\(\int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) [649]

Optimal result

Integrand size = 25, antiderivative size = 214 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}-\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]

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

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3646, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}+\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac {i \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]

[In]

Int[Tan[c + d*x]^(5/2)/(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) - (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*Sqrt[Tan[c + d*x]])/(3*b
*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + (2*a*(a^2 + 7*b^2)*Sqrt[Tan[c + d*x]])/(3*b*(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 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 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])))

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \int \frac {\frac {a^2}{2}-\frac {3}{2} a b \tan (c+d x)+\frac {1}{2} \left (a^2+3 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 b \left (a^2+b^2\right )} \\ & = -\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {4 \int \frac {-\frac {3}{2} a^2 b^2-\frac {3}{4} a b \left (a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a b \left (a^2+b^2\right )^2} \\ & = -\frac {2 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {i \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 {i \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 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {i \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 {i \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 a^2 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {i \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 {i \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 {i \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 {i \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 \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (a^2+7 b^2\right ) \sqrt {\tan (c+d x)}}{3 b \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.90 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {\frac {3 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{5/2}}-\frac {3 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{5/2}}+\frac {2 a \sqrt {\tan (c+d x)} \left (6 a b+\left (a^2+7 b^2\right ) \tan (c+d x)\right )}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^{3/2}}}{3 d} \]

[In]

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

[Out]

((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) - (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) + (2*a*Sqrt[Tan[c + d*x]]*(6*a*b + (a^2 + 7*b^2)*Tan[c + d*x]))/((a^2 + b^2)^2*(a + b*Tan[c + d*x])^(3/2)
))/(3*d)

Maple [B] (warning: unable to verify)

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

Time = 1.27 (sec) , antiderivative size = 1488448, normalized size of antiderivative = 6955.36

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

[In]

int(tan(d*x+c)^(5/2)/(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. 11334 vs. \(2 (174) = 348\).

Time = 2.37 (sec) , antiderivative size = 11334, normalized size of antiderivative = 52.96 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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