3.1.59 \(\int \csc ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [59]

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

[Out]

-25/8*a^(5/2)*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d-25/8*a^3*cot(d*x+c)/d/(a+a*sin(d*x+c))^(1/2
)-13/12*a^3*cot(d*x+c)*csc(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-1/3*a^2*cot(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^(1
/2)/d

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Rubi [A]
time = 0.19, antiderivative size = 144, normalized size of antiderivative = 1.00, 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 = {2841, 3059, 2851, 2852, 212} \begin {gather*} -\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 {25 a^3 \cot (c+d x)}{8 d \sqrt {a \sin (c+d x)+a}}-\frac {13 a^3 \cot (c+d x) \csc (c+d x)}{12 d \sqrt {a \sin (c+d x)+a}}-\frac {a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d} \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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 2851

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 2852

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 3059

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.77, size = 288, normalized size = 2.00 \begin {gather*} \frac {a^2 \csc ^{10}\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (-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 )+228 \sin \left (\frac {1}{2} (c+d x)\right )-225 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+225 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+14 \sin \left (\frac {3}{2} (c+d x)\right )-150 \sin \left (\frac {5}{2} (c+d x)\right )+75 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-75 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))\right )}{24 d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^3} \end {gather*}

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 = 2.12, size = 144, normalized size = 1.00

method	result	size
default	\(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (75 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {5}{2}} a^{\frac {3}{2}}+75 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{4} \left (\sin ^{3}\left (d x +c \right )\right )-184 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {5}{2}}+117 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {7}{2}}\right )}{24 \sin \left (d x +c \right )^{3} a^{\frac {3}{2}} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(144\)

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,method=_RETURNVERBOSE)

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (124) = 248\).
time = 0.35, size = 408, normalized size = 2.83 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 [A]
time = 0.55, size = 196, normalized size = 1.36 \begin {gather*} -\frac {\sqrt {2} {\left (75 \, \sqrt {2} a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {4 \, {\left (300 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 368 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 117 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )} \sqrt {a}}{96 \, d} \end {gather*}

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]

-1/96*sqrt(2)*(75*sqrt(2)*a^2*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(-1/
4*pi + 1/2*d*x + 1/2*c)))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) + 4*(300*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))
*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 368*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)
^3 + 117*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(2*sin(-1/4*pi + 1/2*d*x + 1/
2*c)^2 - 1)^3)*sqrt(a)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\sin \left (c+d\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(5/2)/sin(c + d*x)^4,x)

[Out]

int((a + a*sin(c + d*x))^(5/2)/sin(c + d*x)^4, x)

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