3.9.24 \(\int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\) [824]
Optimal. Leaf size=232 \[ -\frac {(a-b) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a-b) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {2 a^{3/2} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} \left (a^2+b^2\right ) d}+\frac {(a+b) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a+b) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d} \]
[Out]
1/2*(a-b)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)+1/2*(a-b)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))
/(a^2+b^2)/d*2^(1/2)+1/4*(a+b)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)-1/4*(a+b)*ln(1+co
t(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)-2*a^(3/2)*arctan(a^(1/2)*cot(d*x+c)^(1/2)/b^(1/2))/(a^2
+b^2)/d/b^(1/2)
________________________________________________________________________________________
Rubi [A]
time = 0.20, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps
used = 15, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3754, 3655,
3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} -\frac {(a-b) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )}+\frac {(a-b) \text {ArcTan}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )}+\frac {(a+b) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}-\frac {(a+b) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}-\frac {2 a^{3/2} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} d \left (a^2+b^2\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[1/(Cot[c + d*x]^(3/2)*(a + b*Tan[c + d*x])),x]
[Out]
-(((a - b)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d)) + ((a - b)*ArcTan[1 + Sqrt[2]*Sqrt
[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) - (2*a^(3/2)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(Sqrt[b]*(
a^2 + b^2)*d) + ((a + b)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) - ((a +
b)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d)
Rule 65
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
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 211
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
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 3655
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/
(c^2 + d^2), Int[(a + b*Tan[e + f*x])^m*(c - d*Tan[e + f*x]), x], x] + Dist[d^2/(c^2 + d^2), Int[(a + b*Tan[e
+ f*x])^m*((1 + Tan[e + f*x]^2)/(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Rule 3715
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
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 \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx &=\int \frac {1}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx\\ &=\frac {\int \frac {b-a \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{a^2+b^2}+\frac {a^2 \int \frac {1+\cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{a^2+b^2}\\ &=\frac {2 \text {Subst}\left (\int \frac {-b+a x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right ) d}+\frac {a^2 \text {Subst}\left (\int \frac {1}{\sqrt {-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right ) d}+\frac {(a-b) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {(a+b) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right ) d}\\ &=-\frac {2 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} \left (a^2+b^2\right ) d}+\frac {(a-b) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {(a-b) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {(a+b) \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} \left (a^2+b^2\right ) d}+\frac {(a+b) \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} \left (a^2+b^2\right ) d}\\ &=-\frac {2 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} \left (a^2+b^2\right ) d}+\frac {(a+b) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a+b) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a-b) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a-b) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}\\ &=-\frac {(a-b) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a-b) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {2 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} \left (a^2+b^2\right ) d}+\frac {(a+b) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a+b) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 0.17, size = 226, normalized size = 0.97 \begin {gather*} -\frac {\frac {2 a^{3/2} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} \left (a^2+b^2\right )}-\frac {2 a \cot ^{\frac {3}{2}}(c+d x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\cot ^2(c+d x)\right )}{3 \left (a^2+b^2\right )}-\frac {b \left (4 \left (\sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-\sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )\right )+2 \sqrt {2} \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )-2 \sqrt {2} \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )\right )}{8 \left (a^2+b^2\right )}}{d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[1/(Cot[c + d*x]^(3/2)*(a + b*Tan[c + d*x])),x]
[Out]
-(((2*a^(3/2)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(Sqrt[b]*(a^2 + b^2)) - (2*a*Cot[c + d*x]^(3/2)*Hy
pergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2])/(3*(a^2 + b^2)) - (b*(4*(Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c
+ d*x]]] - Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]) + 2*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Co
t[c + d*x]] - 2*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]))/(8*(a^2 + b^2)))/d)
________________________________________________________________________________________
Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 16.66, size = 1900, normalized size = 8.19
| | |
| method |
result |
size |
| | |
| default |
\(\text {Expression too large to display}\) |
\(1900\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
[Out]
-1/2/d*(cos(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))*(a^2
+b^2)^(3/2)*a-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^2+b^2)^(3/2)*
b+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^2+b^2)^(1/2)*a^3+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^2+b^2)^(1/2)*b^3-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^2+b^2)^(3/2)*a-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^2+b^2)^(3/2)*b+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^2+b^2)^(1/2)*a^3+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^2+b^2)^(1/2)*b^3-2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),a/(
b+(a^2+b^2)^(1/2)+a),1/2*2^(1/2))*a^4-I*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^2+b^2)^(1/2)*a^3+I*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^2+b^2)^(3/2)*a+I*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^2+b^2
)^(1/2)*a^3-I*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^2+b^2)^(3/2)*
a-2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),a/(b+(a^2+b^2)^(1/2)+a),1/2*2^(1/2))*a^2*b^2+2*El
lipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(a^2+b^2)^(1/2)*
a^3+2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1/2)-a),1/2*2^(1/2))*a^2*b^2+
2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),a/(b+(a^2+b^2)^(1/2)+a),1/2*2^(1/2))*(a^2+b^2)^(1/2
)*a^3+2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1/2)-a),1/2*2^(1/2))*a^4+I*
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^2+b^2)^(3/2)*b+I*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^2+b^2)^(1/2)*b^3-2*a^2*EllipticPi((-(
cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),a/(b+(a^2+b^2)^(1/2)+a),1/2*2^(1/2))*b*(a^2+b^2)^(1/2)-2*a^2*Ellipt
icPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),-a/(-b+(a^2+b^2)^(1/2)-a),1/2*2^(1/2))*b*(a^2+b^2)^(1/2)+3*
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^2+b^2)^(1/2)*a^2*b+3*Ellipt
icPi((-(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^2+b^2)^(1/2)*a*b^2+3*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^2+b^2)^(1/2)*a^2*b+3*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^2+b^2)^(1/2)*a*b^2-I*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^2+b^2)^(1/2)*b^3-I*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^2+b^2)^(3/2)*b+I*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^2+b^2)^(1/2)*a*b^2+I*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^2+b^2)^(1/2)*a^2*b-I*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^2+b^2)^(1/2)*a^2*b-I*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^2+b^2)^(1/2)*a*b^2)*((-1+cos(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)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))/(cos(d*x+c)/sin(d*x+c))^
(3/2)/sin(d*x+c)^4*2^(1/2)/(a^2+b^2)^(3/2)/(b+(a^2+b^2)^(1/2)+a)/(-b+(a^2+b^2)^(1/2)-a)
________________________________________________________________________________________
Maxima [A]
time = 0.55, size = 175, normalized size = 0.75 \begin {gather*} -\frac {\frac {8 \, a^{2} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{2} + b^{2}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left (a - b\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 - b\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 + b\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 + b\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{2} + b^{2}}}{4 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c)),x, algorithm="maxima")
[Out]
-1/4*(8*a^2*arctan(a/(sqrt(a*b)*sqrt(tan(d*x + c))))/((a^2 + b^2)*sqrt(a*b)) - (2*sqrt(2)*(a - b)*arctan(1/2*s
qrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a - b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c
)))) - sqrt(2)*(a + b)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(2)*(a + b)*log(-sqrt(2)/sqr
t(tan(d*x + c)) + 1/tan(d*x + c) + 1))/(a^2 + b^2))/d
________________________________________________________________________________________
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(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c)),x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right ) \cot ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)**(3/2)/(a+b*tan(d*x+c)),x)
[Out]
Integral(1/((a + b*tan(c + d*x))*cot(c + d*x)**(3/2)), x)
________________________________________________________________________________________
Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: AttributeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c)),x, algorithm="giac")
[Out]
Exception raised: AttributeError >> type
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(cot(c + d*x)^(3/2)*(a + b*tan(c + d*x))),x)
[Out]
int(1/(cot(c + d*x)^(3/2)*(a + b*tan(c + d*x))), x)
________________________________________________________________________________________