3.248 \(\int \frac {\csc ^3(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx\)

Optimal. Leaf size=93 \[ -\frac {3 \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b \sqrt {d}}-\frac {\csc ^2(a+b x) \sqrt {d \cos (a+b x)}}{2 b d}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b \sqrt {d}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2565, 290, 329, 212, 206, 203} \[ -\frac {3 \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b \sqrt {d}}-\frac {\csc ^2(a+b x) \sqrt {d \cos (a+b x)}}{2 b d}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3/Sqrt[d*Cos[a + b*x]],x]

[Out]

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

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

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Mathematica [C]  time = 0.22, size = 69, normalized size = 0.74 \[ \frac {d \left (-\cot ^2(a+b x)\right )^{3/4} \left (\sqrt [4]{-\cot ^2(a+b x)}-\, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};\csc ^2(a+b x)\right )\right )}{2 b (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3/Sqrt[d*Cos[a + b*x]],x]

[Out]

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

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fricas [B]  time = 0.52, size = 334, normalized size = 3.59 \[ \left [\frac {6 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) - 3 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {-d} \log \left (\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )}}{16 \, {\left (b d \cos \left (b x + a\right )^{2} - b d\right )}}, -\frac {6 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) - 3 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) - 8 \, \sqrt {d \cos \left (b x + a\right )}}{16 \, {\left (b d \cos \left (b x + a\right )^{2} - b d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

[1/16*(6*(cos(b*x + a)^2 - 1)*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) + 1)/(d*cos(b*x
+ a))) - 3*(cos(b*x + a)^2 - 1)*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)))/(b*d*cos(b*x +
a)^2 - b*d), -1/16*(6*(cos(b*x + a)^2 - 1)*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)
*cos(b*x + a))) - 3*(cos(b*x + a)^2 - 1)*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)))/(b*d*cos
(b*x + a)^2 - b*d)]

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giac [B]  time = 1.32, size = 181, normalized size = 1.95 \[ \frac {\frac {6 \, \arctan \left (-\frac {\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} - \frac {3 \, \log \left ({\left | -\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d} \right |}\right )}{\sqrt {-d}} + \frac {2 \, \sqrt {-d}}{{\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{2} - d} - \frac {\sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}}{d}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="giac")

[Out]

1/8*(6*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) - 3*
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*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))^2 - d) - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d)
/d)/b

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maple [B]  time = 0.28, size = 297, normalized size = 3.19 \[ -\frac {3 \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right )}{8 \sqrt {d}\, b}-\frac {\sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}}{16 b d \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}-\frac {3 \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right )}{8 \sqrt {d}\, b}+\frac {3 \ln \left (\frac {-2 d +2 \sqrt {-d}\, \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{4 b \sqrt {-d}}-\frac {\sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{8 b d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}+\frac {\sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}}{16 b d \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8/b/d^(1/2)*ln((4*d*cos(1/2*b*x+1/2*a)+2*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)-1))-1/16/b/d/(cos(1/2*b*x+1/2*a)+1)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-3/8/b/d^(1/2)*ln((-4*d*cos(1/2*b*x+
1/2*a)+2*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)+1))+3/4/b/(-d)^(1/2)*ln((-2*d+2*
(-d)^(1/2)*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2))/cos(1/2*b*x+1/2*a))-1/8/b/d/cos(1/2*b*x+1/2*a)^2*(2*cos(1/2*b*x
+1/2*a)^2*d-d)^(1/2)+1/16/b/d/(cos(1/2*b*x+1/2*a)-1)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)

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maxima [A]  time = 0.50, size = 103, normalized size = 1.11 \[ \frac {\frac {4 \, \sqrt {d \cos \left (b x + a\right )} d^{2}}{d^{2} \cos \left (b x + a\right )^{2} - d^{2}} - 6 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) + 3 \, \sqrt {d} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{8 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**3/(d*cos(b*x+a))**(1/2),x)

[Out]

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

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