3.578 \(\int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=326 \[ \frac {2 \left (a^2 A-2 a b B-A b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}-\frac {2 a (5 a B+7 A b) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+b)}{5 d} \]

[Out]

-2/15*a*(7*A*b+5*B*a)*cot(d*x+c)^(3/2)/d-2/5*a*A*cot(d*x+c)^(3/2)*(b+a*cot(d*x+c))/d-1/2*(a^2*(A-B)-b^2*(A-B)-
2*a*b*(A+B))*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)-1/2*(a^2*(A-B)-b^2*(A-B)-2*a*b*(A+B))*arctan(1+2^(1
/2)*cot(d*x+c)^(1/2))/d*2^(1/2)+1/4*(2*a*b*(A-B)+a^2*(A+B)-b^2*(A+B))*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2)
)/d*2^(1/2)-1/4*(2*a*b*(A-B)+a^2*(A+B)-b^2*(A+B))*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)+2*(A*a^2
-A*b^2-2*B*a*b)*cot(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.61, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3581, 3607, 3630, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {2 \left (a^2 A-2 a b B-A b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}-\frac {2 a (5 a B+7 A b) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+b)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(7/2)*(a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

((a^2*(A - B) - b^2*(A - B) - 2*a*b*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) - ((a^2*(A -
B) - b^2*(A - B) - 2*a*b*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) + (2*(a^2*A - A*b^2 - 2*
a*b*B)*Sqrt[Cot[c + d*x]])/d - (2*a*(7*A*b + 5*a*B)*Cot[c + d*x]^(3/2))/(15*d) - (2*a*A*Cot[c + d*x]^(3/2)*(b
+ a*Cot[c + d*x]))/(5*d) + ((2*a*b*(A - B) + a^2*(A + B) - b^2*(A + B))*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + C
ot[c + d*x]])/(2*Sqrt[2]*d) - ((2*a*b*(A - B) + a^2*(A + B) - b^2*(A + B))*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]
+ Cot[c + d*x]])/(2*Sqrt[2]*d)

Rule 204

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3581

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

Rule 3607

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*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*
f*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[a^2*A*d*(m +
n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m
- 1) - b*(A*b + a*B)*d*(m + n))*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] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&
!(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\int \sqrt {\cot (c+d x)} (b+a \cot (c+d x))^2 (B+A \cot (c+d x)) \, dx\\ &=-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {2}{5} \int \sqrt {\cot (c+d x)} \left (\frac {1}{2} b (3 a A-5 b B)+\frac {5}{2} \left (a^2 A-A b^2-2 a b B\right ) \cot (c+d x)-\frac {1}{2} a (7 A b+5 a B) \cot ^2(c+d x)\right ) \, dx\\ &=-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {2}{5} \int \sqrt {\cot (c+d x)} \left (\frac {5}{2} \left (2 a A b+a^2 B-b^2 B\right )+\frac {5}{2} \left (a^2 A-A b^2-2 a b B\right ) \cot (c+d x)\right ) \, dx\\ &=\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {2}{5} \int \frac {-\frac {5}{2} \left (a^2 A-A b^2-2 a b B\right )+\frac {5}{2} \left (2 a A b+a^2 B-b^2 B\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {4 \operatorname {Subst}\left (\int \frac {\frac {5}{2} \left (a^2 A-A b^2-2 a b B\right )-\frac {5}{2} \left (2 a A b+a^2 B-b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{5 d}\\ &=\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}-\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}+\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} d}+\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} d}\\ &=\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}+\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}\\ &=\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}+\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}\\ \end {align*}

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Mathematica [A]  time = 1.86, size = 255, normalized size = 0.78 \[ -\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (30 \sqrt {2} \left (a^2 (A-B)-2 a b (A+B)+b^2 (B-A)\right ) \left (\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )-\frac {120 \left (a^2 A-2 a b B-A b^2\right )}{\sqrt {\tan (c+d x)}}-15 \sqrt {2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \left (\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )+\frac {24 a^2 A}{\tan ^{\frac {5}{2}}(c+d x)}+\frac {40 a (a B+2 A b)}{\tan ^{\frac {3}{2}}(c+d x)}\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(7/2)*(a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

-1/60*(Sqrt[Cot[c + d*x]]*(30*Sqrt[2]*(a^2*(A - B) + b^2*(-A + B) - 2*a*b*(A + B))*(ArcTan[1 - Sqrt[2]*Sqrt[Ta
n[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) - 15*Sqrt[2]*(2*a*b*(A - B) + a^2*(A + B) - b^2*(A + B)
)*(Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]) +
(24*a^2*A)/Tan[c + d*x]^(5/2) + (40*a*(2*A*b + a*B))/Tan[c + d*x]^(3/2) - (120*(a^2*A - A*b^2 - 2*a*b*B))/Sqrt
[Tan[c + d*x]])*Sqrt[Tan[c + d*x]])/d

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2} \cot \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^2*cot(d*x + c)^(7/2), x)

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maple [C]  time = 2.11, size = 13170, normalized size = 40.40 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

result too large to display

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maxima [A]  time = 1.97, size = 279, normalized size = 0.86 \[ -\frac {30 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 30 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \frac {24 \, A a^{2}}{\tan \left (d x + c\right )^{\frac {5}{2}}} - \frac {120 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )}}{\sqrt {\tan \left (d x + c\right )}} + \frac {40 \, {\left (B a^{2} + 2 \, A a b\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(30*sqrt(2)*((A - B)*a^2 - 2*(A + B)*a*b - (A - B)*b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c
)))) + 30*sqrt(2)*((A - B)*a^2 - 2*(A + B)*a*b - (A - B)*b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x +
c)))) + 15*sqrt(2)*((A + B)*a^2 + 2*(A - B)*a*b - (A + B)*b^2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c)
+ 1) - 15*sqrt(2)*((A + B)*a^2 + 2*(A - B)*a*b - (A + B)*b^2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c
) + 1) + 24*A*a^2/tan(d*x + c)^(5/2) - 120*(A*a^2 - 2*B*a*b - A*b^2)/sqrt(tan(d*x + c)) + 40*(B*a^2 + 2*A*a*b)
/tan(d*x + c)^(3/2))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^2,x)

[Out]

int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**(7/2)*(a+b*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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