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

Optimal. Leaf size=98 \[ -\frac{14 a^3 \cos (c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}-\frac{2 a^2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{3 d}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{d} \]

[Out]

(-2*a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (14*a^3*Cos[c + d*x])/(3*d*Sqrt[a +
a*Sin[c + d*x]]) - (2*a^2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d)

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Rubi [A]  time = 0.192339, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2763, 2981, 2773, 206} \[ -\frac{14 a^3 \cos (c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}-\frac{2 a^2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{3 d}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (14*a^3*Cos[c + d*x])/(3*d*Sqrt[a +
a*Sin[c + d*x]]) - (2*a^2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d)

Rule 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, 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 (c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac{2 a^2 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{3 d}+\frac{2}{3} \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \left (\frac{3 a^2}{2}+\frac{7}{2} a^2 \sin (c+d x)\right ) \, dx\\ &=-\frac{14 a^3 \cos (c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a^2 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{3 d}+a^2 \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{14 a^3 \cos (c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a^2 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{3 d}-\frac{\left (2 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{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{d}-\frac{14 a^3 \cos (c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a^2 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.37231, size = 143, normalized size = 1.46 \[ -\frac{(a (\sin (c+d x)+1))^{5/2} \left (-15 \sin \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{3}{2} (c+d x)\right )+15 \cos \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{3}{2} (c+d x)\right )+3 \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(1 + Sin[c + d*x]))^(5/2)*(15*Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2] + 3*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]] - 15*Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2]))/
(3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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Maple [A]  time = 0.615, size = 103, normalized size = 1.1 \begin{align*} -{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) a}{3\,d\cos \left ( dx+c \right ) }\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( 3\,{a}^{3/2}{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( dx+c \right ) }}{\sqrt{a}}} \right ) - \left ( a-a\sin \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}+9\,a\sqrt{a-a\sin \left ( dx+c \right ) } \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)*(a+a*sin(d*x+c))^(5/2),x)

[Out]

-2/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*a*(3*a^(3/2)*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))-(a-a*sin(d*
x+c))^(3/2)+9*a*(a-a*sin(d*x+c))^(1/2))/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{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \csc \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(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), x)

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Fricas [B]  time = 1.44608, size = 733, normalized size = 7.48 \begin{align*} \frac{3 \,{\left (a^{2} \cos \left (d x + c\right ) + a^{2} \sin \left (d x + c\right ) + a^{2}\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 (a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2} \cos \left (d x + c\right ) + 7 \, a^{2} +{\left (a^{2} \cos \left (d x + c\right ) - 7 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{6 \,{\left (d \cos \left (d x + c\right ) + d \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)*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/6*(3*(a^2*cos(d*x + c) + a^2*sin(d*x + c) + a^2)*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 + (co
s(d*x + c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) - 4*(a^2*cos(d*x + c)^2 + 8*a^2*cos(d*x + c) + 7*a^2 + (a^
2*cos(d*x + c) - 7*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a))/(d*cos(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)*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 2.72137, size = 429, normalized size = 4.38 \begin{align*} \frac{\frac{6 \, 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}} - 3 \, 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{{\left (6 \, a^{3} \arctan \left (\frac{\sqrt{2} \sqrt{a} + \sqrt{a}}{\sqrt{-a}}\right ) - 3 \, \sqrt{-a} a^{\frac{5}{2}} \log \left (\sqrt{2} \sqrt{a} + \sqrt{a}\right ) - 14 \, \sqrt{2} \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{-a}} - \frac{4 \,{\left (4 \, a^{4} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) -{\left (3 \, a^{4} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) +{\left (4 \, a^{4} \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 ) - 3 \, a^{4} \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 )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(6*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) - 3*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) - (6*a^3*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 3*sqrt(-a)*a^(5/2)*
log(sqrt(2)*sqrt(a) + sqrt(a)) - 14*sqrt(2)*sqrt(-a)*a^(5/2))*sgn(tan(1/2*d*x + 1/2*c) + 1)/sqrt(-a) - 4*(4*a^
4*sgn(tan(1/2*d*x + 1/2*c) + 1) - (3*a^4*sgn(tan(1/2*d*x + 1/2*c) + 1) + (4*a^4*sgn(tan(1/2*d*x + 1/2*c) + 1)*
tan(1/2*d*x + 1/2*c) - 3*a^4*sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))/(a*tan
(1/2*d*x + 1/2*c)^2 + a)^(3/2))/d