3.7.31 \(\int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx\) [631]

Optimal. Leaf size=318 \[ \frac {(i a-b)^{5/2} \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}} \]

[Out]

(I*a-b)^(5/2)*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)*arctanh((I*a+b)^(1
/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d-2/315*(315*a^4-483*a^2*b^2-10*b^4)*(a+b*tan(d*x+c))^(1/2)/a^2/d
/tan(d*x+c)^(1/2)-2/9*a^2*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(9/2)-38/63*a*b*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x
+c)^(7/2)+2/105*(21*a^2-25*b^2)*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)+2/315*b*(231*a^2-5*b^2)*(a+b*tan(d*x
+c))^(1/2)/a/d/tan(d*x+c)^(3/2)

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Rubi [A]
time = 1.07, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3646, 3730, 3697, 3696, 95, 209, 212} \begin {gather*} \frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}+\frac {(-b+i a)^{5/2} \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {(b+i a)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(11/2),x]

[Out]

((I*a - b)^(5/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - ((I*a + b)^(5/2)*Arc
Tanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (2*a^2*Sqrt[a + b*Tan[c + d*x]])/(9*d*T
an[c + d*x]^(9/2)) - (38*a*b*Sqrt[a + b*Tan[c + d*x]])/(63*d*Tan[c + d*x]^(7/2)) + (2*(21*a^2 - 25*b^2)*Sqrt[a
+ b*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(5/2)) + (2*b*(231*a^2 - 5*b^2)*Sqrt[a + b*Tan[c + d*x]])/(315*a*d*Tan
[c + d*x]^(3/2)) - (2*(315*a^4 - 483*a^2*b^2 - 10*b^4)*Sqrt[a + b*Tan[c + d*x]])/(315*a^2*d*Sqrt[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*} \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2}{9} \int \frac {\frac {19 a^2 b}{2}-\frac {9}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)-\frac {1}{2} b \left (8 a^2-9 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {9}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {4 \int \frac {\frac {3}{4} a^2 \left (21 a^2-25 b^2\right )+\frac {63}{4} a b \left (3 a^2-b^2\right ) \tan (c+d x)+\frac {57}{2} a^2 b^2 \tan ^2(c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{63 a}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 \int \frac {-\frac {3}{8} a^2 b \left (231 a^2-5 b^2\right )+\frac {315}{8} a^3 \left (a^2-3 b^2\right ) \tan (c+d x)+\frac {3}{2} a^2 b \left (21 a^2-25 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{315 a^2}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {16 \int \frac {-\frac {3}{16} a^2 \left (315 a^4-483 a^2 b^2-10 b^4\right )-\frac {945}{16} a^3 b \left (3 a^2-b^2\right ) \tan (c+d x)-\frac {3}{8} a^2 b^2 \left (231 a^2-5 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{945 a^3}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}+\frac {32 \int \frac {\frac {945}{32} a^4 b \left (3 a^2-b^2\right )-\frac {945}{32} a^5 \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{945 a^4}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}+\frac {1}{2} (i a-b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} (i a+b)^3 \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}+\frac {(i a-b)^3 \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 {(i a+b)^3 \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^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}+\frac {(i a-b)^3 \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)^3 \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} \tan ^{-1}\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} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 a b \sqrt {a+b \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2-25 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (231 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4-483 a^2 b^2-10 b^4\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 4.22, size = 300, normalized size = 0.94 \begin {gather*} \frac {-1260 \sqrt [4]{-1} (-a+i b)^{5/2} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-1260 \sqrt [4]{-1} (a+i b)^{5/2} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\frac {\sec ^4(c+d x) \left (-987 a^4+1374 a^2 b^2+30 b^4+4 \left (280 a^4-483 a^2 b^2-10 b^4\right ) \cos (2 (c+d x))+\left (-413 a^4+558 a^2 b^2+10 b^4\right ) \cos (4 (c+d x))+272 a^3 b \sin (2 (c+d x))-10 a b^3 \sin (2 (c+d x))-326 a^3 b \sin (4 (c+d x))+5 a b^3 \sin (4 (c+d x))\right ) \sqrt {a+b \tan (c+d x)}}{a^2 \tan ^{\frac {9}{2}}(c+d x)}}{1260 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(11/2),x]

[Out]

(-1260*(-1)^(1/4)*(-a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*
x]]] - 1260*(-1)^(1/4)*(a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c +
d*x]]] + (Sec[c + d*x]^4*(-987*a^4 + 1374*a^2*b^2 + 30*b^4 + 4*(280*a^4 - 483*a^2*b^2 - 10*b^4)*Cos[2*(c + d*
x)] + (-413*a^4 + 558*a^2*b^2 + 10*b^4)*Cos[4*(c + d*x)] + 272*a^3*b*Sin[2*(c + d*x)] - 10*a*b^3*Sin[2*(c + d*
x)] - 326*a^3*b*Sin[4*(c + d*x)] + 5*a*b^3*Sin[4*(c + d*x)])*Sqrt[a + b*Tan[c + d*x]])/(a^2*Tan[c + d*x]^(9/2)
))/(1260*d)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.90, size = 50542, normalized size = 158.94

method	result	size
default	\(\text {Expression too large to display}\)	\(50542\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),x,method=_RETURNVERBOSE)

[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((a+b*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^(5/2)/tan(d*x + c)^(11/2), 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((a+b*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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((a+b*tan(d*x+c))**(5/2)/tan(d*x+c)**(11/2),x)

[Out]

Timed out

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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((a+b*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),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 {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{11/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^(5/2)/tan(c + d*x)^(11/2),x)

[Out]

int((a + b*tan(c + d*x))^(5/2)/tan(c + d*x)^(11/2), x)

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