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

Optimal. Leaf size=270 \[ \frac {2 a \left (a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}-\frac {8 a^2 b \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 a^2 \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+b)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3673, 3566, 3630, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {2 a \left (a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}-\frac {8 a^2 b \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 a^2 \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+b)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)*(a^2 - 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) - ((a + b)*(a^2 - 4*a*b + b^2
)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) + (2*a*(a^2 - 3*b^2)*Sqrt[Cot[c + d*x]])/d - (8*a^2*b*Co
t[c + d*x]^(3/2))/(5*d) - (2*a^2*Cot[c + d*x]^(3/2)*(b + a*Cot[c + d*x]))/(5*d) + ((a - b)*(a^2 + 4*a*b + b^2)
*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d) - ((a - b)*(a^2 + 4*a*b + b^2)*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 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !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]

Rule 3673

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

Rubi steps

\begin {align*} \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^3 \, dx &=\int \sqrt {\cot (c+d x)} (b+a \cot (c+d x))^3 \, dx\\ &=-\frac {2 a^2 \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 \left (3 a^2-5 b^2\right )+\frac {5}{2} a \left (a^2-3 b^2\right ) \cot (c+d x)-6 a^2 b \cot ^2(c+d x)\right ) \, dx\\ &=-\frac {8 a^2 b \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 a^2 \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} b \left (3 a^2-b^2\right )+\frac {5}{2} a \left (a^2-3 b^2\right ) \cot (c+d x)\right ) \, dx\\ &=\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{d}-\frac {8 a^2 b \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 a^2 \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {2}{5} \int \frac {-\frac {5}{2} a \left (a^2-3 b^2\right )+\frac {5}{2} b \left (3 a^2-b^2\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{d}-\frac {8 a^2 b \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 a^2 \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} a \left (a^2-3 b^2\right )-\frac {5}{2} b \left (3 a^2-b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{5 d}\\ &=\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{d}-\frac {8 a^2 b \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 a^2 \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}-\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{d}-\frac {8 a^2 b \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 a^2 \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\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+b) \left (a^2-4 a b+b^2\right )\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-b) \left (a^2+4 a b+b^2\right )\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 ((a-b) \left (a^2+4 a b+b^2\right )\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 a \left (a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{d}-\frac {8 a^2 b \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 a^2 \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\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-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\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+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}\\ &=\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\cot (c+d x)}}{d}-\frac {8 a^2 b \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 a^2 \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\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 [C]  time = 2.35, size = 225, normalized size = 0.83 \[ -\frac {\frac {2}{5} a^3 \cot ^{\frac {5}{2}}(c+d x)+\frac {2}{3} b \left (b^2-3 a^2\right ) \cot ^{\frac {3}{2}}(c+d x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\cot ^2(c+d x)\right )-\frac {1}{4} a \left (a^2-3 b^2\right ) \left (8 \sqrt {\cot (c+d x)}+\sqrt {2} \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-\sqrt {2} \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )+2 a^2 b \cot ^{\frac {3}{2}}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a^2*b*Cot[c + d*x]^(3/2) + (2*a^3*Cot[c + d*x]^(5/2))/5 + (2*b*(-3*a^2 + b^2)*Cot[c + d*x]^(3/2)*Hypergeo
metric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2])/3 - (a*(a^2 - 3*b^2)*(2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]
] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]] + 8*Sqrt[Cot[c + d*x]] + Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Cot
[c + d*x]] + Cot[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]))/4)/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))^3,x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.53, size = 8630, normalized size = 31.96 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [A]  time = 0.87, size = 240, normalized size = 0.89 \[ -\frac {10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \frac {40 \, a^{2} b}{\tan \left (d x + c\right )^{\frac {3}{2}}} + \frac {8 \, a^{3}}{\tan \left (d x + c\right )^{\frac {5}{2}}} - \frac {40 \, {\left (a^{3} - 3 \, a b^{2}\right )}}{\sqrt {\tan \left (d x + c\right )}}}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(10*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 10*sq
rt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + 5*sqrt(2)*(a^3 +
3*a^2*b - 3*a*b^2 - b^3)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - 5*sqrt(2)*(a^3 + 3*a^2*b - 3*
a*b^2 - b^3)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + 40*a^2*b/tan(d*x + c)^(3/2) + 8*a^3/tan(d
*x + c)^(5/2) - 40*(a^3 - 3*a*b^2)/sqrt(tan(d*x + c)))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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