3.818 \(\int \frac{(a+b \tan (c+d x))^3}{\sqrt{\cot (c+d x)}} \, dx\)

Optimal. Leaf size=272 \[ \frac{2 b \left (3 a^2-b^2\right )}{d \sqrt{\cot (c+d x)}}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 b^2 (a \cot (c+d x)+b)}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{8 a b^2}{5 d \cot ^{\frac{3}{2}}(c+d x)} \]

[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) + (8*a*b^2)/(5*d*Cot[c + d*x]^(3/2)) + (2*b*(3*a^2 - b^2
))/(d*Sqrt[Cot[c + d*x]]) + (2*b^2*(b + a*Cot[c + d*x]))/(5*d*Cot[c + d*x]^(5/2)) + ((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 + Sq
rt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d)

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Rubi [A]  time = 0.392537, antiderivative size = 272, normalized size of antiderivative = 1., 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, 3565, 3628, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{2 b \left (3 a^2-b^2\right )}{d \sqrt{\cot (c+d x)}}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 b^2 (a \cot (c+d x)+b)}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{8 a b^2}{5 d \cot ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^3/Sqrt[Cot[c + d*x]],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) + (8*a*b^2)/(5*d*Cot[c + d*x]^(3/2)) + (2*b*(3*a^2 - b^2
))/(d*Sqrt[Cot[c + d*x]]) + (2*b^2*(b + a*Cot[c + d*x]))/(5*d*Cot[c + d*x]^(5/2)) + ((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 + Sq
rt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d)

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]

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]

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 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 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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.492395, size = 102, normalized size = 0.38 \[ \frac{2 \left (\left (3 b^3-9 a^2 b\right ) \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-\cot ^2(c+d x)\right )+a \left (a (5 a \cot (c+d x)+9 b)-5 \left (a^2-3 b^2\right ) \cot (c+d x) \, _2F_1\left (-\frac{3}{4},1;\frac{1}{4};-\cot ^2(c+d x)\right )\right )\right )}{15 d \cot ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.365, size = 2508, normalized size = 9.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^3/cot(d*x+c)^(1/2),x)

[Out]

-1/10/d*2^(1/2)*(cos(d*x+c)-1)*(30*sin(d*x+c)*a^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))
/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticF((-(cos(d*x+c)-1-sin(d*
x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*b-5*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+
c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(
d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a^3+5*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*
x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((co
s(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*b^3+5*I*sin(d*x+c)*a^3*((cos(d*x+c)-1)/sin(d*
x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x
+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+5*I*sin(d*x+c)*b^3*((cos
(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+
c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-5*I*sin
(d*x+c)*a^3*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin
(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2
*2^(1/2))-5*I*sin(d*x+c)*b^3*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-
(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1
/2),1/2+1/2*I,1/2*2^(1/2))-15*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x
+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin
(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a^2*b+15*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-s
in(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)
-1+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a*b^2-15*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi
((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^
(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a^2*b+15*((cos(d*x+c)-1)/sin(d*x+c)
)^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-(cos(d*x+c)-1-sin(d*
x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a*b^2+12*2^(1/2)*
b^3*cos(d*x+c)^3-12*2^(1/2)*b^3*cos(d*x+c)^2-2*cos(d*x+c)*2^(1/2)*b^3+2*b^3*2^(1/2)+10*cos(d*x+c)*sin(d*x+c)*2
^(1/2)*b^2*a-10*sin(d*x+c)*2^(1/2)*b^2*cos(d*x+c)^2*a-30*2^(1/2)*a^2*cos(d*x+c)^3*b+30*2^(1/2)*a^2*cos(d*x+c)^
2*b-5*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2
*2^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c
)^2*sin(d*x+c)*a^3+5*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2
),1/2+1/2*I,1/2*2^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^
(1/2)*cos(d*x+c)^2*sin(d*x+c)*b^3-15*I*sin(d*x+c)*a^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x
+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-s
in(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*b-15*I*sin(d*x+c)*b^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((
cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*Elliptic
Pi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a+15*I*sin(d*x+c)*a^2*((cos(d*x+c)-1)/
sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*c
os(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b+15*I*sin(d*x+c)*
b^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))
/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2)
)*a-10*sin(d*x+c)*b^3*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*
x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*
2^(1/2)))*(cos(d*x+c)+1)^2/cos(d*x+c)^2/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x+c))^(1/2)

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Maxima [A]  time = 1.72075, size = 328, normalized size = 1.21 \begin{align*} \frac{8 \,{\left (b^{3} + \frac{5 \, a b^{2}}{\tan \left (d x + c\right )} + \frac{5 \,{\left (3 \, a^{2} b - b^{3}\right )}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac{5}{2}} - 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 )}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^3/cot(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/20*(8*(b^3 + 5*a*b^2/tan(d*x + c) + 5*(3*a^2*b - b^3)/tan(d*x + c)^2)*tan(d*x + c)^(5/2) - 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*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)))) - 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))/d

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^3/cot(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \tan{\left (c + d x \right )}\right )^{3}}{\sqrt{\cot{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**3/cot(d*x+c)**(1/2),x)

[Out]

Integral((a + b*tan(c + d*x))**3/sqrt(cot(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (d x + c\right ) + a\right )}^{3}}{\sqrt{\cot \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^3/cot(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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