3.606 \(\int \frac{\cot ^{\frac{5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\)
Optimal. Leaf size=156 \[ -\frac{2 B \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{B \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{B \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{B \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d} \]
[Out]
-((B*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d)) + (B*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2
]*d) - (2*B*Cot[c + d*x]^(3/2))/(3*d) + (B*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d) -
(B*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d)
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Rubi [A] time = 0.105383, antiderivative size = 156, normalized size of antiderivative = 1.,
number of steps used = 13, number of rules used = 10, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used
= {21, 3473, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{2 B \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{B \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{B \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{B \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d} \]
Antiderivative was successfully verified.
[In]
Int[(Cot[c + d*x]^(5/2)*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x]
[Out]
-((B*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d)) + (B*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2
]*d) - (2*B*Cot[c + d*x]^(3/2))/(3*d) + (B*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d) -
(B*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d)
Rule 21
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])
Rule 3473
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Rule 3476
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Rule 329
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 297
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))
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{\cot ^{\frac{5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx &=B \int \cot ^{\frac{5}{2}}(c+d x) \, dx\\ &=-\frac{2 B \cot ^{\frac{3}{2}}(c+d x)}{3 d}-B \int \sqrt{\cot (c+d x)} \, dx\\ &=-\frac{2 B \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{B \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{2 B \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{(2 B) \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=-\frac{2 B \cot ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{B \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}+\frac{B \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=-\frac{2 B \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{B \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 d}+\frac{B \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 d}+\frac{B \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{B \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{2 B \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{B \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}-\frac{B \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}+\frac{B \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{B \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}\\ &=-\frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{B \tan ^{-1}\left (1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{2 B \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{B \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}-\frac{B \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}\\ \end{align*}
Mathematica [C] time = 0.0770606, size = 38, normalized size = 0.24 \[ \frac{2 B \cot ^{\frac{3}{2}}(c+d x) \left (\text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},-\cot ^2(c+d x)\right )-1\right )}{3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cot[c + d*x]^(5/2)*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x]
[Out]
(2*B*Cot[c + d*x]^(3/2)*(-1 + Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2]))/(3*d)
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Maple [C] time = 0.25, size = 1275, normalized size = 8.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)^(5/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x)
[Out]
1/6*B/d*2^(1/2)*(cos(d*x+c)-1)^2*(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))*cos(d*x+c)*((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)*sin(d*x+c)-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))*cos(d*x+c)*((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)*sin(d*x+c)+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))*((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)*sin(d*x+c)-3*(-(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)/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)*sin(d*x+c)-3*I*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/si
n(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((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)*sin(d*x+c)-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))*cos(d*x+c)*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+si
n(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)+6*(-(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)/sin(d*x+c))^(1/2)*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)*sin(d*x+c)-3*(-(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)/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))*sin(d*x+c)-3*((cos(d*x+c)-1)/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)-1-sin(d*x+c))/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))*sin(d*x+c)+6*(-(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)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1
-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*sin(d*x+c)-2*2^(1/2)*cos(d*x+c)^2)*(cos(d*x+c)+1)^2*(cos(d*x+c)/si
n(d*x+c))^(5/2)/cos(d*x+c)^3/sin(d*x+c)^3
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Maxima [A] time = 1.49383, size = 171, normalized size = 1.1 \begin{align*} \frac{3 \,{\left (2 \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) - \sqrt{2} \log \left (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt{2} \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right )\right )} B - \frac{8 \, B}{\tan \left (d x + c\right )^{\frac{3}{2}}}}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(5/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="maxima")
[Out]
1/12*(3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(
2) - 2/sqrt(tan(d*x + c)))) - sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(2)*log(-sqrt
(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*B - 8*B/tan(d*x + c)^(3/2))/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(cot(d*x+c)^(5/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),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(cot(d*x+c)**(5/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B b \tan \left (d x + c\right ) + B a\right )} \cot \left (d x + c\right )^{\frac{5}{2}}}{b \tan \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(5/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*b*tan(d*x + c) + B*a)*cot(d*x + c)^(5/2)/(b*tan(d*x + c) + a), x)