3.39 \(\int \csc ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=102 \[ -\frac{3 a \cot (c+d x)}{4 d \sqrt{a \sin (c+d x)+a}}-\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{4 d}-\frac{a \cot (c+d x) \csc (c+d x)}{2 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(-3*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(4*d) - (3*a*Cot[c + d*x])/(4*d*Sqrt[a +
a*Sin[c + d*x]]) - (a*Cot[c + d*x]*Csc[c + d*x])/(2*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.156279, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2772, 2773, 206} \[ -\frac{3 a \cot (c+d x)}{4 d \sqrt{a \sin (c+d x)+a}}-\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{4 d}-\frac{a \cot (c+d x) \csc (c+d x)}{2 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-3*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(4*d) - (3*a*Cot[c + d*x])/(4*d*Sqrt[a +
a*Sin[c + d*x]]) - (a*Cot[c + d*x]*Csc[c + d*x])/(2*d*Sqrt[a + a*Sin[c + d*x]])

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 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 \csc ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx &=-\frac{a \cot (c+d x) \csc (c+d x)}{2 d \sqrt{a+a \sin (c+d x)}}+\frac{3}{4} \int \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{3 a \cot (c+d x)}{4 d \sqrt{a+a \sin (c+d x)}}-\frac{a \cot (c+d x) \csc (c+d x)}{2 d \sqrt{a+a \sin (c+d x)}}+\frac{3}{8} \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{3 a \cot (c+d x)}{4 d \sqrt{a+a \sin (c+d x)}}-\frac{a \cot (c+d x) \csc (c+d x)}{2 d \sqrt{a+a \sin (c+d x)}}-\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{4 d}\\ &=-\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{4 d}-\frac{3 a \cot (c+d x)}{4 d \sqrt{a+a \sin (c+d x)}}-\frac{a \cot (c+d x) \csc (c+d x)}{2 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 0.699265, size = 249, normalized size = 2.44 \[ \frac{\csc ^7\left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sin (c+d x)+1)} \left (2 \sin \left (\frac{1}{2} (c+d x)\right )-6 \sin \left (\frac{3}{2} (c+d x)\right )-2 \cos \left (\frac{1}{2} (c+d x)\right )-6 \cos \left (\frac{3}{2} (c+d x)\right )+3 \cos (2 (c+d x)) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-3 \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-3 \cos (2 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{4 d \left (\cot \left (\frac{1}{2} (c+d x)\right )+1\right ) \left (\csc ^2\left (\frac{1}{4} (c+d x)\right )-\sec ^2\left (\frac{1}{4} (c+d x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Csc[(c + d*x)/2]^7*Sqrt[a*(1 + Sin[c + d*x])]*(-2*Cos[(c + d*x)/2] - 6*Cos[(3*(c + d*x))/2] - 3*Log[1 + Cos[(
c + d*x)/2] - Sin[(c + d*x)/2]] + 3*Cos[2*(c + d*x)]*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 3*Log[1 -
Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 3*Cos[2*(c + d*x)]*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 2*Sin
[(c + d*x)/2] - 6*Sin[(3*(c + d*x))/2]))/(4*d*(1 + Cot[(c + d*x)/2])*(Csc[(c + d*x)/4]^2 - Sec[(c + d*x)/4]^2)
^2)

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Maple [A]  time = 0.589, size = 132, normalized size = 1.3 \begin{align*} -{\frac{1+\sin \left ( dx+c \right ) }{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( 3\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{3/2}\sin \left ( dx+c \right ) +3\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{2}+2\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{3/2} \right ){a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x)

[Out]

-1/4*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(3*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*sin(d*x+c)+3*arctanh((-a*(s
in(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^2*a^2+2*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2))/sin(d*x+c)^2/a^(3/2)/cos(d*
x+c)/(a+a*sin(d*x+c))^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*csc(d*x + c)^3, x)

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Fricas [B]  time = 1.45284, size = 853, normalized size = 8.36 \begin{align*} \frac{3 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \,{\left (\cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a} - 9 \, a \cos \left (d x + c\right ) +{\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \,{\left (3 \, \cos \left (d x + c\right )^{2} +{\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) - 1\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{16 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) +{\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right ) - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)*sqrt(a)*log((
a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 + (cos(d*x + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) -
3)*sqrt(a*sin(d*x + c) + a)*sqrt(a) - 9*a*cos(d*x + c) + (a*cos(d*x + c)^2 + 8*a*cos(d*x + c) - a)*sin(d*x +
c) - a)/(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) + 4*(3*cos(d
*x + c)^2 + (3*cos(d*x + c) + 1)*sin(d*x + c) + 2*cos(d*x + c) - 1)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^
3 + d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c)^2 - d)*sin(d*x + c) - d)

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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(csc(d*x+c)**3*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 2.54367, size = 721, normalized size = 7.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(6*a*arctan(-(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))/sqrt(-a))*sgn(tan(1/2*d*x
+ 1/2*c) + 1)/sqrt(-a) - 3*sqrt(a)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)
))*sgn(tan(1/2*d*x + 1/2*c) + 1) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d
*x + 1/2*c) + 2*sgn(tan(1/2*d*x + 1/2*c) + 1)) - (12*sqrt(2)*a*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) -
6*sqrt(2)*sqrt(-a)*sqrt(a)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 18*a*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a))
- 9*sqrt(-a)*sqrt(a)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 2*sqrt(2)*sqrt(-a)*sqrt(a) + 2*sqrt(-a)*sqrt(a))*sgn(tan
(1/2*d*x + 1/2*c) + 1)/(2*sqrt(2)*sqrt(-a) + 3*sqrt(-a)) + 2*((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d
*x + 1/2*c)^2 + a))^3*a*sgn(tan(1/2*d*x + 1/2*c) + 1) + 2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x +
1/2*c)^2 + a))^2*a^(3/2)*sgn(tan(1/2*d*x + 1/2*c) + 1) + (sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x +
1/2*c)^2 + a))*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1) - 2*a^(5/2)*sgn(tan(1/2*d*x + 1/2*c) + 1))/((sqrt(a)*tan(1/2
*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)^2)/d