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

Optimal. Leaf size=144 \[ \frac{3 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{a^{3/2} d}-\frac{9 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{3 \cot (c+d x)}{2 a d \sqrt{a \sin (c+d x)+a}}+\frac{\cot (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

(3*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(a^(3/2)*d) - (9*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(
Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + Cot[c + d*x]/(2*d*(a + a*Sin[c + d*x])^(3/2)) - (3
*Cot[c + d*x])/(2*a*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.357679, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2766, 2984, 2985, 2649, 206, 2773} \[ \frac{3 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{a^{3/2} d}-\frac{9 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{3 \cot (c+d x)}{2 a d \sqrt{a \sin (c+d x)+a}}+\frac{\cot (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(a^(3/2)*d) - (9*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(
Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + Cot[c + d*x]/(2*d*(a + a*Sin[c + d*x])^(3/2)) - (3
*Cot[c + d*x])/(2*a*d*Sqrt[a + a*Sin[c + d*x]])

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 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 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 ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac{\cot (c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}+\frac{\int \frac{\csc ^2(c+d x) \left (3 a-\frac{3}{2} a \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{2 a^2}\\ &=\frac{\cot (c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac{3 \cot (c+d x)}{2 a d \sqrt{a+a \sin (c+d x)}}+\frac{\int \frac{\csc (c+d x) \left (-3 a^2+\frac{3}{2} a^2 \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{2 a^3}\\ &=\frac{\cot (c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac{3 \cot (c+d x)}{2 a d \sqrt{a+a \sin (c+d x)}}-\frac{3 \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{2 a^2}+\frac{9 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{4 a}\\ &=\frac{\cot (c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac{3 \cot (c+d x)}{2 a d \sqrt{a+a \sin (c+d x)}}+\frac{3 \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 d}-\frac{9 \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 )}{2 a d}\\ &=\frac{3 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{a^{3/2} d}-\frac{9 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{\cot (c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac{3 \cot (c+d x)}{2 a d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.627292, size = 449, normalized size = 3.12 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (4 \sin \left (\frac{1}{2} (c+d x)\right )+\frac{2 \sin \left (\frac{1}{4} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{\cos \left (\frac{1}{4} (c+d x)\right )-\sin \left (\frac{1}{4} (c+d x)\right )}-\frac{2 \sin \left (\frac{1}{4} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )}+2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2-2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+6 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-6 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-\tan \left (\frac{1}{4} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2-\cot \left (\frac{1}{4} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2+(18+18 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{4 d (a (\sin (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(4*Sin[(c + d*x)/2] - 2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 2*(Cos[
(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (18 + 18*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])^2 - Cot[(c + d*x)/4]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 6*L
og[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 6*Log[1 - Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (2*Sin[(c + d*x)/4]*(Cos[(c + d*x)/2] + Sin[(
c + d*x)/2])^2)/(Cos[(c + d*x)/4] - Sin[(c + d*x)/4]) - (2*Sin[(c + d*x)/4]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])^2)/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4]) - (Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*Tan[(c + d*x)/4]))/(4*d
*(a*(1 + Sin[c + d*x]))^(3/2))

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Maple [A]  time = 0.599, size = 219, normalized size = 1.5 \begin{align*} -{\frac{1}{4\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) d} \left ( 9\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}a-12\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}a+9\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }\sqrt{2}}{\sqrt{a}}} \right ) a\sin \left ( dx+c \right ) +6\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }\sin \left ( dx+c \right ) \sqrt{a}-12\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ) \sin \left ( dx+c \right ) a+4\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }\sqrt{a} \right ) \sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{-{\frac{5}{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)^2/(a+a*sin(d*x+c))^(3/2),x)

[Out]

-1/4/a^(5/2)*(9*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(d*x+c)^2*a-12*arctanh((-a*(
sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^2*a+9*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))*
a*sin(d*x+c)+6*(-a*(sin(d*x+c)-1))^(1/2)*sin(d*x+c)*a^(1/2)-12*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(
d*x+c)*a+4*(-a*(sin(d*x+c)-1))^(1/2)*a^(1/2))*(-a*(sin(d*x+c)-1))^(1/2)/sin(d*x+c)/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 \frac{\csc \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.90998, size = 1434, normalized size = 9.96 \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)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/8*(9*sqrt(2)*(cos(d*x + c)^3 + 2*cos(d*x + c)^2 + (cos(d*x + c)^2 - cos(d*x + c) - 2)*sin(d*x + c) - cos(d*x
+ c) - 2)*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) - sin(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)) + 6*(cos(d*x + c)^3 + 2*cos(d*x + c)^2 + (cos(d*x + c)^2 - cos(d*x + c) -
2)*sin(d*x + c) - cos(d*x + c) - 2)*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*(3*cos(d*x + c)^2 + (3*cos(d*x + c) + 1)*sin(d*x + c) + 2*cos(d*x + c
) - 1)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 - a^2*d*cos(d*x + c) - 2*a^2*d
+ (a^2*d*cos(d*x + c)^2 - a^2*d*cos(d*x + c) - 2*a^2*d)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{2}{\left (c + d x \right )}}{\left (a \left (\sin{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(c + d*x)**2/(a*(sin(c + d*x) + 1))**(3/2), x)

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Giac [B]  time = 2.61023, size = 684, normalized size = 4.75 \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)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/2*(9*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) + sqrt(a
))/sqrt(-a))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 6*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*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 3*log(abs(-sqrt(a)*tan(1/2*d*
x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/(a^(3/2)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + sqrt(a*tan(1/2*d*x
+ 1/2*c)^2 + a)/(a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 2/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x
+ 1/2*c)^2 + a))^2 - a)*sqrt(a)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 2*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t
an(1/2*d*x + 1/2*c)^2 + a))^3 + (sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*sqrt(a)
- (sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a + a^(3/2))/(((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)^2*a*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d