3.59 \(\int \csc ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=144 \[ -\frac{25 a^3 \cot (c+d x)}{8 d \sqrt{a \sin (c+d x)+a}}-\frac{25 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 d}-\frac{a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{a \sin (c+d x)+a}}{3 d}-\frac{13 a^3 \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(-25*a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(8*d) - (25*a^3*Cot[c + d*x])/(8*d*Sqrt
[a + a*Sin[c + d*x]]) - (13*a^3*Cot[c + d*x]*Csc[c + d*x])/(12*d*Sqrt[a + a*Sin[c + d*x]]) - (a^2*Cot[c + d*x]
*Csc[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(3*d)

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Rubi [A]  time = 0.275132, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2762, 2980, 2772, 2773, 206} \[ -\frac{25 a^3 \cot (c+d x)}{8 d \sqrt{a \sin (c+d x)+a}}-\frac{25 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 d}-\frac{a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{a \sin (c+d x)+a}}{3 d}-\frac{13 a^3 \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-25*a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(8*d) - (25*a^3*Cot[c + d*x])/(8*d*Sqrt
[a + a*Sin[c + d*x]]) - (13*a^3*Cot[c + d*x]*Csc[c + d*x])/(12*d*Sqrt[a + a*Sin[c + d*x]]) - (a^2*Cot[c + d*x]
*Csc[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(3*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 2980

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

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 ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac{a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{3 d}-\frac{1}{3} a \int \csc ^3(c+d x) \left (-\frac{13 a}{2}-\frac{9}{2} a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{13 a^3 \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{3 d}+\frac{1}{8} \left (25 a^2\right ) \int \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{25 a^3 \cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}-\frac{13 a^3 \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{3 d}+\frac{1}{16} \left (25 a^2\right ) \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{25 a^3 \cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}-\frac{13 a^3 \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{3 d}-\frac{\left (25 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 )}{8 d}\\ &=-\frac{25 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{8 d}-\frac{25 a^3 \cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}-\frac{13 a^3 \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{3 d}\\ \end{align*}

Mathematica [A]  time = 1.15858, size = 288, normalized size = 2. \[ \frac{a^2 \csc ^{10}\left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sin (c+d x)+1)} \left (228 \sin \left (\frac{1}{2} (c+d x)\right )+14 \sin \left (\frac{3}{2} (c+d x)\right )-150 \sin \left (\frac{5}{2} (c+d x)\right )-228 \cos \left (\frac{1}{2} (c+d x)\right )+14 \cos \left (\frac{3}{2} (c+d x)\right )+150 \cos \left (\frac{5}{2} (c+d x)\right )-225 \sin (c+d x) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+225 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+75 \sin (3 (c+d x)) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-75 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{24 d \left (\cot \left (\frac{1}{2} (c+d x)\right )+1\right ) \left (\csc ^2\left (\frac{1}{4} (c+d x)\right )-\sec ^2\left (\frac{1}{4} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(a^2*Csc[(c + d*x)/2]^10*Sqrt[a*(1 + Sin[c + d*x])]*(-228*Cos[(c + d*x)/2] + 14*Cos[(3*(c + d*x))/2] + 150*Cos
[(5*(c + d*x))/2] + 228*Sin[(c + d*x)/2] - 225*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] + 225
*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] + 14*Sin[(3*(c + d*x))/2] - 150*Sin[(5*(c + d*x))/2
] + 75*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 75*Log[1 - Cos[(c + d*x)/2] + Sin[(c +
d*x)/2]]*Sin[3*(c + d*x)]))/(24*d*(1 + Cot[(c + d*x)/2])*(Csc[(c + d*x)/4]^2 - Sec[(c + d*x)/4]^2)^3)

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Maple [A]  time = 0.732, size = 144, normalized size = 1. \begin{align*} -{\frac{1+\sin \left ( dx+c \right ) }{24\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( 75\, \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{5/2}{a}^{3/2}+75\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ){a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}-184\, \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{3/2}{a}^{5/2}+117\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{7/2} \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)^4*(a+a*sin(d*x+c))^(5/2),x)

[Out]

-1/24*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(75*(-a*(sin(d*x+c)-1))^(5/2)*a^(3/2)+75*arctanh((-a*(sin(d*x+c
)-1))^(1/2)/a^(1/2))*a^4*sin(d*x+c)^3-184*(-a*(sin(d*x+c)-1))^(3/2)*a^(5/2)+117*(-a*(sin(d*x+c)-1))^(1/2)*a^(7
/2))/sin(d*x+c)^3/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{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \csc \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(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)^4, x)

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Fricas [B]  time = 1.91238, size = 1040, normalized size = 7.22 \begin{align*} \frac{75 \,{\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} -{\left (a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2} - 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 (75 \, a^{2} \cos \left (d x + c\right )^{3} + 41 \, a^{2} \cos \left (d x + c\right )^{2} - 83 \, a^{2} \cos \left (d x + c\right ) - 49 \, a^{2} -{\left (75 \, a^{2} \cos \left (d x + c\right )^{2} + 34 \, a^{2} \cos \left (d x + c\right ) - 49 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{96 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} -{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - 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)^4*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/96*(75*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2 - (a^2*cos(d*x + c)^3 + a^2*cos(d*x + c)^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)*si
n(d*x + c) - cos(d*x + c) - 1)) + 4*(75*a^2*cos(d*x + c)^3 + 41*a^2*cos(d*x + c)^2 - 83*a^2*cos(d*x + c) - 49*
a^2 - (75*a^2*cos(d*x + c)^2 + 34*a^2*cos(d*x + c) - 49*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a))/(d*cos(d*
x + c)^4 - 2*d*cos(d*x + c)^2 - (d*cos(d*x + c)^3 + d*cos(d*x + c)^2 - 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)**4*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 2.64298, size = 852, normalized size = 5.92 \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)^4*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/48*(150*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) - 75*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) + (62*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1) + (2*a^2*sgn(tan(1/2*d*x + 1/2*
c) + 1)*tan(1/2*d*x + 1/2*c) + 15*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c))*sqrt(a*tan(1/2*d*x
+ 1/2*c)^2 + a) - (750*sqrt(2)*a^3*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 375*sqrt(2)*sqrt(-a)*a^(5/2)
*log(sqrt(2)*sqrt(a) + sqrt(a)) + 1050*a^3*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 525*sqrt(-a)*a^(5/2)
*log(sqrt(2)*sqrt(a) + sqrt(a)) + 376*sqrt(2)*sqrt(-a)*a^(5/2) + 546*sqrt(-a)*a^(5/2))*sgn(tan(1/2*d*x + 1/2*c
) + 1)/(5*sqrt(2)*sqrt(-a) + 7*sqrt(-a)) + 2*(15*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2
+ a))^5*a^3*sgn(tan(1/2*d*x + 1/2*c) + 1) + 66*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2
+ a))^4*a^(7/2)*sgn(tan(1/2*d*x + 1/2*c) + 1) - 120*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c
)^2 + a))^2*a^(9/2)*sgn(tan(1/2*d*x + 1/2*c) + 1) - 15*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/
2*c)^2 + a))*a^5*sgn(tan(1/2*d*x + 1/2*c) + 1) + 62*a^(11/2)*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)^3)/d