3.67 \(\int \frac {(a+b \cot (c+d x))^3}{(e \cot (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=343 \[ -\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{7/2}}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{7/2}}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{7/2}}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{7/2}}-\frac {2 a \left (a^2-3 b^2\right )}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {8 a^2 b}{5 d e^2 (e \cot (c+d x))^{3/2}}+\frac {2 a^2 (a+b \cot (c+d x))}{5 d e (e \cot (c+d x))^{5/2}} \]

[Out]

8/5*a^2*b/d/e^2/(e*cot(d*x+c))^(3/2)+2/5*a^2*(a+b*cot(d*x+c))/d/e/(e*cot(d*x+c))^(5/2)+1/2*(a+b)*(a^2-4*a*b+b^
2)*arctan(1-2^(1/2)*(e*cot(d*x+c))^(1/2)/e^(1/2))/d/e^(7/2)*2^(1/2)-1/2*(a+b)*(a^2-4*a*b+b^2)*arctan(1+2^(1/2)
*(e*cot(d*x+c))^(1/2)/e^(1/2))/d/e^(7/2)*2^(1/2)-1/4*(a-b)*(a^2+4*a*b+b^2)*ln(e^(1/2)+cot(d*x+c)*e^(1/2)-2^(1/
2)*(e*cot(d*x+c))^(1/2))/d/e^(7/2)*2^(1/2)+1/4*(a-b)*(a^2+4*a*b+b^2)*ln(e^(1/2)+cot(d*x+c)*e^(1/2)+2^(1/2)*(e*
cot(d*x+c))^(1/2))/d/e^(7/2)*2^(1/2)-2*a*(a^2-3*b^2)/d/e^3/(e*cot(d*x+c))^(1/2)

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Rubi [A]  time = 0.56, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3565, 3628, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {2 a \left (a^2-3 b^2\right )}{d e^3 \sqrt {e \cot (c+d x)}}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{7/2}}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{7/2}}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{7/2}}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{7/2}}+\frac {8 a^2 b}{5 d e^2 (e \cot (c+d x))^{3/2}}+\frac {2 a^2 (a+b \cot (c+d x))}{5 d e (e \cot (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cot[c + d*x])^3/(e*Cot[c + d*x])^(7/2),x]

[Out]

((a + b)*(a^2 - 4*a*b + b^2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(7/2)) - ((a + b
)*(a^2 - 4*a*b + b^2)*ArcTan[1 + (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(7/2)) + (8*a^2*b)/(5*d
*e^2*(e*Cot[c + d*x])^(3/2)) - (2*a*(a^2 - 3*b^2))/(d*e^3*Sqrt[e*Cot[c + d*x]]) + (2*a^2*(a + b*Cot[c + d*x]))
/(5*d*e*(e*Cot[c + d*x])^(5/2)) - ((a - b)*(a^2 + 4*a*b + b^2)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*Sq
rt[e*Cot[c + d*x]]])/(2*Sqrt[2]*d*e^(7/2)) + ((a - b)*(a^2 + 4*a*b + b^2)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] +
Sqrt[2]*Sqrt[e*Cot[c + d*x]]])/(2*Sqrt[2]*d*e^(7/2))

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 3529

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

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 3565

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 3628

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

Rubi steps

\begin {align*} \int \frac {(a+b \cot (c+d x))^3}{(e \cot (c+d x))^{7/2}} \, dx &=\frac {2 a^2 (a+b \cot (c+d x))}{5 d e (e \cot (c+d x))^{5/2}}-\frac {2 \int \frac {-6 a^2 b e^2+\frac {5}{2} a \left (a^2-3 b^2\right ) e^2 \cot (c+d x)+\frac {1}{2} b \left (3 a^2-5 b^2\right ) e^2 \cot ^2(c+d x)}{(e \cot (c+d x))^{5/2}} \, dx}{5 e^3}\\ &=\frac {8 a^2 b}{5 d e^2 (e \cot (c+d x))^{3/2}}+\frac {2 a^2 (a+b \cot (c+d x))}{5 d e (e \cot (c+d x))^{5/2}}-\frac {2 \int \frac {\frac {5}{2} a \left (a^2-3 b^2\right ) e^3+\frac {5}{2} b \left (3 a^2-b^2\right ) e^3 \cot (c+d x)}{(e \cot (c+d x))^{3/2}} \, dx}{5 e^5}\\ &=\frac {8 a^2 b}{5 d e^2 (e \cot (c+d x))^{3/2}}-\frac {2 a \left (a^2-3 b^2\right )}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {2 a^2 (a+b \cot (c+d x))}{5 d e (e \cot (c+d x))^{5/2}}-\frac {2 \int \frac {\frac {5}{2} b \left (3 a^2-b^2\right ) e^4-\frac {5}{2} a \left (a^2-3 b^2\right ) e^4 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{5 e^7}\\ &=\frac {8 a^2 b}{5 d e^2 (e \cot (c+d x))^{3/2}}-\frac {2 a \left (a^2-3 b^2\right )}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {2 a^2 (a+b \cot (c+d x))}{5 d e (e \cot (c+d x))^{5/2}}-\frac {4 \operatorname {Subst}\left (\int \frac {-\frac {5}{2} b \left (3 a^2-b^2\right ) e^5+\frac {5}{2} a \left (a^2-3 b^2\right ) e^4 x^2}{e^2+x^4} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{5 d e^7}\\ &=\frac {8 a^2 b}{5 d e^2 (e \cot (c+d x))^{3/2}}-\frac {2 a \left (a^2-3 b^2\right )}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {2 a^2 (a+b \cot (c+d x))}{5 d e (e \cot (c+d x))^{5/2}}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{d e^3}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{d e^3}\\ &=\frac {8 a^2 b}{5 d e^2 (e \cot (c+d x))^{3/2}}-\frac {2 a \left (a^2-3 b^2\right )}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {2 a^2 (a+b \cot (c+d x))}{5 d e (e \cot (c+d x))^{5/2}}-\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{7/2}}-\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{7/2}}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{2 d e^3}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{2 d e^3}\\ &=\frac {8 a^2 b}{5 d e^2 (e \cot (c+d x))^{3/2}}-\frac {2 a \left (a^2-3 b^2\right )}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {2 a^2 (a+b \cot (c+d x))}{5 d e (e \cot (c+d x))^{5/2}}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{7/2}}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{7/2}}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{7/2}}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{7/2}}\\ &=\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{7/2}}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{7/2}}+\frac {8 a^2 b}{5 d e^2 (e \cot (c+d x))^{3/2}}-\frac {2 a \left (a^2-3 b^2\right )}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {2 a^2 (a+b \cot (c+d x))}{5 d e (e \cot (c+d x))^{5/2}}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{7/2}}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{2 \sqrt {2} d e^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.61, size = 108, normalized size = 0.31 \[ \frac {2 \left (3 a \left (a^2-3 b^2\right ) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\cot ^2(c+d x)\right )+b \left (5 \left (3 a^2-b^2\right ) \cot (c+d x) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\cot ^2(c+d x)\right )+b (9 a+5 b \cot (c+d x))\right )\right )}{15 d e (e \cot (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cot[c + d*x])^3/(e*Cot[c + d*x])^(7/2),x]

[Out]

(2*(3*a*(a^2 - 3*b^2)*Hypergeometric2F1[-5/4, 1, -1/4, -Cot[c + d*x]^2] + b*(b*(9*a + 5*b*Cot[c + d*x]) + 5*(3
*a^2 - b^2)*Cot[c + d*x]*Hypergeometric2F1[-3/4, 1, 1/4, -Cot[c + d*x]^2])))/(15*d*e*(e*Cot[c + d*x])^(5/2))

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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((a+b*cot(d*x+c))^3/(e*cot(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \cot \left (d x + c\right ) + a\right )}^{3}}{\left (e \cot \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(d*x+c))^3/(e*cot(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*cot(d*x + c) + a)^3/(e*cot(d*x + c))^(7/2), x)

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maple [B]  time = 0.46, size = 786, normalized size = 2.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cot(d*x+c))^3/(e*cot(d*x+c))^(7/2),x)

[Out]

-3/2/d/e^4*(e^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)*a^2*b+1/2/d/e^4*(e^2)^(1/4)
*2^(1/2)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)*b^3+3/4/d/e^4*(e^2)^(1/4)*2^(1/2)*ln((e*cot(d*x+c
)+(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x+c)-(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)
+(e^2)^(1/2)))*a^2*b-1/4/d/e^4*(e^2)^(1/4)*2^(1/2)*ln((e*cot(d*x+c)+(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(
e^2)^(1/2))/(e*cot(d*x+c)-(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))*b^3+3/2/d/e^4*(e^2)^(1/4)*2^(
1/2)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)*a^2*b-1/2/d/e^4*(e^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^
2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)*b^3-1/4/d*a^3/e^3*2^(1/2)/(e^2)^(1/4)*ln((e*cot(d*x+c)-(e^2)^(1/4)*(e*cot(d*x
+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x+c)+(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+3/4/d/e^3
*2^(1/2)/(e^2)^(1/4)*ln((e*cot(d*x+c)-(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x+c)+(e^2
)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))*a*b^2+1/2/d*a^3/e^3*2^(1/2)/(e^2)^(1/4)*arctan(-2^(1/2)/(e^
2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-3/2/d/e^3*2^(1/2)/(e^2)^(1/4)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2
)+1)*a*b^2-1/2/d*a^3/e^3*2^(1/2)/(e^2)^(1/4)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+3/2/d/e^3*2^(1
/2)/(e^2)^(1/4)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)*a*b^2+2/5/d*a^3/e/(e*cot(d*x+c))^(5/2)+2*a^
2*b/d/e^2/(e*cot(d*x+c))^(3/2)-2*a^3/d/e^3/(e*cot(d*x+c))^(1/2)+6/d/e^3*a/(e*cot(d*x+c))^(1/2)*b^2

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maxima [A]  time = 0.71, size = 316, normalized size = 0.92 \[ -\frac {e {\left (\frac {5 \, {\left (\frac {2 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {2 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} - \frac {\sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} + \frac {\sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}}\right )}}{e^{4}} - \frac {8 \, {\left (a^{3} e^{2} + \frac {5 \, a^{2} b e^{2}}{\tan \left (d x + c\right )} - \frac {5 \, {\left (a^{3} - 3 \, a b^{2}\right )} e^{2}}{\tan \left (d x + c\right )^{2}}\right )}}{e^{4} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {5}{2}}}\right )}}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(d*x+c))^3/(e*cot(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-1/20*e*(5*(2*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) + 2*sqrt(e/tan(d*x +
c)))/sqrt(e))/sqrt(e) + 2*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sq
rt(e/tan(d*x + c)))/sqrt(e))/sqrt(e) - sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*log(sqrt(2)*sqrt(e)*sqrt(e/tan(
d*x + c)) + e + e/tan(d*x + c))/sqrt(e) + sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*log(-sqrt(2)*sqrt(e)*sqrt(e/
tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e))/e^4 - 8*(a^3*e^2 + 5*a^2*b*e^2/tan(d*x + c) - 5*(a^3 - 3*a*b^2)*e
^2/tan(d*x + c)^2)/(e^4*(e/tan(d*x + c))^(5/2)))/d

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mupad [B]  time = 3.06, size = 1969, normalized size = 5.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cot(c + d*x))^3/(e*cot(c + d*x))^(7/2),x)

[Out]

atan((((e*cot(c + d*x))^(1/2)*(16*a^6*d^3*e^11 - 16*b^6*d^3*e^11 + 240*a^2*b^4*d^3*e^11 - 240*a^4*b^2*d^3*e^11
) + (32*b^3*d^4*e^15 - 96*a^2*b*d^4*e^15)*(-((a*b^5*6i + a^5*b*6i - a^6 + b^6 - 15*a^2*b^4 - a^3*b^3*20i + 15*
a^4*b^2)*1i)/(4*d^2*e^7))^(1/2))*(-((a*b^5*6i + a^5*b*6i - a^6 + b^6 - 15*a^2*b^4 - a^3*b^3*20i + 15*a^4*b^2)*
1i)/(4*d^2*e^7))^(1/2)*1i + ((e*cot(c + d*x))^(1/2)*(16*a^6*d^3*e^11 - 16*b^6*d^3*e^11 + 240*a^2*b^4*d^3*e^11
- 240*a^4*b^2*d^3*e^11) - (32*b^3*d^4*e^15 - 96*a^2*b*d^4*e^15)*(-((a*b^5*6i + a^5*b*6i - a^6 + b^6 - 15*a^2*b
^4 - a^3*b^3*20i + 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2))*(-((a*b^5*6i + a^5*b*6i - a^6 + b^6 - 15*a^2*b^4 - a^3*
b^3*20i + 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2)*1i)/(((e*cot(c + d*x))^(1/2)*(16*a^6*d^3*e^11 - 16*b^6*d^3*e^11 +
240*a^2*b^4*d^3*e^11 - 240*a^4*b^2*d^3*e^11) - (32*b^3*d^4*e^15 - 96*a^2*b*d^4*e^15)*(-((a*b^5*6i + a^5*b*6i
- a^6 + b^6 - 15*a^2*b^4 - a^3*b^3*20i + 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2))*(-((a*b^5*6i + a^5*b*6i - a^6 + b
^6 - 15*a^2*b^4 - a^3*b^3*20i + 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2) - ((e*cot(c + d*x))^(1/2)*(16*a^6*d^3*e^11
- 16*b^6*d^3*e^11 + 240*a^2*b^4*d^3*e^11 - 240*a^4*b^2*d^3*e^11) + (32*b^3*d^4*e^15 - 96*a^2*b*d^4*e^15)*(-((a
*b^5*6i + a^5*b*6i - a^6 + b^6 - 15*a^2*b^4 - a^3*b^3*20i + 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2))*(-((a*b^5*6i +
a^5*b*6i - a^6 + b^6 - 15*a^2*b^4 - a^3*b^3*20i + 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2) - 16*a^9*d^2*e^8 + 48*a*
b^8*d^2*e^8 + 128*a^3*b^6*d^2*e^8 + 96*a^5*b^4*d^2*e^8))*(-((a*b^5*6i + a^5*b*6i - a^6 + b^6 - 15*a^2*b^4 - a^
3*b^3*20i + 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2)*2i + atan((((e*cot(c + d*x))^(1/2)*(16*a^6*d^3*e^11 - 16*b^6*d^
3*e^11 + 240*a^2*b^4*d^3*e^11 - 240*a^4*b^2*d^3*e^11) + (32*b^3*d^4*e^15 - 96*a^2*b*d^4*e^15)*(-((a*b^5*6i + a
^5*b*6i + a^6 - b^6 + 15*a^2*b^4 - a^3*b^3*20i - 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2))*(-((a*b^5*6i + a^5*b*6i +
a^6 - b^6 + 15*a^2*b^4 - a^3*b^3*20i - 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2)*1i + ((e*cot(c + d*x))^(1/2)*(16*a^
6*d^3*e^11 - 16*b^6*d^3*e^11 + 240*a^2*b^4*d^3*e^11 - 240*a^4*b^2*d^3*e^11) - (32*b^3*d^4*e^15 - 96*a^2*b*d^4*
e^15)*(-((a*b^5*6i + a^5*b*6i + a^6 - b^6 + 15*a^2*b^4 - a^3*b^3*20i - 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2))*(-(
(a*b^5*6i + a^5*b*6i + a^6 - b^6 + 15*a^2*b^4 - a^3*b^3*20i - 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2)*1i)/(((e*cot(
c + d*x))^(1/2)*(16*a^6*d^3*e^11 - 16*b^6*d^3*e^11 + 240*a^2*b^4*d^3*e^11 - 240*a^4*b^2*d^3*e^11) - (32*b^3*d^
4*e^15 - 96*a^2*b*d^4*e^15)*(-((a*b^5*6i + a^5*b*6i + a^6 - b^6 + 15*a^2*b^4 - a^3*b^3*20i - 15*a^4*b^2)*1i)/(
4*d^2*e^7))^(1/2))*(-((a*b^5*6i + a^5*b*6i + a^6 - b^6 + 15*a^2*b^4 - a^3*b^3*20i - 15*a^4*b^2)*1i)/(4*d^2*e^7
))^(1/2) - ((e*cot(c + d*x))^(1/2)*(16*a^6*d^3*e^11 - 16*b^6*d^3*e^11 + 240*a^2*b^4*d^3*e^11 - 240*a^4*b^2*d^3
*e^11) + (32*b^3*d^4*e^15 - 96*a^2*b*d^4*e^15)*(-((a*b^5*6i + a^5*b*6i + a^6 - b^6 + 15*a^2*b^4 - a^3*b^3*20i
- 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2))*(-((a*b^5*6i + a^5*b*6i + a^6 - b^6 + 15*a^2*b^4 - a^3*b^3*20i - 15*a^4*
b^2)*1i)/(4*d^2*e^7))^(1/2) - 16*a^9*d^2*e^8 + 48*a*b^8*d^2*e^8 + 128*a^3*b^6*d^2*e^8 + 96*a^5*b^4*d^2*e^8))*(
-((a*b^5*6i + a^5*b*6i + a^6 - b^6 + 15*a^2*b^4 - a^3*b^3*20i - 15*a^4*b^2)*1i)/(4*d^2*e^7))^(1/2)*2i + ((2*a^
3*e)/5 + 2*e*cot(c + d*x)^2*(3*a*b^2 - a^3) + 2*a^2*b*e*cot(c + d*x))/(d*e^2*(e*cot(c + d*x))^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \cot {\left (c + d x \right )}\right )^{3}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(d*x+c))**3/(e*cot(d*x+c))**(7/2),x)

[Out]

Integral((a + b*cot(c + d*x))**3/(e*cot(c + d*x))**(7/2), x)

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