3.231 \(\int \frac {\csc (a+b x)}{(d \cos (a+b x))^{9/2}} \, dx\)

Optimal. Leaf size=103 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{9/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{9/2}}+\frac {2}{3 b d^3 (d \cos (a+b x))^{3/2}}+\frac {2}{7 b d (d \cos (a+b x))^{7/2}} \]

[Out]

-arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(9/2)-arctanh((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(9/2)+2/7/b/d/(d*cos
(b*x+a))^(7/2)+2/3/b/d^3/(d*cos(b*x+a))^(3/2)

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Rubi [A]  time = 0.08, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2565, 325, 329, 212, 206, 203} \[ \frac {2}{3 b d^3 (d \cos (a+b x))^{3/2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{9/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{9/2}}+\frac {2}{7 b d (d \cos (a+b x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]/(d*Cos[a + b*x])^(9/2),x]

[Out]

-(ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*d^(9/2))) - ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*d^(9/2)) + 2/(7
*b*d*(d*Cos[a + b*x])^(7/2)) + 2/(3*b*d^3*(d*Cos[a + b*x])^(3/2))

Rule 203

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

Rule 206

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

Rule 212

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

Rule 325

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 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 2565

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*} \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{9/2}} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {1}{x^{9/2} \left (1-\frac {x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=\frac {2}{7 b d (d \cos (a+b x))^{7/2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{x^{5/2} \left (1-\frac {x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{b d^3}\\ &=\frac {2}{7 b d (d \cos (a+b x))^{7/2}}+\frac {2}{3 b d^3 (d \cos (a+b x))^{3/2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-\frac {x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{b d^5}\\ &=\frac {2}{7 b d (d \cos (a+b x))^{7/2}}+\frac {2}{3 b d^3 (d \cos (a+b x))^{3/2}}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{1-\frac {x^4}{d^2}} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d^5}\\ &=\frac {2}{7 b d (d \cos (a+b x))^{7/2}}+\frac {2}{3 b d^3 (d \cos (a+b x))^{3/2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d^4}-\frac {\operatorname {Subst}\left (\int \frac {1}{d+x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d^4}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{9/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{9/2}}+\frac {2}{7 b d (d \cos (a+b x))^{7/2}}+\frac {2}{3 b d^3 (d \cos (a+b x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 38, normalized size = 0.37 \[ \frac {2 \, _2F_1\left (-\frac {7}{4},1;-\frac {3}{4};\cos ^2(a+b x)\right )}{7 b d (d \cos (a+b x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]/(d*Cos[a + b*x])^(9/2),x]

[Out]

(2*Hypergeometric2F1[-7/4, 1, -3/4, Cos[a + b*x]^2])/(7*b*d*(d*Cos[a + b*x])^(7/2))

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fricas [A]  time = 0.52, size = 342, normalized size = 3.32 \[ \left [\frac {42 \, \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right )^{4} - 21 \, \sqrt {-d} \cos \left (b x + a\right )^{4} \log \left (\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (7 \, \cos \left (b x + a\right )^{2} + 3\right )}}{84 \, b d^{5} \cos \left (b x + a\right )^{4}}, -\frac {42 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right )^{4} - 21 \, \sqrt {d} \cos \left (b x + a\right )^{4} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) - 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (7 \, \cos \left (b x + a\right )^{2} + 3\right )}}{84 \, b d^{5} \cos \left (b x + a\right )^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(9/2),x, algorithm="fricas")

[Out]

[1/84*(42*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)^
4 - 21*sqrt(-d)*cos(b*x + a)^4*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))*(7*cos(b*x + a)^2 + 3))/(b
*d^5*cos(b*x + a)^4), -1/84*(42*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)^4 - 21*sqrt(d)*cos(b*x + a)^4*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))*(7*cos(b*x
+ a)^2 + 3))/(b*d^5*cos(b*x + a)^4)]

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giac [B]  time = 1.38, size = 436, normalized size = 4.23 \[ \frac {\frac {42 \, \arctan \left (-\frac {\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} - \frac {21 \, \log \left ({\left | -\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d} \right |}\right )}{\sqrt {-d}} + \frac {16 \, {\left (21 \, {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{6} - 42 \, {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{5} \sqrt {-d} - 119 \, {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{4} d + 56 \, {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{3} \sqrt {-d} d + 63 \, {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{2} d^{2} - 14 \, {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )} \sqrt {-d} d^{2} - 5 \, d^{3}\right )}}{{\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d} - \sqrt {-d}\right )}^{7}}}{42 \, b d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(9/2),x, algorithm="giac")

[Out]

1/42*(42*arctan(-(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))/sqrt(-d))/sqrt(-d) -
21*log(abs(-sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 + sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d)))/sqrt(-d) + 16*(21*(sqrt(-d
)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))^6 - 42*(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt
(-d*tan(1/2*b*x + 1/2*a)^4 + d))^5*sqrt(-d) - 119*(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2
*a)^4 + d))^4*d + 56*(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))^3*sqrt(-d)*d + 63
*(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))^2*d^2 - 14*(sqrt(-d)*tan(1/2*b*x + 1/
2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))*sqrt(-d)*d^2 - 5*d^3)/(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d
*tan(1/2*b*x + 1/2*a)^4 + d) - sqrt(-d))^7)/(b*d^4)

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maple [B]  time = 0.37, size = 1086, normalized size = 10.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)/(d*cos(b*x+a))^(9/2),x)

[Out]

1/42/d^(19/2)/(-d)^(1/2)/(16*sin(1/2*b*x+1/2*a)^8-32*sin(1/2*b*x+1/2*a)^6+24*sin(1/2*b*x+1/2*a)^4-8*sin(1/2*b*
x+1/2*a)^2+1)*(42*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(11/2)+40*(-2*
sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(9/2)*(-d)^(1/2)-21*ln(2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2
*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^5-21*ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*sin(1/2
*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^5+336*(2*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*
(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(11/2)-ln(2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2
*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^5-ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2
*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^5)*sin(1/2*b*x+1/2*a)^8-672*(2*ln(2/cos(1/2*b*x+1/2*a
)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(11/2)-ln(2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/
2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^5-ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*s
in(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^5)*sin(1/2*b*x+1/2*a)^6-56*(6*ln(2/cos(
1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(11/2)+2*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1
/2)*d^(9/2)*(-d)^(1/2)-3*ln(2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*
b*x+1/2*a)-d))*(-d)^(1/2)*d^5-3*ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*c
os(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^5)*sin(1/2*b*x+1/2*a)^2+56*(18*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(
1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(11/2)+2*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(9/2)*(-d)^(1/2)-9*ln(2/(cos(
1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^5-9*ln(
2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^
5)*sin(1/2*b*x+1/2*a)^4)/b

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maxima [A]  time = 0.58, size = 102, normalized size = 0.99 \[ -\frac {\frac {42 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {7}{2}}} - \frac {21 \, \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{d^{\frac {7}{2}}} - \frac {4 \, {\left (7 \, d^{2} \cos \left (b x + a\right )^{2} + 3 \, d^{2}\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}} d^{2}}}{42 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(9/2),x, algorithm="maxima")

[Out]

-1/42*(42*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(7/2) - 21*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*cos(b
*x + a)) + sqrt(d)))/d^(7/2) - 4*(7*d^2*cos(b*x + a)^2 + 3*d^2)/((d*cos(b*x + a))^(7/2)*d^2))/(b*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sin \left (a+b\,x\right )\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)*(d*cos(a + b*x))^(9/2)),x)

[Out]

int(1/(sin(a + b*x)*(d*cos(a + b*x))^(9/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))**(9/2),x)

[Out]

Timed out

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