3.3.51 \(\int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx\) [251]

Optimal. Leaf size=137 \[ \frac {9 \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}-\frac {9 \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}+\frac {9}{10 b d (d \cos (a+b x))^{5/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}} \]

[Out]

9/4*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(7/2)-9/4*arctanh((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(7/2)+9/10/b/
d/(d*cos(b*x+a))^(5/2)-1/2*csc(b*x+a)^2/b/d/(d*cos(b*x+a))^(5/2)+9/2/b/d^3/(d*cos(b*x+a))^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2645, 296, 331, 335, 304, 209, 212} \begin {gather*} \frac {9 \text {ArcTan}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}-\frac {9 \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}+\frac {9}{10 b d (d \cos (a+b x))^{5/2}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3/(d*Cos[a + b*x])^(7/2),x]

[Out]

(9*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(7/2)) - (9*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(7/2
)) + 9/(10*b*d*(d*Cos[a + b*x])^(5/2)) + 9/(2*b*d^3*Sqrt[d*Cos[a + b*x]]) - Csc[a + b*x]^2/(2*b*d*(d*Cos[a + b
*x])^(5/2))

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 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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*} \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx &=-\frac {\text {Subst}\left (\int \frac {1}{x^{7/2} \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))^{5/2}}-\frac {9 \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 )}{4 b d}\\ &=\frac {9}{10 b d (d \cos (a+b x))^{5/2}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}}-\frac {9 \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 )}{4 b d^3}\\ &=\frac {9}{10 b d (d \cos (a+b x))^{5/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}}-\frac {9 \text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{4 b d^5}\\ &=\frac {9}{10 b d (d \cos (a+b x))^{5/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}}-\frac {9 \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{d^2}} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{2 b d^5}\\ &=\frac {9}{10 b d (d \cos (a+b x))^{5/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}}-\frac {9 \text {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{4 b d^3}+\frac {9 \text {Subst}\left (\int \frac {1}{d+x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{4 b d^3}\\ &=\frac {9 \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}-\frac {9 \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}+\frac {9}{10 b d (d \cos (a+b x))^{5/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.34, size = 102, normalized size = 0.74 \begin {gather*} \frac {45 \cot ^2(a+b x) \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\csc ^2(a+b x)\right )+\left (-\cot ^2(a+b x)\right )^{3/4} \left (40-5 \cot ^2(a+b x)+4 \sec ^2(a+b x)\right )}{10 b d^3 \sqrt {d \cos (a+b x)} \left (-\cot ^2(a+b x)\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3/(d*Cos[a + b*x])^(7/2),x]

[Out]

(45*Cot[a + b*x]^2*Hypergeometric2F1[1/4, 1/4, 5/4, Csc[a + b*x]^2] + (-Cot[a + b*x]^2)^(3/4)*(40 - 5*Cot[a +
b*x]^2 + 4*Sec[a + b*x]^2))/(10*b*d^3*Sqrt[d*Cos[a + b*x]]*(-Cot[a + b*x]^2)^(3/4))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1164\) vs. \(2(109)=218\).
time = 1.34, size = 1165, normalized size = 8.50

method	result	size
default	\(\text {Expression too large to display}\)	\(1165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^3/(d*cos(b*x+a))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/40/(-d)^(1/2)/d^(15/2)/sin(1/2*b*x+1/2*a)^2/(8*sin(1/2*b*x+1/2*a)^8-20*sin(1/2*b*x+1/2*a)^6+18*sin(1/2*b*x+1
/2*a)^4-7*sin(1/2*b*x+1/2*a)^2+1)*(-5*(-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(7/2)-360*(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)+(-d)^(1/2)*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^4+(-d)^(1/2)*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^4)*sin(1/2*b*x+1/2*a)^10-180
*(-10*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*(-d)^(1/2)*(-2*sin
(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(7/2)-5*(-d)^(1/2)*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^4-5*(-d)^(1/2)*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^4)*sin(1/2*b*x+1/2*a)^8+90*(-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^(9/2)+16*(-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(7/
2)-9*(-d)^(1/2)*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^4-9*(-d)^(1/2)*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^4)*sin(1/2*b*x+1/2*a)^6-9*(-70*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)+104*(-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(7/2)-35*(-d)^(1/2)*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^4-35*(-d)^(1/2)*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^4)*sin(1/2*b*x+1/
2*a)^4+9*(-10*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)+24*(-d)^(1/2
)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(7/2)-5*(-d)^(1/2)*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^4-5*(-d)^(1/2)*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^4)*sin(1/2*b*x+1/2*a)^2)/b

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Maxima [A]
time = 0.49, size = 134, normalized size = 0.98 \begin {gather*} \frac {\frac {4 \, {\left (45 \, d^{4} \cos \left (b x + a\right )^{4} - 36 \, d^{4} \cos \left (b x + a\right )^{2} - 4 \, d^{4}\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} d^{2} - \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} d^{4}} + \frac {90 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {45 \, \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}}}}{40 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

1/40*(4*(45*d^4*cos(b*x + a)^4 - 36*d^4*cos(b*x + a)^2 - 4*d^4)/((d*cos(b*x + a))^(9/2)*d^2 - (d*cos(b*x + a))
^(5/2)*d^4) + 90*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(5/2) + 45*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(
d*cos(b*x + a)) + sqrt(d)))/d^(5/2))/(b*d)

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Fricas [A]
time = 0.45, size = 438, normalized size = 3.20 \begin {gather*} \left [\frac {90 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\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 ) - 45 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\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 \, {\left (45 \, \cos \left (b x + a\right )^{4} - 36 \, \cos \left (b x + a\right )^{2} - 4\right )} \sqrt {d \cos \left (b x + a\right )}}{80 \, {\left (b d^{4} \cos \left (b x + a\right )^{5} - b d^{4} \cos \left (b x + a\right )^{3}\right )}}, \frac {90 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\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 ) + 45 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\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 \, {\left (45 \, \cos \left (b x + a\right )^{4} - 36 \, \cos \left (b x + a\right )^{2} - 4\right )} \sqrt {d \cos \left (b x + a\right )}}{80 \, {\left (b d^{4} \cos \left (b x + a\right )^{5} - b d^{4} \cos \left (b x + a\right )^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(7/2),x, algorithm="fricas")

[Out]

[1/80*(90*(cos(b*x + a)^5 - cos(b*x + a)^3)*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) +
1)/(d*cos(b*x + a))) - 45*(cos(b*x + a)^5 - cos(b*x + a)^3)*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*(45*cos(b
*x + a)^4 - 36*cos(b*x + a)^2 - 4)*sqrt(d*cos(b*x + a)))/(b*d^4*cos(b*x + a)^5 - b*d^4*cos(b*x + a)^3), 1/80*(
90*(cos(b*x + a)^5 - cos(b*x + a)^3)*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*cos(b
*x + a))) + 45*(cos(b*x + a)^5 - cos(b*x + a)^3)*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*(45*cos(b*x + a)^4 - 3
6*cos(b*x + a)^2 - 4)*sqrt(d*cos(b*x + a)))/(b*d^4*cos(b*x + a)^5 - b*d^4*cos(b*x + a)^3)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**3/(d*cos(b*x+a))**(7/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(7/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^3/(d*cos(b*x + a))^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\sin \left (a+b\,x\right )}^3\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)^3*(d*cos(a + b*x))^(7/2)),x)

[Out]

int(1/(sin(a + b*x)^3*(d*cos(a + b*x))^(7/2)), x)

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