\(\int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx\) [230]

Optimal result

Integrand size = 19, antiderivative size = 100 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{7/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{7/2}}+\frac {2}{5 b d (d \cos (a+b x))^{5/2}}+\frac {2}{b d^3 \sqrt {d \cos (a+b x)}} \]

[Out]

arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(7/2)-arctanh((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(7/2)+2/5/b/d/(d*cos(
b*x+a))^(5/2)+2/b/d^3/(d*cos(b*x+a))^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2645, 331, 335, 304, 209, 212} \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{7/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{7/2}}+\frac {2}{b d^3 \sqrt {d \cos (a+b x)}}+\frac {2}{5 b d (d \cos (a+b x))^{5/2}} \]

[In]

Int[Csc[a + b*x]/(d*Cos[a + b*x])^(7/2),x]

[Out]

ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*d^(7/2)) - ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*d^(7/2)) + 2/(5*b*
d*(d*Cos[a + b*x])^(5/2)) + 2/(b*d^3*Sqrt[d*Cos[a + b*x]])

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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 {1}{x^{7/2} \left (1-\frac {x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{b d} \\ & = \frac {2}{5 b d (d \cos (a+b x))^{5/2}}-\frac {\text {Subst}\left (\int \frac {1}{x^{3/2} \left (1-\frac {x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{b d^3} \\ & = \frac {2}{5 b d (d \cos (a+b x))^{5/2}}+\frac {2}{b d^3 \sqrt {d \cos (a+b x)}}-\frac {\text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{b d^5} \\ & = \frac {2}{5 b d (d \cos (a+b x))^{5/2}}+\frac {2}{b d^3 \sqrt {d \cos (a+b x)}}-\frac {2 \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{d^2}} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d^5} \\ & = \frac {2}{5 b d (d \cos (a+b x))^{5/2}}+\frac {2}{b d^3 \sqrt {d \cos (a+b x)}}-\frac {\text {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d^3}+\frac {\text {Subst}\left (\int \frac {1}{d+x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d^3} \\ & = \frac {\arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{7/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{7/2}}+\frac {2}{5 b d (d \cos (a+b x))^{5/2}}+\frac {2}{b d^3 \sqrt {d \cos (a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.81 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {5 \arctan \left (\sqrt {\cos (a+b x)}\right ) \sqrt {\cos (a+b x)}-5 \text {arctanh}\left (\sqrt {\cos (a+b x)}\right ) \sqrt {\cos (a+b x)}+2 \left (5+\sec ^2(a+b x)\right )}{5 b d^3 \sqrt {d \cos (a+b x)}} \]

[In]

Integrate[Csc[a + b*x]/(d*Cos[a + b*x])^(7/2),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(861\) vs. \(2(82)=164\).

Time = 0.14 (sec) , antiderivative size = 862, normalized size of antiderivative = 8.62

method	result	size
default	\(\text {Expression too large to display}\)	\(862\)

[In]

int(csc(b*x+a)/(d*cos(b*x+a))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/10/d^(9/2)/(-d)^(1/2)/(8*sin(1/2*b*x+1/2*a)^6-12*sin(1/2*b*x+1/2*a)^4+6*sin(1/2*b*x+1/2*a)^2-1)*(10*d^(3/2)*
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))-24*(-d)^(1/2)*d^(1/2)*(-2*d*sin(1/
2*b*x+1/2*a)^2+d)^(1/2)+5*(-d)^(1/2)*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+5*(-d)^(1/2)*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*s
in(1/2*b*x+1/2*a)^2+d)^(1/2)+d))*d-40*(2*d^(3/2)*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))+(-d)^(1/2)*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+(-d)^(1/2)*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)*sin(1/2*b*x+1/2*a)^6-20*(-6*d^(3/2)*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))+4*(-d)^(1/2)*d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-3*(-d)^(1/2)*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-3*(-d)^(1/2)*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)*sin(1/2*b*x+1/2*
a)^4+10*(-6*d^(3/2)*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))+8*(-d)^(1/2)*d
^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-3*(-d)^(1/2)*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-3*(-d)^(1/2)*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)*sin(1/2*b*x+1/2*a)^2)/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (82) = 164\).

Time = 0.35 (sec) , antiderivative size = 342, normalized size of antiderivative = 3.42 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\left [\frac {10 \, \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right )^{3} - 5 \, \sqrt {-d} \cos \left (b x + a\right )^{3} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} + 1\right )}}{20 \, b d^{4} \cos \left (b x + a\right )^{3}}, \frac {10 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right )^{3} + 5 \, \sqrt {d} \cos \left (b x + a\right )^{3} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} + 1\right )}}{20 \, b d^{4} \cos \left (b x + a\right )^{3}}\right ] \]

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(7/2),x, algorithm="fricas")

[Out]

[1/20*(10*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) + 1)/(d*cos(b*x + a)))*cos(b*x + a)^
3 - 5*sqrt(-d)*cos(b*x + a)^3*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)) + 8*sqrt(d*cos(b*x + a))*(5*cos(b*x + a)^2 + 1))/(b*
d^4*cos(b*x + a)^3), 1/20*(10*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*cos(b*x + a)
))*cos(b*x + a)^3 + 5*sqrt(d)*cos(b*x + a)^3*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)) + 8*sqrt(d*cos(b*x + a))*(5*cos(b*x +
a)^2 + 1))/(b*d^4*cos(b*x + a)^3)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {\frac {10 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {5 \, \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {4 \, {\left (5 \, d^{2} \cos \left (b x + a\right )^{2} + d^{2}\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} d^{2}}}{10 \, b d} \]

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

1/10*(10*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(5/2) + 5*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x
+ a)) + sqrt(d)))/d^(5/2) + 4*(5*d^2*cos(b*x + a)^2 + d^2)/((d*cos(b*x + a))^(5/2)*d^2))/(b*d)

Giac [F]

\[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\int { \frac {\csc \left (b x + a\right )}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(7/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)/(d*cos(b*x + a))^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\int \frac {1}{\sin \left (a+b\,x\right )\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2}} \,d x \]

[In]

int(1/(sin(a + b*x)*(d*cos(a + b*x))^(7/2)),x)

[Out]

int(1/(sin(a + b*x)*(d*cos(a + b*x))^(7/2)), x)