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

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

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 115, normalized size of antiderivative = 1.00, 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, 298, 203, 206} \[ \frac {5 \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{3/2}}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{3/2}}+\frac {5}{2 b d \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.24, size = 91, normalized size = 0.79 \[ \frac {5 \cot ^2(a+b x) \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\csc ^2(a+b x)\right )-\left (-\cot ^2(a+b x)\right )^{3/4} \left (\cot ^2(a+b x)-4\right )}{2 b d \left (-\cot ^2(a+b x)\right )^{3/4} \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.50, size = 406, normalized size = 3.53 \[ \left [\frac {10 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) - 5 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {-d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} - 4\right )}}{16 \, {\left (b d^{2} \cos \left (b x + a\right )^{3} - b d^{2} \cos \left (b x + a\right )\right )}}, \frac {10 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) + 5 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} - 4\right )}}{16 \, {\left (b d^{2} \cos \left (b x + a\right )^{3} - b d^{2} \cos \left (b x + a\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(10*(cos(b*x + a)^3 - cos(b*x + a))*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) + 1)
/(d*cos(b*x + a))) - 5*(cos(b*x + a)^3 - cos(b*x + a))*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))*(5*cos(b*x + a)^2 - 4))/(b*d^2*cos(b*x + a)^3 - b*d^2*cos(b*x + a)), 1/16*(10*(cos(b*x + a)^3 - cos(b*x
+ a))*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*cos(b*x + a))) + 5*(cos(b*x + a)^3
- cos(b*x + a))*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))*(5*cos(b*x + a)^2 - 4))/(b*d^2*cos
(b*x + a)^3 - b*d^2*cos(b*x + a))]

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giac [B]  time = 1.18, size = 361, normalized size = 3.14 \[ \frac {\frac {10 \, \arctan \left (-\frac {\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} + \frac {5 \, \log \left ({\left | -\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d} \right |}\right )}{\sqrt {-d}} - \frac {2 \, {\left (16 \, {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{2} - {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )} \sqrt {-d} - 17 \, d\right )}}{{\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{3} - {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{2} \sqrt {-d} - {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )} d + \sqrt {-d} d} + \frac {\sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}}{d}}{8 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(10*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) + 5
*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) - 2*(16*(sqrt(-d)*t
an(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))^2 - (sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*ta
n(1/2*b*x + 1/2*a)^4 + d))*sqrt(-d) - 17*d)/((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)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))^2*sqrt(-d) - (sqrt(-d)*tan(
1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))*d + sqrt(-d)*d) + sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d
)/d)/(b*d)

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maple [B]  time = 0.54, size = 705, normalized size = 6.13 \[ \frac {-\sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, d^{\frac {3}{2}} \sqrt {-d}-\left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (20 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) d^{\frac {5}{2}}+10 \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}\, d^{2}+10 \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}\, d^{2}\right )+5 \left (6 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) d^{\frac {5}{2}}-4 \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, d^{\frac {3}{2}} \sqrt {-d}+3 \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}\, d^{2}+3 \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}\, d^{2}\right ) \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-5 \left (2 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) d^{\frac {5}{2}}-4 \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, d^{\frac {3}{2}} \sqrt {-d}+\ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}\, d^{2}+\ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}\, d^{2}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 d^{\frac {7}{2}} \sqrt {-d}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-3 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1\right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d^(7/2)/(-d)^(1/2)/sin(1/2*b*x+1/2*a)^2/(2*sin(1/2*b*x+1/2*a)^4-3*sin(1/2*b*x+1/2*a)^2+1)*(-(-2*sin(1/2*b*
x+1/2*a)^2*d+d)^(1/2)*d^(3/2)*(-d)^(1/2)-sin(1/2*b*x+1/2*a)^6*(20*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^(5/2)+10*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^2+10*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^2)+5*(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^(5/2)-4*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(3/2)*(-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^2+3*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^2)*si
n(1/2*b*x+1/2*a)^4-5*(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^(5/2)-4*
(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(3/2)*(-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)^(1/2)*d^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^2)*sin(1/2*b*x+1/2*a)^2)/b

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maxima [A]  time = 0.87, size = 117, normalized size = 1.02 \[ \frac {\frac {10 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{\sqrt {d}} + \frac {5 \, \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{\sqrt {d}} + \frac {4 \, {\left (5 \, d^{2} \cos \left (b x + a\right )^{2} - 4 \, d^{2}\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} - \sqrt {d \cos \left (b x + a\right )} d^{2}}}{8 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(10*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/sqrt(d) + 5*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x
+ a)) + sqrt(d)))/sqrt(d) + 4*(5*d^2*cos(b*x + a)^2 - 4*d^2)/((d*cos(b*x + a))^(5/2) - sqrt(d*cos(b*x + a))*d^
2))/(b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\left (a + b x \right )}}{\left (d \cos {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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