3.51 \(\int \csc ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=144 \[ -\frac{11 a^2 \cot (c+d x)}{8 d \sqrt{a \sin (c+d x)+a}}-\frac{11 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 d}-\frac{a^2 \cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}-\frac{11 a^2 \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(-11*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(8*d) - (11*a^2*Cot[c + d*x])/(8*d*Sqrt
[a + a*Sin[c + d*x]]) - (11*a^2*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)/(3*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.231238, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2762, 21, 2772, 2773, 206} \[ -\frac{11 a^2 \cot (c+d x)}{8 d \sqrt{a \sin (c+d x)+a}}-\frac{11 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 d}-\frac{a^2 \cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}-\frac{11 a^2 \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-11*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(8*d) - (11*a^2*Cot[c + d*x])/(8*d*Sqrt
[a + a*Sin[c + d*x]]) - (11*a^2*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)/(3*d*Sqrt[a + a*Sin[c + d*x]])

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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac{a^2 \cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}-\frac{1}{3} a \int \frac{\csc ^3(c+d x) \left (-\frac{11 a}{2}-\frac{11}{2} a \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{a^2 \cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}+\frac{1}{6} (11 a) \int \csc ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{11 a^2 \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)}{3 d \sqrt{a+a \sin (c+d x)}}+\frac{1}{8} (11 a) \int \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{11 a^2 \cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}-\frac{11 a^2 \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)}{3 d \sqrt{a+a \sin (c+d x)}}+\frac{1}{16} (11 a) \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{11 a^2 \cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}-\frac{11 a^2 \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)}{3 d \sqrt{a+a \sin (c+d x)}}-\frac{\left (11 a^2\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{11 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{8 d}-\frac{11 a^2 \cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}-\frac{11 a^2 \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)}{3 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.928437, size = 286, normalized size = 1.99 \[ \frac{a \csc ^{10}\left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sin (c+d x)+1)} \left (108 \sin \left (\frac{1}{2} (c+d x)\right )-22 \sin \left (\frac{3}{2} (c+d x)\right )-66 \sin \left (\frac{5}{2} (c+d x)\right )-108 \cos \left (\frac{1}{2} (c+d x)\right )-22 \cos \left (\frac{3}{2} (c+d x)\right )+66 \cos \left (\frac{5}{2} (c+d x)\right )-99 \sin (c+d x) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+99 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+33 \sin (3 (c+d x)) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-33 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{24 d \left (\cot \left (\frac{1}{2} (c+d x)\right )+1\right ) \left (\csc ^2\left (\frac{1}{4} (c+d x)\right )-\sec ^2\left (\frac{1}{4} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(a*Csc[(c + d*x)/2]^10*Sqrt[a*(1 + Sin[c + d*x])]*(-108*Cos[(c + d*x)/2] - 22*Cos[(3*(c + d*x))/2] + 66*Cos[(5
*(c + d*x))/2] + 108*Sin[(c + d*x)/2] - 99*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] + 99*Log[
1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 22*Sin[(3*(c + d*x))/2] - 66*Sin[(5*(c + d*x))/2] + 33
*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 33*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 = 0.744, 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 ( 33\, \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{5/2}{a}^{5/2}-88\, \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{3/2}{a}^{7/2}+33\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ){a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{3}+63\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{9/2} \right ){a}^{-{\frac{7}{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))^(3/2),x)

[Out]

-1/24*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(7/2)*(33*(-a*(sin(d*x+c)-1))^(5/2)*a^(5/2)-88*(-a*(sin(d*x+c
)-1))^(3/2)*a^(7/2)+33*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^5*sin(d*x+c)^3+63*(-a*(sin(d*x+c)-1))^(1/2
)*a^(9/2))/sin(d*x+c)^3/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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(3/2)*csc(d*x + c)^4, x)

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Fricas [B]  time = 1.5301, size = 1002, normalized size = 6.96 \begin{align*} \frac{33 \,{\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} -{\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) + a\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 (33 \, a \cos \left (d x + c\right )^{3} + 11 \, a \cos \left (d x + c\right )^{2} - 41 \, a \cos \left (d x + c\right ) -{\left (33 \, a \cos \left (d x + c\right )^{2} + 22 \, a \cos \left (d x + c\right ) - 19 \, a\right )} \sin \left (d x + c\right ) - 19 \, a\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))^(3/2),x, algorithm="fricas")

[Out]

1/96*(33*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 - (a*cos(d*x + c)^3 + a*cos(d*x + c)^2 - a*cos(d*x + c) - a)*s
in(d*x + c) + a)*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)*s
in(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) - c
os(d*x + c) - 1)) + 4*(33*a*cos(d*x + c)^3 + 11*a*cos(d*x + c)^2 - 41*a*cos(d*x + c) - (33*a*cos(d*x + c)^2 +
22*a*cos(d*x + c) - 19*a)*sin(d*x + c) - 19*a)*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))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 3.18432, size = 844, normalized size = 5.86 \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))^(3/2),x, algorithm="giac")

[Out]

1/48*(66*a^2*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) - 33*a^(3/2)*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) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(26*a*sgn(tan(1/2*d*x + 1/2*c) + 1)
+ (2*a*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c) + 9*a*sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x +
1/2*c)) - (330*sqrt(2)*a^2*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 165*sqrt(2)*sqrt(-a)*a^(3/2)*log(sqr
t(2)*sqrt(a) + sqrt(a)) + 462*a^2*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 231*sqrt(-a)*a^(3/2)*log(sqrt
(2)*sqrt(a) + sqrt(a)) + 136*sqrt(2)*sqrt(-a)*a^(3/2) + 198*sqrt(-a)*a^(3/2))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(5
*sqrt(2)*sqrt(-a) + 7*sqrt(-a)) + 2*(9*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^5*a
^2*sgn(tan(1/2*d*x + 1/2*c) + 1) + 30*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*a^
(5/2)*sgn(tan(1/2*d*x + 1/2*c) + 1) - 48*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2
*a^(7/2)*sgn(tan(1/2*d*x + 1/2*c) + 1) - 9*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))
*a^4*sgn(tan(1/2*d*x + 1/2*c) + 1) + 26*a^(9/2)*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