\(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx\) [468]

Optimal result

Integrand size = 35, antiderivative size = 359 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\frac {(A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]

[Out]

(A+I*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+(A-I*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+2/3*b*(8*A*a^4*b+30*A*a^2*b^3+16*A*b^5-3*B*a^5-
17*B*a^3*b^2-8*B*a*b^4)*tan(d*x+c)^(1/2)/a^4/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)+2*(2*A*b-B*a)/a^2/d/tan(d*x+
c)^(1/2)/(a+b*tan(d*x+c))^(3/2)+2/3*b*(7*A*a^2*b+8*A*b^3-3*B*a^3-4*B*a*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/a/d/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 2.22 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3690, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (-3 a^3 B+7 a^2 A b-4 a b^2 B+8 A b^3\right ) \sqrt {\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (-3 a^5 B+8 a^4 A b-17 a^3 b^2 B+30 a^2 A b^3-8 a b^4 B+16 A b^5\right ) \sqrt {\tan (c+d x)}}{3 a^4 d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}+\frac {(A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

((A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a - b)^(5/2)*d) + ((A - I*
B)*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)/(3*a*d*Ta
n[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2)) + (2*(2*A*b - a*B))/(a^2*d*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x]
)^(3/2)) + (2*b*(7*a^2*A*b + 8*A*b^3 - 3*a^3*B - 4*a*b^2*B)*Sqrt[Tan[c + d*x]])/(3*a^3*(a^2 + b^2)*d*(a + b*Ta
n[c + d*x])^(3/2)) + (2*b*(8*a^4*A*b + 30*a^2*A*b^3 + 16*A*b^5 - 3*a^5*B - 17*a^3*b^2*B - 8*a*b^4*B)*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 3690

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*(A*b - a*B)*(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[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(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])))

Rubi steps \begin{align*} \text {integral}& = -\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}-\frac {2 \int \frac {\frac {3}{2} (2 A b-a B)+\frac {3}{2} a A \tan (c+d x)+3 A 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 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a 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 A-8 A b^2+4 a b B\right )-\frac {3}{4} a^2 B \tan (c+d x)+3 b (2 A b-a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}} \, dx}{3 a^2} \\ & = -\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\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 A-14 a^2 A b^2-16 A b^4+9 a^3 b B+8 a b^3 B\right )+\frac {9}{8} a^3 (A b-a B) \tan (c+d x)+\frac {3}{4} b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\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 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\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 {2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\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 A-A b^2+2 a b B\right )+\frac {9}{16} a^4 \left (2 a A b-a^2 B+b^2 B\right ) \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 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\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 {2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\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 {(A-i B) \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 {(A+i B) \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}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\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 {2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\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 {(A-i B) \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 {(A+i B) \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}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\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 {2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\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 {(A-i B) \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 {(A+i B) \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 {(A+i 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 {(A-i 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}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\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 {2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\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*}

Mathematica [A] (verified)

Time = 3.89 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.07 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\frac {-\frac {6 A}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {6 (6 A b-3 a B)}{a \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {6 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {9 (-1)^{3/4} a^4 \left (\frac {(a+i b)^2 (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {(a-i b)^2 (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )+\frac {6 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{a^3 \left (a^2+b^2\right )^2}}{9 a d} \]

[In]

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

[Out]

((-6*A)/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2)) + (6*(6*A*b - 3*a*B))/(a*Sqrt[Tan[c + d*x]]*(a + b*Tan
[c + d*x])^(3/2)) + (6*b*(7*a^2*A*b + 8*A*b^3 - 3*a^3*B - 4*a*b^2*B)*Sqrt[Tan[c + d*x]])/(a^2*(a^2 + b^2)*(a +
b*Tan[c + d*x])^(3/2)) + (9*(-1)^(3/4)*a^4*(((a + I*b)^2*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan
[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] + ((a - I*b)^2*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b
]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b]) + (6*b*(8*a^4*A*b + 30*a^2*A*b^3 + 16*A*b^5 -
3*a^5*B - 17*a^3*b^2*B - 8*a*b^4*B)*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]])/(a^3*(a^2 + b^2)^2))/(9*a*d)

Maple [B] (warning: unable to verify)

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

Time = 5.47 (sec) , antiderivative size = 2982515, normalized size of antiderivative = 8307.84

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

[In]

int((A+B*tan(d*x+c))/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. 27865 vs. \(2 (308) = 616\).

Time = 14.52 (sec) , antiderivative size = 27865, normalized size of antiderivative = 77.62 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F(-1)]

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

[In]

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

[In]

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

[Out]

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