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

Optimal. Leaf size=144 \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{a^{5/2} d}+\frac{43 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{11 \cos (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}}+\frac{\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}} \]

[Out]

(-2*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(a^(5/2)*d) + (43*ArcTanh[(Sqrt[a]*Cos[c + d*x])
/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + Cos[c + d*x]/(4*d*(a + a*Sin[c + d*x])^(5/2)) +
(11*Cos[c + d*x])/(16*a*d*(a + a*Sin[c + d*x])^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.325342, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2766, 2978, 2985, 2649, 206, 2773} \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{a^{5/2} d}+\frac{43 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{11 \cos (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}}+\frac{\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(a^(5/2)*d) + (43*ArcTanh[(Sqrt[a]*Cos[c + d*x])
/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + Cos[c + d*x]/(4*d*(a + a*Sin[c + d*x])^(5/2)) +
(11*Cos[c + d*x])/(16*a*d*(a + a*Sin[c + d*x])^(3/2))

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + 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] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2978

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

Rule 2985

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), 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]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^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])

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]

Rubi steps

\begin{align*} \int \frac{\csc (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\frac{\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{\int \frac{\csc (c+d x) \left (4 a-\frac{3}{2} a \sin (c+d x)\right )}{(a+a \sin (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{11 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac{\int \frac{\csc (c+d x) \left (8 a^2-\frac{11}{4} a^2 \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{8 a^4}\\ &=\frac{\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{11 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac{\int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{a^3}-\frac{43 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{32 a^2}\\ &=\frac{\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{11 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac{2 \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 )}{a^2 d}+\frac{43 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{16 a^2 d}\\ &=-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{a^{5/2} d}+\frac{43 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{11 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.275614, size = 296, normalized size = 2.06 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (-8 \sin \left (\frac{1}{2} (c+d x)\right )+11 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3-22 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2+4 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-16 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+16 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+(-43-43 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{16 d (a (\sin (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-8*Sin[(c + d*x)/2] + 4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) - 22*Sin
[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 11*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (43 + 43*
I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4
- 16*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + 16*Log[1 - Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4))/(16*d*(a*(1 + Sin[c + d*x]))^(5/2))

________________________________________________________________________________________

Maple [B]  time = 0.696, size = 262, normalized size = 1.8 \begin{align*}{\frac{1}{ \left ( 32+32\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) d} \left ( 2\,\sin \left ( dx+c \right ){a}^{5} \left ( 43\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) -64\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( dx+c \right ) }}{\sqrt{a}}} \right ) \right ) -{a}^{5} \left ( 43\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) -64\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( dx+c \right ) }}{\sqrt{a}}} \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+52\,\sqrt{a-a\sin \left ( dx+c \right ) }{a}^{9/2}-22\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}{a}^{7/2}+86\,{a}^{5}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) -128\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( dx+c \right ) }}{\sqrt{a}}} \right ){a}^{5} \right ) \sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{-{\frac{15}{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)/(a+a*sin(d*x+c))^(5/2),x)

[Out]

1/32/a^(15/2)*(2*sin(d*x+c)*a^5*(43*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-64*arctanh((a-
a*sin(d*x+c))^(1/2)/a^(1/2)))-a^5*(43*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-64*arctanh((
a-a*sin(d*x+c))^(1/2)/a^(1/2)))*cos(d*x+c)^2+52*(a-a*sin(d*x+c))^(1/2)*a^(9/2)-22*(a-a*sin(d*x+c))^(3/2)*a^(7/
2)+86*a^5*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-128*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/
2))*a^5)*(-a*(sin(d*x+c)-1))^(1/2)/(1+sin(d*x+c))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{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(csc(d*x + c)/(a*sin(d*x + c) + a)^(5/2), x)

________________________________________________________________________________________

Fricas [B]  time = 1.98025, size = 1458, normalized size = 10.12 \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)/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/64*(43*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 2*cos(d*x + c) - 4)*sin(d*x + c) - 2*c
os(d*x + c) - 4)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - s
in(d*x + c) + 1) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x +
c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) + 32*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 2*cos(d*
x + c) - 4)*sin(d*x + c) - 2*cos(d*x + c) - 4)*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*(11*cos(d*x + c)^2 + (11*cos(d*x + c) - 4)*sin(d*x + c) +
15*cos(d*x + c) + 4)*sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*
x + c) - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))

________________________________________________________________________________________

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

________________________________________________________________________________________

Giac [B]  time = 3.17235, size = 776, normalized size = 5.39 \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)/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/16*(43*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) + sqr
t(a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 32*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))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 16*log(abs(-sqrt(a)*t
an(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/(a^(5/2)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 2*(53*(sq
rt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^7 + 179*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(
a*tan(1/2*d*x + 1/2*c)^2 + a))^6*sqrt(a) + 127*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 +
a))^5*a - 195*(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) + 7*(sqrt(a)*tan(
1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3*a^2 + 121*(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) - 67*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a^
3 + 15*a^(7/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 2*(sqrt(a)*tan(1/2*d
*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*sqrt(a) - a)^4*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d