {6 b}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {3 \left (\frac {a b \sqrt {\tan (c+d x)}}{3 (i a-b) (a+b \tan (c+d x))^{3/2}}+\frac {16 b^2 \sqrt {\tan (c+d x)}}{3 a (a+b \tan (c+d x))^{3/2}}-\frac {a b \sqrt {\tan (c+d x)}}{3 (i a+b) (a+b \tan (c+d x))^{3/2}}+\frac {32 b^2 \sqrt {\tan (c+d x)}}{3 a^2 \sqrt {a+b \tan (c+d x)}}-\frac {-\frac {3 \sqrt [4]{-1} a^2 \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)^{3/2}}+\frac {(5 a-2 i b) b \sqrt {\tan (c+d x)}}{(a-i b) \sqrt {a+b \tan (c+d x)}}}{3 (i a+b)}+\frac {-\frac {3 \sqrt [4]{-1} a^2 \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)^{3/2}}+\frac {(5 a+2 i b) b \sqrt {\tan (c+d x)}}{(a+i b) \sqrt {a+b \tan (c+d x)}}}{3 (i a-b)}\right )}{2 a d}\right )}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^(5/2)),x]

[Out]

-2/(3*a*d*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2)) - (2*((-6*b)/(a*d*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d
*x])^(3/2)) - (3*((a*b*Sqrt[Tan[c + d*x]])/(3*(I*a - b)*(a + b*Tan[c + d*x])^(3/2)) + (16*b^2*Sqrt[Tan[c + d*x
]])/(3*a*(a + b*Tan[c + d*x])^(3/2)) - (a*b*Sqrt[Tan[c + d*x]])/(3*(I*a + b)*(a + b*Tan[c + d*x])^(3/2)) + (32
*b^2*Sqrt[Tan[c + d*x]])/(3*a^2*Sqrt[a + b*Tan[c + d*x]]) - ((-3*(-1)^(1/4)*a^2*ArcTan[((-1)^(1/4)*Sqrt[-a + I
*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(-a + I*b)^(3/2) + ((5*a - (2*I)*b)*b*Sqrt[Tan[c + d*x]])/(
(a - I*b)*Sqrt[a + b*Tan[c + d*x]]))/(3*(I*a + b)) + ((-3*(-1)^(1/4)*a^2*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt
[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(a + I*b)^(3/2) + ((5*a + (2*I)*b)*b*Sqrt[Tan[c + d*x]])/((a + I*b)
*Sqrt[a + b*Tan[c + d*x]]))/(3*(I*a - b))))/(2*a*d)))/(3*a)

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 3.45, size = 1491406, normalized size = 5004.72 \[\text {output too large to display}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/tan(d*x+c)^(5/2)/(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(1/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^(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(1/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)**(5/2)/(a+b*tan(d*x+c))**(5/2),x)

[Out]

Integral(1/((a + b*tan(c + d*x))**(5/2)*tan(c + d*x)**(5/2)), x)

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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(1/tan(d*x+c)^(5/2)/(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 {1}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,{\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(1/(tan(c + d*x)^(5/2)*(a + b*tan(c + d*x))^(5/2)),x)

[Out]

int(1/(tan(c + d*x)^(5/2)*(a + b*tan(c + d*x))^(5/2)), x)

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