\(\int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx\) [450]

Optimal result

Integrand size = 35, antiderivative size = 460 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=-\frac {(i a-b)^{5/2} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (14 A b+11 a B) \sqrt {a+b \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 \left (99 a^2 A-113 A b^2-209 a b B\right ) \sqrt {a+b \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (495 a^2 A b-5 A b^3+231 a^3 B-275 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{1155 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (1155 a^4 A-1485 a^2 A b^2-20 A b^4-2541 a^3 b B+55 a b^3 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^2 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (8085 a^4 A b-495 a^2 A b^3+40 A b^5+3465 a^5 B-5313 a^3 b^2 B-110 a b^4 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^3 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)} \]

[Out]

-(I*a-b)^(5/2)*(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d-(I*a+b)^(5/2)*(I*A+B)*a
rctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d-2/3465*(8085*A*a^4*b-495*A*a^2*b^3+40*A*b^5+34
65*B*a^5-5313*B*a^3*b^2-110*B*a*b^4)*(a+b*tan(d*x+c))^(1/2)/a^3/d/tan(d*x+c)^(1/2)-2/99*a*(14*A*b+11*B*a)*(a+b
*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(9/2)+2/693*(99*A*a^2-113*A*b^2-209*B*a*b)*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)
^(7/2)+2/1155*(495*A*a^2*b-5*A*b^3+231*B*a^3-275*B*a*b^2)*(a+b*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(5/2)-2/3465*(
1155*A*a^4-1485*A*a^2*b^2-20*A*b^4-2541*B*a^3*b+55*B*a*b^3)*(a+b*tan(d*x+c))^(1/2)/a^2/d/tan(d*x+c)^(3/2)-2/11
*a*A*(a+b*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(11/2)

Rubi [A] (verified)

Time = 2.74 (sec) , antiderivative size = 460, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3686, 3726, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\frac {2 \left (99 a^2 A-209 a b B-113 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (231 a^3 B+495 a^2 A b-275 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{1155 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (1155 a^4 A-2541 a^3 b B-1485 a^2 A b^2+55 a b^3 B-20 A b^4\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^2 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (3465 a^5 B+8085 a^4 A b-5313 a^3 b^2 B-495 a^2 A b^3-110 a b^4 B+40 A b^5\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^3 d \sqrt {\tan (c+d x)}}-\frac {(-b+i a)^{5/2} (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(b+i a)^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (11 a B+14 A b) \sqrt {a+b \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)} \]

[In]

Int[((a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(13/2),x]

[Out]

-(((I*a - b)^(5/2)*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) - ((I*a +
b)^(5/2)*(I*A + B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (2*a*(14*A*b + 1
1*a*B)*Sqrt[a + b*Tan[c + d*x]])/(99*d*Tan[c + d*x]^(9/2)) + (2*(99*a^2*A - 113*A*b^2 - 209*a*b*B)*Sqrt[a + b*
Tan[c + d*x]])/(693*d*Tan[c + d*x]^(7/2)) + (2*(495*a^2*A*b - 5*A*b^3 + 231*a^3*B - 275*a*b^2*B)*Sqrt[a + b*Ta
n[c + d*x]])/(1155*a*d*Tan[c + d*x]^(5/2)) - (2*(1155*a^4*A - 1485*a^2*A*b^2 - 20*A*b^4 - 2541*a^3*b*B + 55*a*
b^3*B)*Sqrt[a + b*Tan[c + d*x]])/(3465*a^2*d*Tan[c + d*x]^(3/2)) - (2*(8085*a^4*A*b - 495*a^2*A*b^3 + 40*A*b^5
+ 3465*a^5*B - 5313*a^3*b^2*B - 110*a*b^4*B)*Sqrt[a + b*Tan[c + d*x]])/(3465*a^3*d*Sqrt[Tan[c + d*x]]) - (2*a
*A*(a + b*Tan[c + d*x])^(3/2))/(11*d*Tan[c + d*x]^(11/2))

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 3686

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

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 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 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 A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {2}{11} \int \frac {\sqrt {a+b \tan (c+d x)} \left (\frac {1}{2} a (14 A b+11 a B)-\frac {11}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac {1}{2} b (8 a A-11 b B) \tan ^2(c+d x)\right )}{\tan ^{\frac {11}{2}}(c+d x)} \, dx \\ & = -\frac {2 a (14 A b+11 a B) \sqrt {a+b \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {4}{99} \int \frac {-\frac {1}{4} a \left (99 a^2 A-113 A b^2-209 a b B\right )-\frac {99}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-\frac {1}{4} b \left (184 a A b+88 a^2 B-99 b^2 B\right ) \tan ^2(c+d x)}{\tan ^{\frac {9}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 a (14 A b+11 a B) \sqrt {a+b \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 \left (99 a^2 A-113 A b^2-209 a b B\right ) \sqrt {a+b \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}-\frac {8 \int \frac {\frac {3}{8} a \left (495 a^2 A b-5 A b^3+231 a^3 B-275 a b^2 B\right )-\frac {693}{8} a \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)-\frac {3}{4} a b \left (99 a^2 A-113 A b^2-209 a b B\right ) \tan ^2(c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{693 a} \\ & = -\frac {2 a (14 A b+11 a B) \sqrt {a+b \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 \left (99 a^2 A-113 A b^2-209 a b B\right ) \sqrt {a+b \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (495 a^2 A b-5 A b^3+231 a^3 B-275 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{1155 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {16 \int \frac {\frac {3}{16} a \left (1155 a^4 A-1485 a^2 A b^2-20 A b^4-2541 a^3 b B+55 a b^3 B\right )+\frac {3465}{16} a^2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)+\frac {3}{4} a b \left (495 a^2 A b-5 A b^3+231 a^3 B-275 a b^2 B\right ) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{3465 a^2} \\ & = -\frac {2 a (14 A b+11 a B) \sqrt {a+b \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 \left (99 a^2 A-113 A b^2-209 a b B\right ) \sqrt {a+b \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (495 a^2 A b-5 A b^3+231 a^3 B-275 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{1155 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (1155 a^4 A-1485 a^2 A b^2-20 A b^4-2541 a^3 b B+55 a b^3 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^2 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}-\frac {32 \int \frac {-\frac {3}{32} a \left (8085 a^4 A b-495 a^2 A b^3+40 A b^5+3465 a^5 B-5313 a^3 b^2 B-110 a b^4 B\right )+\frac {10395}{32} a^3 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)+\frac {3}{16} a b \left (1155 a^4 A-1485 a^2 A b^2-20 A b^4-2541 a^3 b B+55 a b^3 B\right ) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{10395 a^3} \\ & = -\frac {2 a (14 A b+11 a B) \sqrt {a+b \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 \left (99 a^2 A-113 A b^2-209 a b B\right ) \sqrt {a+b \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (495 a^2 A b-5 A b^3+231 a^3 B-275 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{1155 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (1155 a^4 A-1485 a^2 A b^2-20 A b^4-2541 a^3 b B+55 a b^3 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^2 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (8085 a^4 A b-495 a^2 A b^3+40 A b^5+3465 a^5 B-5313 a^3 b^2 B-110 a b^4 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^3 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {64 \int \frac {-\frac {10395}{64} a^4 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )-\frac {10395}{64} a^4 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{10395 a^4} \\ & = -\frac {2 a (14 A b+11 a B) \sqrt {a+b \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 \left (99 a^2 A-113 A b^2-209 a b B\right ) \sqrt {a+b \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (495 a^2 A b-5 A b^3+231 a^3 B-275 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{1155 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (1155 a^4 A-1485 a^2 A b^2-20 A b^4-2541 a^3 b B+55 a b^3 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^2 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (8085 a^4 A b-495 a^2 A b^3+40 A b^5+3465 a^5 B-5313 a^3 b^2 B-110 a b^4 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^3 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}-\frac {1}{2} \left ((a-i b)^3 (A-i B)\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} \left ((a+i b)^3 (A+i B)\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 a (14 A b+11 a B) \sqrt {a+b \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 \left (99 a^2 A-113 A b^2-209 a b B\right ) \sqrt {a+b \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (495 a^2 A b-5 A b^3+231 a^3 B-275 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{1155 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (1155 a^4 A-1485 a^2 A b^2-20 A b^4-2541 a^3 b B+55 a b^3 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^2 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (8085 a^4 A b-495 a^2 A b^3+40 A b^5+3465 a^5 B-5313 a^3 b^2 B-110 a b^4 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^3 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}-\frac {\left ((a-i b)^3 (A-i B)\right ) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {\left ((a+i b)^3 (A+i B)\right ) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = -\frac {2 a (14 A b+11 a B) \sqrt {a+b \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 \left (99 a^2 A-113 A b^2-209 a b B\right ) \sqrt {a+b \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (495 a^2 A b-5 A b^3+231 a^3 B-275 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{1155 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (1155 a^4 A-1485 a^2 A b^2-20 A b^4-2541 a^3 b B+55 a b^3 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^2 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (8085 a^4 A b-495 a^2 A b^3+40 A b^5+3465 a^5 B-5313 a^3 b^2 B-110 a b^4 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^3 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}-\frac {\left ((a-i b)^3 (A-i B)\right ) \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 )}{d}-\frac {\left ((a+i b)^3 (A+i B)\right ) \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 )}{d} \\ & = -\frac {(i a-b)^{5/2} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (14 A b+11 a B) \sqrt {a+b \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 \left (99 a^2 A-113 A b^2-209 a b B\right ) \sqrt {a+b \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (495 a^2 A b-5 A b^3+231 a^3 B-275 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{1155 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (1155 a^4 A-1485 a^2 A b^2-20 A b^4-2541 a^3 b B+55 a b^3 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^2 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (8085 a^4 A b-495 a^2 A b^3+40 A b^5+3465 a^5 B-5313 a^3 b^2 B-110 a b^4 B\right ) \sqrt {a+b \tan (c+d x)}}{3465 a^3 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.31 (sec) , antiderivative size = 632, normalized size of antiderivative = 1.37 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=-\frac {b B (a+b \tan (c+d x))^{3/2}}{4 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {1}{4} \left (-\frac {b (8 A b+5 a B) \sqrt {a+b \tan (c+d x)}}{10 d \tan ^{\frac {11}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {\left (80 a^2 A-88 A b^2-165 a b B\right ) \sqrt {a+b \tan (c+d x)}}{22 d \tan ^{\frac {11}{2}}(c+d x)}-\frac {2 \left (\frac {5 a \left (184 a A b+88 a^2 B-99 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{18 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 \left (\frac {10 a^2 \left (99 a^2 A-113 A b^2-209 a b B\right ) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 \left (-\frac {3 a^2 \left (495 a^2 A b-5 A b^3+231 a^3 B-275 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (-\frac {5 a^2 \left (1155 a^4 A-1485 a^2 A b^2-20 A b^4-2541 a^3 b B+55 a b^3 B\right ) \sqrt {a+b \tan (c+d x)}}{2 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (\frac {51975 a^5 \left ((-1)^{3/4} (-a+i b)^{5/2} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(-1)^{3/4} (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )}{8 d}+\frac {15 a^2 \left (8085 a^4 A b-495 a^2 A b^3+40 A b^5+3465 a^5 B-5313 a^3 b^2 B-110 a b^4 B\right ) \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\tan (c+d x)}}\right )}{3 a}\right )}{5 a}\right )}{7 a}\right )}{9 a}\right )}{11 a}\right )\right ) \]

[In]

Integrate[((a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(13/2),x]

[Out]

-1/4*(b*B*(a + b*Tan[c + d*x])^(3/2))/(d*Tan[c + d*x]^(11/2)) + (-1/10*(b*(8*A*b + 5*a*B)*Sqrt[a + b*Tan[c + d
*x]])/(d*Tan[c + d*x]^(11/2)) + (-1/22*((80*a^2*A - 88*A*b^2 - 165*a*b*B)*Sqrt[a + b*Tan[c + d*x]])/(d*Tan[c +
d*x]^(11/2)) - (2*((5*a*(184*a*A*b + 88*a^2*B - 99*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(18*d*Tan[c + d*x]^(9/2))
- (2*((10*a^2*(99*a^2*A - 113*A*b^2 - 209*a*b*B)*Sqrt[a + b*Tan[c + d*x]])/(7*d*Tan[c + d*x]^(7/2)) - (2*((-3
*a^2*(495*a^2*A*b - 5*A*b^3 + 231*a^3*B - 275*a*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(d*Tan[c + d*x]^(5/2)) - (2*(
(-5*a^2*(1155*a^4*A - 1485*a^2*A*b^2 - 20*A*b^4 - 2541*a^3*b*B + 55*a*b^3*B)*Sqrt[a + b*Tan[c + d*x]])/(2*d*Ta
n[c + d*x]^(3/2)) - (2*((51975*a^5*((-1)^(3/4)*(-a + I*b)^(5/2)*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sq
rt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - (-1)^(3/4)*(a + I*b)^(5/2)*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a +
I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]))/(8*d) + (15*a^2*(8085*a^4*A*b - 495*a^2*A*b^3 + 40*A*b^5
+ 3465*a^5*B - 5313*a^3*b^2*B - 110*a*b^4*B)*Sqrt[a + b*Tan[c + d*x]])/(4*d*Sqrt[Tan[c + d*x]])))/(3*a)))/(5*
a)))/(7*a)))/(9*a)))/(11*a))/5)/4

Maple [B] (warning: unable to verify)

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

Time = 5.88 (sec) , antiderivative size = 2660696, normalized size of antiderivative = 5784.12

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

[In]

int((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),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. 18981 vs. \(2 (396) = 792\).

Time = 4.91 (sec) , antiderivative size = 18981, normalized size of antiderivative = 41.26 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(13/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{13/2}} \,d x \]

[In]

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2))/tan(c + d*x)^(13/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2))/tan(c + d*x)^(13/2), x)