3.9.16 \(\int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^3 \, dx\) [816]
Optimal. Leaf size=245 \[ -\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {2 a \left (a^2+b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {2 b^2 (b+a \cot (c+d x))}{d \sqrt {\cot (c+d x)}}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d} \]
[Out]
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*b^2*(b+a*cot(d*x+c))/d/cot(d*
x+c)^(1/2)-2*a*(a^2+b^2)*cot(d*x+c)^(1/2)/d
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Rubi [A]
time = 0.24, 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 = {3754, 3646,
3711, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\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 ) \text {ArcTan}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}-\frac {2 a \left (a^2+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 {2 b^2 (a \cot (c+d x)+b)}{d \sqrt {\cot (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^(3/2)*(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 + b^2)*Sqrt[Cot[c + d*x]])/d + (2*b^2*(b
+ a*Cot[c + d*x]))/(d*Sqrt[Cot[c + d*x]]) - ((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 210
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 631
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 642
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 1176
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 1179
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 1182
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 3615
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 3646
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 3711
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 3754
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 {3}{2}}(c+d x) (a+b \tan (c+d x))^3 \, dx &=\int \frac {(b+a \cot (c+d x))^3}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 (b+a \cot (c+d x))}{d \sqrt {\cot (c+d x)}}-2 \int \frac {-2 a b^2-\frac {1}{2} b \left (3 a^2-b^2\right ) \cot (c+d x)-\frac {1}{2} a \left (a^2+b^2\right ) \cot ^2(c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=-\frac {2 a \left (a^2+b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {2 b^2 (b+a \cot (c+d x))}{d \sqrt {\cot (c+d x)}}-2 \int \frac {\frac {1}{2} a \left (a^2-3 b^2\right )-\frac {1}{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+b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {2 b^2 (b+a \cot (c+d x))}{d \sqrt {\cot (c+d x)}}-\frac {4 \text {Subst}\left (\int \frac {-\frac {1}{2} a \left (a^2-3 b^2\right )+\frac {1}{2} b \left (3 a^2-b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=-\frac {2 a \left (a^2+b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {2 b^2 (b+a \cot (c+d x))}{d \sqrt {\cot (c+d x)}}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {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 ) \text {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+b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {2 b^2 (b+a \cot (c+d x))}{d \sqrt {\cot (c+d x)}}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {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 ) \text {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 ) \text {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 ) \text {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+b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {2 b^2 (b+a \cot (c+d x))}{d \sqrt {\cot (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 ) \text {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 ) \text {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+b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {2 b^2 (b+a \cot (c+d x))}{d \sqrt {\cot (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] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 3.20, size = 188, normalized size = 0.77 \begin {gather*} -\frac {2 \left (-3 a^2 b+a^3 \cot (c+d x)-b \left (-3 a^2+b^2\right ) \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\cot ^2(c+d x)\right )+\frac {a \left (a^2-3 b^2\right ) \sqrt {\cot (c+d x)} \left (2 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )+\log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )\right )}{4 \sqrt {2}}\right )}{d \sqrt {\cot (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^3,x]
[Out]
(-2*(-3*a^2*b + a^3*Cot[c + d*x] - b*(-3*a^2 + b^2)*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2] + (a*(a^2
- 3*b^2)*Sqrt[Cot[c + d*x]]*(2*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x
]]] + Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])
)/(4*Sqrt[2])))/(d*Sqrt[Cot[c + d*x]])
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 52.28, size = 4227, normalized size = 17.25
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\(\text {Expression too large to display}\) |
\(4227\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
[Out]
1/2/d*(-3*I*cos(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))*((cos(d*
x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))
^(1/2)*a^2*b-2*cos(d*x+c)*2^(1/2)*a^3-3*I*cos(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))*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(
d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*a*b^2+3*I*cos(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+co
s(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))*((1
-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*a^2*b+3*I*cos(d*x+c)*((cos(d*x+c)-1+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)
)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*a*b^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)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(
d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*a^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))*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d
*x+c))/sin(d*x+c))^(1/2)*b^3+((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*E
llipticPi(((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)*a^3-((cos(d*x+c)-1+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))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)
*b^3+2*((cos(d*x+c)-1+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))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*b^3-I*((cos(d*x+c)-1
+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))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*a^3-I*((cos(d*x+c)-1+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))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*b^3+I*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)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin
(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*a^3+I*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)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*b^3+cos(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))*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*a^3-cos(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))*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*b^3+cos(d*x+c)*((cos(d*x+c)-1+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)
)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*a^3-cos(d*x+c)*((cos(d*x+c)-1+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))
*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*b^3+2*cos(d*x+c)*((cos(d*x+c)-1+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))*((1-cos(d
*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*b^3+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))*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(
d*x+c))/sin(d*x+c))^(1/2)*a^2*b-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
))*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/
sin(d*x+c))^(1/2)*a*b^2+3*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*Elli
pticPi(((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)*a^2*b-3*((cos(d*x+c)-1+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))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2
)*a*b^2-6*((cos(d*x+c)-1+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))*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*a^2*b-3*I*Ellipti
cPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2...
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Maxima [A]
time = 0.50, size = 218, normalized size = 0.89 \begin {gather*} \frac {8 \, b^{3} \sqrt {\tan \left (d x + c\right )} + 2 \, \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 ) + 2 \, \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 ) + \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 ) - \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 {8 \, a^{3}}{\sqrt {\tan \left (d x + c\right )}}}{4 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^3,x, algorithm="maxima")
[Out]
1/4*(8*b^3*sqrt(tan(d*x + c)) + 2*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)))) + 2*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
)))) + 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) - 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*a^3/sqrt(tan(d*x + c
)))/d
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^3,x, algorithm="fricas")
[Out]
Timed out
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)**(3/2)*(a+b*tan(d*x+c))**3,x)
[Out]
Integral((a + b*tan(c + d*x))**3*cot(c + d*x)**(3/2), x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((b*tan(d*x + c) + a)^3*cot(d*x + c)^(3/2), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(c + d*x)^(3/2)*(a + b*tan(c + d*x))^3,x)
[Out]
int(cot(c + d*x)^(3/2)*(a + b*tan(c + d*x))^3, x)
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