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

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

[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)

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Rubi [A]  time = 0.065534, antiderivative size = 93, normalized size of antiderivative = 1., 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, 298, 203, 206} \[ \frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b}-\frac{\csc ^2(a+b x) (d \cos (a+b x))^{3/2}}{2 b d}-\frac{\sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b} \]

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 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 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 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 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])

Rubi steps

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

Mathematica [C]  time = 0.250396, size = 62, normalized size = 0.67 \[ -\frac{d \left (\sqrt [4]{-\cot ^2(a+b x)} \, _2F_1\left (\frac{1}{4},\frac{1}{4};\frac{5}{4};\csc ^2(a+b x)\right )+\cot ^2(a+b x)\right )}{2 b \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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]))/(2*b*Sqrt[d*C
os[a + b*x]])

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Maple [B]  time = 0.256, size = 289, normalized size = 3.1 \begin{align*}{\frac{1}{16\,b}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{8\,b}\sqrt{d}\ln \left ({ \left ( 4\,d\cos \left ( 1/2\,bx+a/2 \right ) +2\,\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-2\,d \right ) \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) -1 \right ) ^{-1}} \right ) }-{\frac{1}{16\,b}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{8\,b}\sqrt{d}\ln \left ({ \left ( -4\,d\cos \left ( 1/2\,bx+a/2 \right ) +2\,\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-2\,d \right ) \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1 \right ) ^{-1}} \right ) }+{\frac{1}{8\,b}\sqrt{2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d-d} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-2}}-{\frac{d}{4\,b}\ln \left ({ \left ( -2\,d+2\,\sqrt{-d}\sqrt{2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d-d} \right ) \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}} \right ){\frac{1}{\sqrt{-d}}}} \end{align*}

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)

[Out]

1/16/b/(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/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/(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/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))+1/8/b/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/b*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))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.7607, size = 940, normalized size = 10.11 \begin{align*} \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{align*}

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

[Out]

Timed out

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

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]

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