3.3.47 \(\int \sqrt {d \cos (a+b x)} \csc ^3(a+b x) \, dx\) [247]

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

[Out]

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

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Rubi [A]
time = 0.04, 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 = {2645, 296, 335, 304, 209, 212} \begin {gather*} \frac {\sqrt {d} \text {ArcTan}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}-\frac {\csc ^2(a+b x) (d \cos (a+b x))^{3/2}}{2 b d}-\frac {\sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.19, size = 62, normalized size = 0.67 \begin {gather*} -\frac {d \left (\cot ^2(a+b x)+\sqrt [4]{-\cot ^2(a+b x)} \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\csc ^2(a+b x)\right )\right )}{2 b \sqrt {d \cos (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(274\) vs. \(2(73)=146\).
time = 0.72, size = 275, normalized size = 2.96

method	result	size
default	\(\frac {-\frac {\sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}}{16 \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}-\frac {\sqrt {d}\, \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right )}{8}+\frac {\sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{8 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}-\frac {d \ln \left (\frac {-2 d +2 \sqrt {-d}\, \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{4 \sqrt {-d}}+\frac {\sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}}{16 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-16}-\frac {\sqrt {d}\, \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right )}{8}}{b}\)	\(275\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 102, normalized size = 1.10 \begin {gather*} \frac {\frac {4 \, \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} d^{2}}{d^{2} \cos \left (b x + a\right )^{2} - d^{2}} + 2 \, d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) + d^{\frac {3}{2}} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (73) = 146\).
time = 0.43, size = 340, normalized size = 3.66 \begin {gather*} \left [\frac {2 \, {\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 ) + {\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 )} \cos \left (b x + a\right )}{16 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}, \frac {2 \, {\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 ) + {\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 )} \cos \left (b x + a\right )}{16 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(2*(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))) + (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))*cos(b*x + a))/(b*c
os(b*x + a)^2 - b), 1/16*(2*(cos(b*x + a)^2 - 1)*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(s
qrt(d)*cos(b*x + a))) + (cos(b*x + a)^2 - 1)*sqrt(d)*log((d*cos(b*x + a)^2 - 4*sqrt(d*cos(b*x + a))*sqrt(d)*(c
os(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))*cos(b
*x + a))/(b*cos(b*x + a)^2 - b)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 5.22, size = 95, normalized size = 1.02 \begin {gather*} \frac {d^{3} {\left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \cos \left (b x + a\right )}{{\left (d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} d} + \frac {\arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {-d}}\right )}{\sqrt {-d} d^{2}} + \frac {\arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {5}{2}}}\right )}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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