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

Optimal. Leaf size=100 \[ \frac{2}{b d^3 \sqrt{d \cos (a+b x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b d^{7/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b d^{7/2}}+\frac{2}{5 b d (d \cos (a+b x))^{5/2}} \]

[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]])

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Rubi [A]  time = 0.0778657, antiderivative size = 100, normalized size of antiderivative = 1., 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, 298, 203, 206} \[ \frac{2}{b d^3 \sqrt{d \cos (a+b x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b d^{7/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b d^{7/2}}+\frac{2}{5 b d (d \cos (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[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 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])

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 298

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 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])

Rubi steps

\begin{align*} \int \frac{\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx &=-\frac{\operatorname{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{\operatorname{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{\operatorname{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 \operatorname{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{\operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b d^3}+\frac{\operatorname{Subst}\left (\int \frac{1}{d+x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b d^3}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b d^{7/2}}-\frac{\tanh ^{-1}\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 [C]  time = 0.0682588, size = 38, normalized size = 0.38 \[ \frac{2 \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};\cos ^2(a+b x)\right )}{5 b d (d \cos (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Hypergeometric2F1[-5/4, 1, -1/4, Cos[a + b*x]^2])/(5*b*d*(d*Cos[a + b*x])^(5/2))

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Maple [B]  time = 0.22, size = 882, normalized size = 8.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10/d^(15/2)/(-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*ln(2/co
s(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^(9/2)-24*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^
(1/2)*d^(7/2)*(-d)^(1/2)+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^4+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^4-40*(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^(9/2)+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^4+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^4)*sin(1/2*b*x+1/2*a)^6+20*(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^(9/2)-4*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(7/2)*(-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^4+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^4)*sin(1/2
*b*x+1/2*a)^4-10*(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^(9/2)-8*(-2*
sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(7/2)*(-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^4+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^4)*sin(1/2*b*x+1/2*a)^2)/b

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.70545, size = 942, normalized size = 9.42 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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

[Out]

Timed out

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Giac [A]  time = 1.13774, size = 127, normalized size = 1.27 \begin{align*} \frac{d{\left (\frac{5 \, \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{-d}}\right )}{\sqrt{-d} d^{4}} + \frac{5 \, \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{d}}\right )}{d^{\frac{9}{2}}} + \frac{2 \,{\left (5 \, d^{2} \cos \left (b x + a\right )^{2} + d^{2}\right )}}{\sqrt{d \cos \left (b x + a\right )} d^{6} \cos \left (b x + a\right )^{2}}\right )}}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*d*(5*arctan(sqrt(d*cos(b*x + a))/sqrt(-d))/(sqrt(-d)*d^4) + 5*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(9/2)
+ 2*(5*d^2*cos(b*x + a)^2 + d^2)/(sqrt(d*cos(b*x + a))*d^6*cos(b*x + a)^2))/b