3.611 \(\int \frac{a B+b B \tan (c+d x)}{\cot ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\)

Optimal. Leaf size=156 \[ \frac{2 B}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{B \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{B \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{B \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d} \]

[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)/(3*d*Cot[c + d*x]^(3/2)) - (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 = 0.101635, 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, 3474, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{2 B}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{B \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{B \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{B \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Int[(a*B + b*B*Tan[c + d*x])/(Cot[c + d*x]^(5/2)*(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)/(3*d*Cot[c + d*x]^(3/2)) - (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 3474

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

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]

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])

Rubi steps

\begin{align*} \int \frac{a B+b B \tan (c+d x)}{\cot ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx &=B \int \frac{1}{\cot ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 B}{3 d \cot ^{\frac{3}{2}}(c+d x)}-B \int \frac{1}{\sqrt{\cot (c+d x)}} \, dx\\ &=\frac{2 B}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{B \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1+x^2\right )} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{2 B}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{(2 B) \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=\frac{2 B}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\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}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\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}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\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}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\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 = 0.0320397, size = 36, normalized size = 0.23 \[ \frac{2 B \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\cot ^2(c+d x)\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*B + b*B*Tan[c + d*x])/(Cot[c + d*x]^(5/2)*(a + b*Tan[c + d*x])),x]

[Out]

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

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Maple [C]  time = 0.263, size = 546, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*B+b*B*tan(d*x+c))/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x)

[Out]

1/6*B/d*2^(1/2)*(cos(d*x+c)-1)*(3*I*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/s
in(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)-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)-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))^(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))*cos(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)+2*2^(1/2)*cos(d*x+c)-2*2^(1/2))*cos(d*x+c)*(cos(d*x+c)+1)^2/sin(d*x+c)^5/(cos(d*x+
c)/sin(d*x+c))^(5/2)

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Maxima [A]  time = 1.51639, size = 173, normalized size = 1.11 \begin{align*} \frac{6 \, \sqrt{2} B \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + 6 \, \sqrt{2} B \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + 3 \, \sqrt{2} B \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} B \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) + 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((a*B+b*B*tan(d*x+c))/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(6*sqrt(2)*B*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 6*sqrt(2)*B*arctan(-1/2*sqrt(2)*(sqrt
(2) - 2/sqrt(tan(d*x + c)))) + 3*sqrt(2)*B*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - 3*sqrt(2)*B*
log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + 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((a*B+b*B*tan(d*x+c))/cot(d*x+c)^(5/2)/(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((a*B+b*B*tan(d*x+c))/cot(d*x+c)**(5/2)/(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{B b \tan \left (d x + c\right ) + B a}{{\left (b \tan \left (d x + c\right ) + a\right )} \cot \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*tan(d*x+c))/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*b*tan(d*x + c) + B*a)/((b*tan(d*x + c) + a)*cot(d*x + c)^(5/2)), x)