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

Optimal. Leaf size=245 \[ -\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 b^2 (a \cot (c+d x)+b)}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {16 a b^2}{3 d \sqrt {\cot (c+d x)}} \]

[Out]

2/3*b^2*(b+a*cot(d*x+c))/d/cot(d*x+c)^(3/2)-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)-1/2*(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+co
t(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)+16/3*a*b^2/d/cot(d*x+c)^(1/2)

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Rubi [A]  time = 0.34, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3673, 3565, 3628, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 b^2 (a \cot (c+d x)+b)}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {16 a b^2}{3 d \sqrt {\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cot[c + d*x]]*(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) + (16*a*b^2)/(3*d*Sqrt[Cot[c + d*x]]) + (2*b^2*(b + a*Co
t[c + d*x]))/(3*d*Cot[c + d*x]^(3/2)) - ((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 + 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 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]

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 \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^3 \, dx &=\int \frac {(b+a \cot (c+d x))^3}{\cot ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 (b+a \cot (c+d x))}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {2}{3} \int \frac {-4 a b^2-\frac {3}{2} b \left (3 a^2-b^2\right ) \cot (c+d x)-\frac {1}{2} a \left (3 a^2-b^2\right ) \cot ^2(c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {16 a b^2}{3 d \sqrt {\cot (c+d x)}}+\frac {2 b^2 (b+a \cot (c+d x))}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {2}{3} \int \frac {-\frac {3}{2} b \left (3 a^2-b^2\right )-\frac {3}{2} a \left (a^2-3 b^2\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=\frac {16 a b^2}{3 d \sqrt {\cot (c+d x)}}+\frac {2 b^2 (b+a \cot (c+d x))}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {4 \operatorname {Subst}\left (\int \frac {\frac {3}{2} b \left (3 a^2-b^2\right )+\frac {3}{2} a \left (a^2-3 b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{3 d}\\ &=\frac {16 a b^2}{3 d \sqrt {\cot (c+d x)}}+\frac {2 b^2 (b+a \cot (c+d x))}{3 d \cot ^{\frac {3}{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 {16 a b^2}{3 d \sqrt {\cot (c+d x)}}+\frac {2 b^2 (b+a \cot (c+d x))}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\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 {\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 {16 a b^2}{3 d \sqrt {\cot (c+d x)}}+\frac {2 b^2 (b+a \cot (c+d x))}{3 d \cot ^{\frac {3}{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 {16 a b^2}{3 d \sqrt {\cot (c+d x)}}+\frac {2 b^2 (b+a \cot (c+d x))}{3 d \cot ^{\frac {3}{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*}

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Mathematica [C]  time = 0.39, size = 100, normalized size = 0.41 \[ \frac {\left (2 b^3-6 a^2 b\right ) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\cot ^2(c+d x)\right )+6 a \left (a (a \cot (c+d x)+b)-\left (a^2-3 b^2\right ) \cot (c+d x) \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\cot ^2(c+d x)\right )\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*a^2*b + 2*b^3)*Hypergeometric2F1[-3/4, 1, 1/4, -Cot[c + d*x]^2] + 6*a*(a*(b + a*Cot[c + d*x]) - (a^2 - 3*
b^2)*Cot[c + d*x]*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2]))/(3*d*Cot[c + d*x]^(3/2))

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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(cot(d*x+c)^(1/2)*(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 \[ \int {\left (b \tan \left (d x + c\right ) + a\right )}^{3} \sqrt {\cot \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.52, size = 2370, normalized size = 9.67 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d*(-1+cos(d*x+c))*(3*I*cos(d*x+c)*sin(d*x+c)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1
/2*I,1/2*2^(1/2))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*a^3+9*I*cos(d*x+c)*sin(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(
(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(
d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a*b^2-3*I*cos(d*x+c)*sin(d*x+c)*EllipticPi(((1-cos(d*x+c)+sin
(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)
+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*b^3+9*I*cos(d*x+c)*sin(d*x+c)*EllipticPi(((1
-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*
((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*a^2*b-3*I*cos(d*x+c)*sin(d*x+
c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))
/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a^3-9*I*cos(
d*x+c)*sin(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2
))*a^2*b+3*I*cos(d*x+c)*sin(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/si
n(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2
-1/2*I,1/2*2^(1/2))*b^3-9*I*cos(d*x+c)*sin(d*x+c)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+
1/2*I,1/2*2^(1/2))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*
((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*a*b^2-3*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1
/2*2^(1/2))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+co
s(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*sin(d*x+c)*a^3+9*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2
),1/2-1/2*I,1/2*2^(1/2))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^
(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*sin(d*x+c)*a^2*b+9*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/s
in(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c
))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*sin(d*x+c)*a*b^2-3*EllipticPi(((1-cos(d*x+c
)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d
*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*sin(d*x+c)*b^3+6*((1-cos(d*x
+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a^3-18*((1-cos(
d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*a*b^2-3*((1-
cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*
x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+
c)*a^3+9*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d
*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*
x+c)*sin(d*x+c)*a^2*b+9*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(
1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*
2^(1/2))*cos(d*x+c)*sin(d*x+c)*a*b^2-3*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c)
)/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)
,1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*b^3+18*2^(1/2)*cos(d*x+c)^2*a*b^2+2*2^(1/2)*cos(d*x+c)*sin(d*x+c
)*b^3-18*cos(d*x+c)*2^(1/2)*a*b^2-2*sin(d*x+c)*2^(1/2)*b^3)*(1+cos(d*x+c))^2*(cos(d*x+c)/sin(d*x+c))^(1/2)/cos
(d*x+c)^2/sin(d*x+c)^3*2^(1/2)

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maxima [A]  time = 1.01, size = 221, normalized size = 0.90 \[ -\frac {6 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 6 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - 3 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + 3 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 8 \, {\left (b^{3} + \frac {9 \, a b^{2}}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*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)))) + 6*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)))) - 3*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) + 3*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) - 8*(b^3 + 9*a*b^2/tan(d*x + c))*tan(d*x + c)
^(3/2))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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