3.98 \(\int (-a+b \cot (c+d x)) (a+b \cot (c+d x))^{5/2} \, dx\)
Optimal. Leaf size=151 \[ \frac{2 b \left (a^2+b^2\right ) \sqrt{a+b \cot (c+d x)}}{d}-\frac{2 b (a+b \cot (c+d x))^{5/2}}{5 d}-\frac{(-b+i a) (a-i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{(a+i b)^{5/2} (b+i a) \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a+i b}}\right )}{d} \]
[Out]
-(((I*a - b)*(a - I*b)^(5/2)*ArcTanh[Sqrt[a + b*Cot[c + d*x]]/Sqrt[a - I*b]])/d) + ((a + I*b)^(5/2)*(I*a + b)*
ArcTanh[Sqrt[a + b*Cot[c + d*x]]/Sqrt[a + I*b]])/d + (2*b*(a^2 + b^2)*Sqrt[a + b*Cot[c + d*x]])/d - (2*b*(a +
b*Cot[c + d*x])^(5/2))/(5*d)
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Rubi [A] time = 0.277705, antiderivative size = 151, normalized size of antiderivative = 1.,
number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used
= {3528, 12, 3482, 3539, 3537, 63, 208} \[ \frac{2 b \left (a^2+b^2\right ) \sqrt{a+b \cot (c+d x)}}{d}-\frac{2 b (a+b \cot (c+d x))^{5/2}}{5 d}-\frac{(-b+i a) (a-i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{(a+i b)^{5/2} (b+i a) \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a+i b}}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[(-a + b*Cot[c + d*x])*(a + b*Cot[c + d*x])^(5/2),x]
[Out]
-(((I*a - b)*(a - I*b)^(5/2)*ArcTanh[Sqrt[a + b*Cot[c + d*x]]/Sqrt[a - I*b]])/d) + ((a + I*b)^(5/2)*(I*a + b)*
ArcTanh[Sqrt[a + b*Cot[c + d*x]]/Sqrt[a + I*b]])/d + (2*b*(a^2 + b^2)*Sqrt[a + b*Cot[c + d*x]])/d - (2*b*(a +
b*Cot[c + d*x])^(5/2))/(5*d)
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 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 3482
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]
Rule 3539
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*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 3537
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Rule 63
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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int (-a+b \cot (c+d x)) (a+b \cot (c+d x))^{5/2} \, dx &=-\frac{2 b (a+b \cot (c+d x))^{5/2}}{5 d}+\int \left (-a^2-b^2\right ) (a+b \cot (c+d x))^{3/2} \, dx\\ &=-\frac{2 b (a+b \cot (c+d x))^{5/2}}{5 d}+\left (-a^2-b^2\right ) \int (a+b \cot (c+d x))^{3/2} \, dx\\ &=\frac{2 b \left (a^2+b^2\right ) \sqrt{a+b \cot (c+d x)}}{d}-\frac{2 b (a+b \cot (c+d x))^{5/2}}{5 d}+\left (-a^2-b^2\right ) \int \frac{a^2-b^2+2 a b \cot (c+d x)}{\sqrt{a+b \cot (c+d x)}} \, dx\\ &=\frac{2 b \left (a^2+b^2\right ) \sqrt{a+b \cot (c+d x)}}{d}-\frac{2 b (a+b \cot (c+d x))^{5/2}}{5 d}-\frac{1}{2} \left ((a-i b)^2 \left (a^2+b^2\right )\right ) \int \frac{1+i \cot (c+d x)}{\sqrt{a+b \cot (c+d x)}} \, dx-\frac{1}{2} \left ((a+i b)^2 \left (a^2+b^2\right )\right ) \int \frac{1-i \cot (c+d x)}{\sqrt{a+b \cot (c+d x)}} \, dx\\ &=\frac{2 b \left (a^2+b^2\right ) \sqrt{a+b \cot (c+d x)}}{d}-\frac{2 b (a+b \cot (c+d x))^{5/2}}{5 d}-\frac{\left ((a+i b)^3 (i a+b)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \cot (c+d x)\right )}{2 d}-\frac{\left ((a+i b) (i a+b)^3\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \cot (c+d x)\right )}{2 d}\\ &=\frac{2 b \left (a^2+b^2\right ) \sqrt{a+b \cot (c+d x)}}{d}-\frac{2 b (a+b \cot (c+d x))^{5/2}}{5 d}-\frac{\left ((a-i b)^3 (a+i b)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i a}{b}+\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \cot (c+d x)}\right )}{b d}-\frac{\left ((a-i b) (a+i b)^3\right ) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i a}{b}-\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \cot (c+d x)}\right )}{b d}\\ &=-\frac{(i a-b) (a-i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{(a+i b)^{5/2} (i a+b) \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a+i b}}\right )}{d}+\frac{2 b \left (a^2+b^2\right ) \sqrt{a+b \cot (c+d x)}}{d}-\frac{2 b (a+b \cot (c+d x))^{5/2}}{5 d}\\ \end{align*}
Mathematica [A] time = 3.88035, size = 253, normalized size = 1.68 \[ \frac{\sin (c+d x) (b \cot (c+d x)-a) (a+b \cot (c+d x))^{5/2} \left (\frac{2 b \left (-4 a^2+2 a b \cot (c+d x)+b^2 \csc ^2(c+d x)-6 b^2\right )}{(a+b \cot (c+d x))^2}+\frac{5 i \left (a^2+b^2\right ) \left ((a-i b)^2 \sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a-i b}}\right )-\sqrt{a-i b} (a+i b)^2 \tanh ^{-1}\left (\frac{\sqrt{a+b \cot (c+d x)}}{\sqrt{a+i b}}\right )\right )}{\sqrt{a-i b} \sqrt{a+i b} (a+b \cot (c+d x))^{5/2}}\right )}{5 d (a \sin (c+d x)-b \cos (c+d x))} \]
Antiderivative was successfully verified.
[In]
Integrate[(-a + b*Cot[c + d*x])*(a + b*Cot[c + d*x])^(5/2),x]
[Out]
((-a + b*Cot[c + d*x])*(a + b*Cot[c + d*x])^(5/2)*(((5*I)*(a^2 + b^2)*((a - I*b)^2*Sqrt[a + I*b]*ArcTanh[Sqrt[
a + b*Cot[c + d*x]]/Sqrt[a - I*b]] - Sqrt[a - I*b]*(a + I*b)^2*ArcTanh[Sqrt[a + b*Cot[c + d*x]]/Sqrt[a + I*b]]
))/(Sqrt[a - I*b]*Sqrt[a + I*b]*(a + b*Cot[c + d*x])^(5/2)) + (2*b*(-4*a^2 - 6*b^2 + 2*a*b*Cot[c + d*x] + b^2*
Csc[c + d*x]^2))/(a + b*Cot[c + d*x])^2)*Sin[c + d*x])/(5*d*(-(b*Cos[c + d*x]) + a*Sin[c + d*x]))
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Maple [B] time = 0.049, size = 1375, normalized size = 9.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((-a+b*cot(d*x+c))*(a+b*cot(d*x+c))^(5/2),x)
[Out]
-2/5*b*(a+b*cot(d*x+c))^(5/2)/d+2/d*b*a^2*(a+b*cot(d*x+c))^(1/2)+2/d*b^3*(a+b*cot(d*x+c))^(1/2)-1/4/d/b*ln(b*c
ot(d*x+c)+a+(a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2
)*(a^2+b^2)^(1/2)*a^3-1/4/d*b*ln(b*cot(d*x+c)+a+(a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)
^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a+1/4/d/b*ln(b*cot(d*x+c)+a+(a+b*cot(d*x+c))^(1/2)*(2*(a
^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^4-1/4/d*b^3*ln(b*cot(d*x+c)+a+(a+b*c
ot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+2/d*b/(2*(a^2+b^
2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*cot(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1
/2))*a^3+2/d*b^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*cot(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))
/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*cot(d*x+c))^(1/2)+(2*(a^2
+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(a^2+b^2)^(1/2)*a^2-1/d*b^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/
2)*arctan((2*(a+b*cot(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(a^2+b^2)^(1
/2)+1/4/d/b*ln((a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*cot(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^
2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a^3+1/4/d*b*ln((a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*cot(
d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a-1/4/d/b*ln((a+b*cot(d*x+c))^(1/2)*(2
*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*cot(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^4+1/4/d*b^3*ln((a+
b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*cot(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2
)-2/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*cot(d*x+c))^(1/2))/(2*(a^2+
b^2)^(1/2)-2*a)^(1/2))*a^3-2/d*b^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*
cot(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(
1/2)+2*a)^(1/2)-2*(a+b*cot(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(a^2+b^2)^(1/2)*a^2+1/d*b^3/(2*(a^2+b
^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*cot(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(
1/2))*(a^2+b^2)^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cot \left (d x + c\right ) + a\right )}^{\frac{5}{2}}{\left (b \cot \left (d x + c\right ) - a\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-a+b*cot(d*x+c))*(a+b*cot(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
integrate((b*cot(d*x + c) + a)^(5/2)*(b*cot(d*x + c) - a), x)
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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*cot(d*x+c))*(a+b*cot(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
Timed out
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Sympy [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*cot(d*x+c))*(a+b*cot(d*x+c))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cot \left (d x + c\right ) + a\right )}^{\frac{5}{2}}{\left (b \cot \left (d x + c\right ) - a\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-a+b*cot(d*x+c))*(a+b*cot(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((b*cot(d*x + c) + a)^(5/2)*(b*cot(d*x + c) - a), x)