3.40 \(\int \csc ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=138 \[ -\frac{5 a \cot (c+d x)}{8 d \sqrt{a \sin (c+d x)+a}}-\frac{5 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 d}-\frac{a \cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}-\frac{5 a \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(-5*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(8*d) - (5*a*Cot[c + d*x])/(8*d*Sqrt[a +
a*Sin[c + d*x]]) - (5*a*Cot[c + d*x]*Csc[c + d*x])/(12*d*Sqrt[a + a*Sin[c + d*x]]) - (a*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 = 0.212697, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2772, 2773, 206} \[ -\frac{5 a \cot (c+d x)}{8 d \sqrt{a \sin (c+d x)+a}}-\frac{5 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 d}-\frac{a \cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}-\frac{5 a \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-5*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(8*d) - (5*a*Cot[c + d*x])/(8*d*Sqrt[a +
a*Sin[c + d*x]]) - (5*a*Cot[c + d*x]*Csc[c + d*x])/(12*d*Sqrt[a + a*Sin[c + d*x]]) - (a*Cot[c + d*x]*Csc[c +
d*x]^2)/(3*d*Sqrt[a + a*Sin[c + d*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) \sqrt{a+a \sin (c+d x)} \, dx &=-\frac{a \cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}+\frac{5}{6} \int \csc ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{5 a \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{a \cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}+\frac{5}{8} \int \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{5 a \cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}-\frac{5 a \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{a \cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}+\frac{5}{16} \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{5 a \cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}-\frac{5 a \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{a \cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}-\frac{(5 a) \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{5 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{8 d}-\frac{5 a \cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}-\frac{5 a \cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{a \cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 1.28357, size = 285, normalized size = 2.07 \[ \frac{\csc ^{10}\left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sin (c+d x)+1)} \left (84 \sin \left (\frac{1}{2} (c+d x)\right )-10 \sin \left (\frac{3}{2} (c+d x)\right )-30 \sin \left (\frac{5}{2} (c+d x)\right )-84 \cos \left (\frac{1}{2} (c+d x)\right )-10 \cos \left (\frac{3}{2} (c+d x)\right )+30 \cos \left (\frac{5}{2} (c+d x)\right )-45 \sin (c+d x) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+45 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+15 \sin (3 (c+d x)) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-15 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{24 d \left (\cot \left (\frac{1}{2} (c+d x)\right )+1\right ) \left (\csc ^2\left (\frac{1}{4} (c+d x)\right )-\sec ^2\left (\frac{1}{4} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Csc[(c + d*x)/2]^10*Sqrt[a*(1 + Sin[c + d*x])]*(-84*Cos[(c + d*x)/2] - 10*Cos[(3*(c + d*x))/2] + 30*Cos[(5*(c
+ d*x))/2] + 84*Sin[(c + d*x)/2] - 45*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] + 45*Log[1 -
Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 10*Sin[(3*(c + d*x))/2] - 30*Sin[(5*(c + d*x))/2] + 15*Log
[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 15*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*S
in[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 = 0.58, size = 158, normalized size = 1.1 \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 ( 15\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{3/2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}+15\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}{a}^{2}+10\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{3/2}\sin \left ( dx+c \right ) +8\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{3/2} \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))^(1/2),x)

[Out]

-1/24*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(15*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*sin(d*x+c)^2+15*arctanh((
-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^3*a^2+10*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*sin(d*x+c)+8*(-a*(sin(
d*x+c)-1))^(1/2)*a^(3/2))/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 \sqrt{a \sin \left (d x + c\right ) + a} \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))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*csc(d*x + c)^4, x)

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Fricas [B]  time = 1.61838, size = 968, normalized size = 7.01 \begin{align*} \frac{15 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1\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 (15 \, \cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )^{2} -{\left (15 \, \cos \left (d x + c\right )^{2} + 10 \, \cos \left (d x + c\right ) - 13\right )} \sin \left (d x + c\right ) - 23 \, \cos \left (d x + c\right ) - 13\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))^(1/2),x, algorithm="fricas")

[Out]

1/96*(15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*sin(d*x + c
) + 1)*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*(15*cos(d*x + c)^3 + 5*cos(d*x + c)^2 - (15*cos(d*x + c)^2 + 10*cos(d*x + c) - 13)*sin(d*x + c) -
23*cos(d*x + c) - 13)*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))**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 2.44771, size = 829, normalized size = 6.01 \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))^(1/2),x, algorithm="giac")

[Out]

1/48*(30*a*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) - 15*sqrt(a)*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)*((2*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan
(1/2*d*x + 1/2*c) + 3*sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 14*sgn(tan(1/2*d*x + 1/2*c) + 1))
- (150*sqrt(2)*a*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 75*sqrt(2)*sqrt(-a)*sqrt(a)*log(sqrt(2)*sqrt(a
) + sqrt(a)) + 210*a*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 105*sqrt(-a)*sqrt(a)*log(sqrt(2)*sqrt(a) +
sqrt(a)) + 112*sqrt(2)*sqrt(-a)*sqrt(a) + 162*sqrt(-a)*sqrt(a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(5*sqrt(2)*sqrt
(-a) + 7*sqrt(-a)) + 2*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^5*a*sgn(tan(1/2*
d*x + 1/2*c) + 1) + 18*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*a^(3/2)*sgn(tan(1
/2*d*x + 1/2*c) + 1) - 24*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(5/2)*sgn(ta
n(1/2*d*x + 1/2*c) + 1) - 3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a^3*sgn(tan(1/
2*d*x + 1/2*c) + 1) + 14*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)^3)/d