\(\int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx\) [224]

Optimal result

Integrand size = 19, antiderivative size = 78 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\frac {d^{5/2} \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}-\frac {d^{5/2} \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}+\frac {2 d (d \cos (a+b x))^{3/2}}{3 b} \]

[Out]

d^(5/2)*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b-d^(5/2)*arctanh((d*cos(b*x+a))^(1/2)/d^(1/2))/b+2/3*d*(d*cos(b*
x+a))^(3/2)/b

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2645, 327, 335, 304, 209, 212} \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\frac {d^{5/2} \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}-\frac {d^{5/2} \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}+\frac {2 d (d \cos (a+b x))^{3/2}}{3 b} \]

[In]

Int[(d*Cos[a + b*x])^(5/2)*Csc[a + b*x],x]

[Out]

(d^(5/2)*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/b - (d^(5/2)*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/b + (2*d*(d
*Cos[a + b*x])^(3/2))/(3*b)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

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

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

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 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {x^{5/2}}{1-\frac {x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{b d} \\ & = \frac {2 d (d \cos (a+b x))^{3/2}}{3 b}-\frac {d \text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{b} \\ & = \frac {2 d (d \cos (a+b x))^{3/2}}{3 b}-\frac {(2 d) \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{d^2}} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b} \\ & = \frac {2 d (d \cos (a+b x))^{3/2}}{3 b}-\frac {d^3 \text {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b}+\frac {d^3 \text {Subst}\left (\int \frac {1}{d+x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b} \\ & = \frac {d^{5/2} \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}-\frac {d^{5/2} \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}+\frac {2 d (d \cos (a+b x))^{3/2}}{3 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.87 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\frac {(d \cos (a+b x))^{5/2} \left (3 \arctan \left (\sqrt {\cos (a+b x)}\right )-3 \text {arctanh}\left (\sqrt {\cos (a+b x)}\right )+2 \cos ^{\frac {3}{2}}(a+b x)\right )}{3 b \cos ^{\frac {5}{2}}(a+b x)} \]

[In]

Integrate[(d*Cos[a + b*x])^(5/2)*Csc[a + b*x],x]

[Out]

((d*Cos[a + b*x])^(5/2)*(3*ArcTan[Sqrt[Cos[a + b*x]]] - 3*ArcTanh[Sqrt[Cos[a + b*x]]] + 2*Cos[a + b*x]^(3/2)))
/(3*b*Cos[a + b*x]^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(252\) vs. \(2(62)=124\).

Time = 0.36 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.24

method	result	size
default	\(-\frac {3 d^{\frac {5}{2}} \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}+3 d^{\frac {5}{2}} \ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}+8 d^{2} \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}\, \sqrt {-d}\, \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-4 d^{2} \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}\, \sqrt {-d}+6 d^{3} \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{6 \sqrt {-d}\, b}\)	\(253\)

[In]

int((d*cos(b*x+a))^(5/2)*csc(b*x+a),x,method=_RETURNVERBOSE)

[Out]

-1/6/(-d)^(1/2)*(3*d^(5/2)*ln(2/(cos(1/2*b*x+1/2*a)-1)*(2*d*cos(1/2*b*x+1/2*a)+d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a
)^2+d)^(1/2)-d))*(-d)^(1/2)+3*d^(5/2)*ln(-2/(cos(1/2*b*x+1/2*a)+1)*(2*d*cos(1/2*b*x+1/2*a)-d^(1/2)*(-2*d*sin(1
/2*b*x+1/2*a)^2+d)^(1/2)+d))*(-d)^(1/2)+8*d^2*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)*(-d)^(1/2)*sin(1/2*b*x+1/2*a
)^2-4*d^2*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)*(-d)^(1/2)+6*d^3*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*d*sin(1
/2*b*x+1/2*a)^2+d)^(1/2)-d)))/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (62) = 124\).

Time = 0.42 (sec) , antiderivative size = 281, normalized size of antiderivative = 3.60 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\left [\frac {6 \, \sqrt {-d} d^{2} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d}}{d \cos \left (b x + a\right ) + d}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right ) + 3 \, \sqrt {-d} d^{2} \log \left (-\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right )}{12 \, b}, -\frac {6 \, d^{\frac {5}{2}} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d}}{d \cos \left (b x + a\right ) - d}\right ) - 8 \, \sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right ) - 3 \, d^{\frac {5}{2}} \log \left (-\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right )}{12 \, b}\right ] \]

[In]

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a),x, algorithm="fricas")

[Out]

[1/12*(6*sqrt(-d)*d^2*arctan(2*sqrt(d*cos(b*x + a))*sqrt(-d)/(d*cos(b*x + a) + d)) + 8*sqrt(d*cos(b*x + a))*d^
2*cos(b*x + a) + 3*sqrt(-d)*d^2*log(-(d*cos(b*x + a)^2 + 4*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) - 1) -
6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 + 2*cos(b*x + a) + 1)))/b, -1/12*(6*d^(5/2)*arctan(2*sqrt(d*cos(b*x + a)
)*sqrt(d)/(d*cos(b*x + a) - d)) - 8*sqrt(d*cos(b*x + a))*d^2*cos(b*x + a) - 3*d^(5/2)*log(-(d*cos(b*x + a)^2 -
4*sqrt(d*cos(b*x + a))*sqrt(d)*(cos(b*x + a) + 1) + 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 - 2*cos(b*x + a) +
1)))/b]

Sympy [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\text {Timed out} \]

[In]

integrate((d*cos(b*x+a))**(5/2)*csc(b*x+a),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.06 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\frac {6 \, d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) + 3 \, d^{\frac {7}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right ) + 4 \, \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} d^{2}}{6 \, b d} \]

[In]

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a),x, algorithm="maxima")

[Out]

1/6*(6*d^(7/2)*arctan(sqrt(d*cos(b*x + a))/sqrt(d)) + 3*d^(7/2)*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*c
os(b*x + a)) + sqrt(d))) + 4*(d*cos(b*x + a))^(3/2)*d^2)/(b*d)

Giac [F]

\[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right ) \,d x } \]

[In]

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a),x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(5/2)*csc(b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}}{\sin \left (a+b\,x\right )} \,d x \]

[In]

int((d*cos(a + b*x))^(5/2)/sin(a + b*x),x)

[Out]

int((d*cos(a + b*x))^(5/2)/sin(a + b*x), x)