3.57 \(\int \csc ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=94 \[ -\frac{a^3 \cos (c+d x)}{d \sqrt{a \sin (c+d x)+a}}-\frac{a^2 \cot (c+d x) \sqrt{a \sin (c+d x)+a}}{d}-\frac{5 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{d} \]

[Out]

(-5*a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (a^3*Cos[c + d*x])/(d*Sqrt[a + a*Sin
[c + d*x]]) - (a^2*Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/d

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Rubi [A]  time = 0.193435, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2762, 2981, 2773, 206} \[ -\frac{a^3 \cos (c+d x)}{d \sqrt{a \sin (c+d x)+a}}-\frac{a^2 \cot (c+d x) \sqrt{a \sin (c+d x)+a}}{d}-\frac{5 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^2*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-5*a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (a^3*Cos[c + d*x])/(d*Sqrt[a + a*Sin
[c + d*x]]) - (a^2*Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/d

Rule 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2981

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

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 ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac{a^2 \cot (c+d x) \sqrt{a+a \sin (c+d x)}}{d}-a \int \csc (c+d x) \left (-\frac{5 a}{2}-\frac{1}{2} a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{a^3 \cos (c+d x)}{d \sqrt{a+a \sin (c+d x)}}-\frac{a^2 \cot (c+d x) \sqrt{a+a \sin (c+d x)}}{d}+\frac{1}{2} \left (5 a^2\right ) \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{a^3 \cos (c+d x)}{d \sqrt{a+a \sin (c+d x)}}-\frac{a^2 \cot (c+d x) \sqrt{a+a \sin (c+d x)}}{d}-\frac{\left (5 a^3\right ) \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 )}{d}\\ &=-\frac{5 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{d}-\frac{a^3 \cos (c+d x)}{d \sqrt{a+a \sin (c+d x)}}-\frac{a^2 \cot (c+d x) \sqrt{a+a \sin (c+d x)}}{d}\\ \end{align*}

Mathematica [A]  time = 0.801497, size = 182, normalized size = 1.94 \[ -\frac{a^2 \csc ^4\left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sin (c+d x)+1)} \left (2 \sin \left (\frac{3}{2} (c+d x)\right )+2 \cos \left (\frac{3}{2} (c+d x)\right )+5 \sin (c+d x) \left (\log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )\right )}{d \left (\cot \left (\frac{1}{2} (c+d x)\right )+1\right ) \left (\csc \left (\frac{1}{4} (c+d x)\right )-\sec \left (\frac{1}{4} (c+d x)\right )\right ) \left (\csc \left (\frac{1}{4} (c+d x)\right )+\sec \left (\frac{1}{4} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^2*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-((a^2*Csc[(c + d*x)/2]^4*Sqrt[a*(1 + Sin[c + d*x])]*(2*Cos[(3*(c + d*x))/2] + 5*(Log[1 + Cos[(c + d*x)/2] - S
in[(c + d*x)/2]] - Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*Sin[c + d*x] + 2*Sin[(3*(c + d*x))/2]))/(d*(1
+ Cot[(c + d*x)/2])*(Csc[(c + d*x)/4] - Sec[(c + d*x)/4])*(Csc[(c + d*x)/4] + Sec[(c + d*x)/4])))

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Maple [A]  time = 0.703, size = 123, normalized size = 1.3 \begin{align*} -{\frac{1+\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{{\frac{3}{2}}} \left ( \sin \left ( dx+c \right ) \left ( 2\,\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{a}+5\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( dx+c \right ) }}{\sqrt{a}}} \right ) a \right ) +\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{a} \right ){\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)^2*(a+a*sin(d*x+c))^(5/2),x)

[Out]

-(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*(sin(d*x+c)*(2*(a-a*sin(d*x+c))^(1/2)*a^(1/2)+5*arctanh((a-a
*sin(d*x+c))^(1/2)/a^(1/2))*a)+(a-a*sin(d*x+c))^(1/2)*a^(1/2))/sin(d*x+c)/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{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \csc \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(5/2)*csc(d*x + c)^2, x)

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Fricas [B]  time = 1.4929, size = 790, normalized size = 8.4 \begin{align*} \frac{5 \,{\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} -{\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )\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 (2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right ) - a^{2} +{\left (2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{4 \,{\left (d \cos \left (d x + c\right )^{2} -{\left (d \cos \left (d x + c\right ) + 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)^2*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/4*(5*(a^2*cos(d*x + c)^2 - a^2 - (a^2*cos(d*x + c) + a^2)*sin(d*x + c))*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*(2*a^2*cos(d*x + c)^2 + a^2*cos
(d*x + c) - a^2 + (2*a^2*cos(d*x + c) + a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^2 - (d*co
s(d*x + c) + 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)**2*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 2.77994, size = 567, normalized size = 6.03 \begin{align*} \frac{\frac{10 \, a^{3} \arctan \left (-\frac{\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{\sqrt{-a}}\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{\sqrt{-a}} - 5 \, a^{\frac{5}{2}} \log \left ({\left | -\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) + \frac{2 \, a^{\frac{7}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a} - \frac{{\left (10 \, \sqrt{2} a^{3} \arctan \left (\frac{\sqrt{2} \sqrt{a} + \sqrt{a}}{\sqrt{-a}}\right ) - 5 \, \sqrt{2} \sqrt{-a} a^{\frac{5}{2}} \log \left (\sqrt{2} \sqrt{a} + \sqrt{a}\right ) + 10 \, a^{3} \arctan \left (\frac{\sqrt{2} \sqrt{a} + \sqrt{a}}{\sqrt{-a}}\right ) - 5 \, \sqrt{-a} a^{\frac{5}{2}} \log \left (\sqrt{2} \sqrt{a} + \sqrt{a}\right ) - 3 \, \sqrt{2} \sqrt{-a} a^{\frac{5}{2}} - 5 \, \sqrt{-a} a^{\frac{5}{2}}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{\sqrt{2} \sqrt{-a} + \sqrt{-a}} - \frac{3 \, a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) -{\left (a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/2*(10*a^3*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) - 5*a^(5/2)*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) + 2*a^(7/2)*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) - (10*sqrt(2)*a^3*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 5
*sqrt(2)*sqrt(-a)*a^(5/2)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 10*a^3*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a))
- 5*sqrt(-a)*a^(5/2)*log(sqrt(2)*sqrt(a) + sqrt(a)) - 3*sqrt(2)*sqrt(-a)*a^(5/2) - 5*sqrt(-a)*a^(5/2))*sgn(ta
n(1/2*d*x + 1/2*c) + 1)/(sqrt(2)*sqrt(-a) + sqrt(-a)) - (3*a^3*sgn(tan(1/2*d*x + 1/2*c) + 1) - (a^3*sgn(tan(1/
2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c) + 4*a^3*sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c))/sqrt(a*t
an(1/2*d*x + 1/2*c)^2 + a))/d