3.250 \(\int \frac{\csc ^3(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=115 \[ -\frac{7 \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{5/2}}-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{5/2}}+\frac{7}{6 b d (d \cos (a+b x))^{3/2}}-\frac{\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{3/2}} \]

[Out]

(-7*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(5/2)) - (7*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(5/
2)) + 7/(6*b*d*(d*Cos[a + b*x])^(3/2)) - Csc[a + b*x]^2/(2*b*d*(d*Cos[a + b*x])^(3/2))

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Rubi [A]  time = 0.0833959, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2565, 290, 325, 329, 212, 206, 203} \[ -\frac{7 \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{5/2}}-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{5/2}}+\frac{7}{6 b d (d \cos (a+b x))^{3/2}}-\frac{\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3/(d*Cos[a + b*x])^(5/2),x]

[Out]

(-7*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(5/2)) - (7*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(5/
2)) + 7/(6*b*d*(d*Cos[a + b*x])^(3/2)) - Csc[a + b*x]^2/(2*b*d*(d*Cos[a + b*x])^(3/2))

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 290

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

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 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 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 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])

Rubi steps

\begin{align*} \int \frac{\csc ^3(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{x^{5/2} \left (1-\frac{x^2}{d^2}\right )^2} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=-\frac{\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{3/2}}-\frac{7 \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 )}{4 b d}\\ &=\frac{7}{6 b d (d \cos (a+b x))^{3/2}}-\frac{\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{3/2}}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-\frac{x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{4 b d^3}\\ &=\frac{7}{6 b d (d \cos (a+b x))^{3/2}}-\frac{\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{3/2}}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^4}{d^2}} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{2 b d^3}\\ &=\frac{7}{6 b d (d \cos (a+b x))^{3/2}}-\frac{\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{3/2}}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{4 b d^2}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{d+x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{4 b d^2}\\ &=-\frac{7 \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{5/2}}-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{5/2}}+\frac{7}{6 b d (d \cos (a+b x))^{3/2}}-\frac{\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.366873, size = 92, normalized size = 0.8 \[ \frac{7 \cot ^2(a+b x) \, _2F_1\left (\frac{3}{4},\frac{3}{4};\frac{7}{4};\csc ^2(a+b x)\right )+\sqrt [4]{-\cot ^2(a+b x)} \left (4-3 \cot ^2(a+b x)\right )}{6 b d \sqrt [4]{-\cot ^2(a+b x)} (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3/(d*Cos[a + b*x])^(5/2),x]

[Out]

((-Cot[a + b*x]^2)^(1/4)*(4 - 3*Cot[a + b*x]^2) + 7*Cot[a + b*x]^2*Hypergeometric2F1[3/4, 3/4, 7/4, Csc[a + b*
x]^2])/(6*b*d*(d*Cos[a + b*x])^(3/2)*(-Cot[a + b*x]^2)^(1/4))

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Maple [B]  time = 0.345, size = 909, normalized size = 7.9 \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)^3/(d*cos(b*x+a))^(5/2),x)

[Out]

1/24/d^(11/2)/(-d)^(1/2)/sin(1/2*b*x+1/2*a)^2/(4*sin(1/2*b*x+1/2*a)^6-8*sin(1/2*b*x+1/2*a)^4+5*sin(1/2*b*x+1/2
*a)^2-1)*(3*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(5/2)*(-d)^(1/2)+84*(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^(7/2)-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^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^3)*sin(1/2*b*x+1/2*a)^8-168*(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^(7/2)-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^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^3)*sin(1/2*b*x+1/2*a)^6-7*(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^(7/2)+4*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d
^(5/2)*(-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^3-3*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^3)*sin(1/2*b*x+1/2*a)^2+7*(30*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^(7/2)+4*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(5/2)*(-d)^(1/2)-15*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^3-15*ln(2/(co
s(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^3)*si
n(1/2*b*x+1/2*a)^4)/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)^3/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.80907, size = 1111, normalized size = 9.66 \begin{align*} \left [\frac{42 \,{\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \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 ) - 21 \,{\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sqrt{-d} \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} - 4\right )}}{48 \,{\left (b d^{3} \cos \left (b x + a\right )^{4} - b d^{3} \cos \left (b x + a\right )^{2}\right )}}, -\frac{42 \,{\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \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 ) - 21 \,{\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sqrt{d} \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} - 4\right )}}{48 \,{\left (b d^{3} \cos \left (b x + a\right )^{4} - b d^{3} \cos \left (b x + a\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

[1/48*(42*(cos(b*x + a)^4 - cos(b*x + a)^2)*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) +
1)/(d*cos(b*x + a))) - 21*(cos(b*x + a)^4 - cos(b*x + a)^2)*sqrt(-d)*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*co
s(b*x + a))*(7*cos(b*x + a)^2 - 4))/(b*d^3*cos(b*x + a)^4 - b*d^3*cos(b*x + a)^2), -1/48*(42*(cos(b*x + a)^4 -
cos(b*x + a)^2)*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*cos(b*x + a))) - 21*(cos(
b*x + a)^4 - cos(b*x + a)^2)*sqrt(d)*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 - 4
))/(b*d^3*cos(b*x + a)^4 - b*d^3*cos(b*x + a)^2)]

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

[Out]

Timed out

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Giac [A]  time = 1.14238, size = 154, normalized size = 1.34 \begin{align*} \frac{d^{3}{\left (\frac{6 \, \sqrt{d \cos \left (b x + a\right )}}{{\left (d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} d^{4}} + \frac{21 \, \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{-d}}\right )}{\sqrt{-d} d^{5}} - \frac{21 \, \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{d}}\right )}{d^{\frac{11}{2}}} + \frac{8}{\sqrt{d \cos \left (b x + a\right )} d^{5} \cos \left (b x + a\right )}\right )}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(5/2),x, algorithm="giac")

[Out]

1/12*d^3*(6*sqrt(d*cos(b*x + a))/((d^2*cos(b*x + a)^2 - d^2)*d^4) + 21*arctan(sqrt(d*cos(b*x + a))/sqrt(-d))/(
sqrt(-d)*d^5) - 21*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(11/2) + 8/(sqrt(d*cos(b*x + a))*d^5*cos(b*x + a)))/
b