3.13 \(\int \frac{1}{(a+b \csc ^2(c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=77 \[ \frac{b \cot (c+d x)}{a d (a+b) \sqrt{a+b \cot ^2(c+d x)+b}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)+b}}\right )}{a^{3/2} d} \]
[Out]
-(ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]]/(a^(3/2)*d)) + (b*Cot[c + d*x])/(a*(a + b)*d*S
qrt[a + b + b*Cot[c + d*x]^2])
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Rubi [A] time = 0.0494553, antiderivative size = 77, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used =
{4128, 382, 377, 203} \[ \frac{b \cot (c+d x)}{a d (a+b) \sqrt{a+b \cot ^2(c+d x)+b}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)+b}}\right )}{a^{3/2} d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Csc[c + d*x]^2)^(-3/2),x]
[Out]
-(ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]]/(a^(3/2)*d)) + (b*Cot[c + d*x])/(a*(a + b)*d*S
qrt[a + b + b*Cot[c + d*x]^2])
Rule 4128
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]
Rule 382
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1]) && NeQ[p, -1]
Rule 377
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{1}{\left (a+b \csc ^2(c+d x)\right )^{3/2}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{b \cot (c+d x)}{a (a+b) d \sqrt{a+b+b \cot ^2(c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\cot (c+d x)\right )}{a d}\\ &=\frac{b \cot (c+d x)}{a (a+b) d \sqrt{a+b+b \cot ^2(c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{\cot (c+d x)}{\sqrt{a+b+b \cot ^2(c+d x)}}\right )}{a d}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \cot (c+d x)}{\sqrt{a+b+b \cot ^2(c+d x)}}\right )}{a^{3/2} d}+\frac{b \cot (c+d x)}{a (a+b) d \sqrt{a+b+b \cot ^2(c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.35066, size = 146, normalized size = 1.9 \[ \frac{\csc ^2(c+d x) \left (\sqrt{2} \csc (c+d x) (a \cos (2 (c+d x))-a-2 b)^{3/2} \log \left (\sqrt{a \cos (2 (c+d x))-a-2 b}+\sqrt{2} \sqrt{a} \cos (c+d x)\right )-\frac{2 \sqrt{a} b \cot (c+d x) (a \cos (2 (c+d x))-a-2 b)}{a+b}\right )}{4 a^{3/2} d \left (a+b \csc ^2(c+d x)\right )^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Csc[c + d*x]^2)^(-3/2),x]
[Out]
(Csc[c + d*x]^2*((-2*Sqrt[a]*b*(-a - 2*b + a*Cos[2*(c + d*x)])*Cot[c + d*x])/(a + b) + Sqrt[2]*(-a - 2*b + a*C
os[2*(c + d*x)])^(3/2)*Csc[c + d*x]*Log[Sqrt[2]*Sqrt[a]*Cos[c + d*x] + Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]]))/
(4*a^(3/2)*d*(a + b*Csc[c + d*x]^2)^(3/2))
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Maple [B] time = 0.38, size = 648, normalized size = 8.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+b*csc(d*x+c)^2)^(3/2),x)
[Out]
1/d*b/(a+b)/(((a+b)*a)^(1/2)+a)/(((a+b)*a)^(1/2)-a)/(-a)^(1/2)*(-1+cos(d*x+c))^2*(cos(d*x+c)+1)^2*(a*cos(d*x+c
)^2-a-b)*(cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*ln(4*cos(d*x+c)*(-a)^(1/2)*(-(a*cos(d*x+c)
^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*a*cos(d*x+c)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2))*a+
cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*ln(4*cos(d*x+c)*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(c
os(d*x+c)+1)^2)^(1/2)-4*a*cos(d*x+c)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2))*b-cos(d*x+c)
*(-a)^(1/2)*b+(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*ln(4*cos(d*x+c)*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)
/(cos(d*x+c)+1)^2)^(1/2)-4*a*cos(d*x+c)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2))*a+(-(a*co
s(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*ln(4*cos(d*x+c)*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1
/2)-4*a*cos(d*x+c)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2))*b)/((a*cos(d*x+c)^2-a-b)/(cos(
d*x+c)^2-1))^(3/2)/sin(d*x+c)^7
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*csc(d*x+c)^2)^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 0.917379, size = 1528, normalized size = 19.84 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*csc(d*x+c)^2)^(3/2),x, algorithm="fricas")
[Out]
[-1/8*(8*a*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*cos(d*x + c)*sin(d*x + c) + ((a^2 + a*b)*co
s(d*x + c)^2 - a^2 - 2*a*b - b^2)*sqrt(-a)*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 + a^3*b)*cos(d*x + c)^6 + 160
*(a^4 + 2*a^3*b + a^2*b^2)*cos(d*x + c)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^4 + 3*a^3*b + 3*
a^2*b^2 + a*b^3)*cos(d*x + c)^2 + 8*(16*a^3*cos(d*x + c)^7 - 24*(a^3 + a^2*b)*cos(d*x + c)^5 + 10*(a^3 + 2*a^2
*b + a*b^2)*cos(d*x + c)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c)^2 - a
- b)/(cos(d*x + c)^2 - 1))*sin(d*x + c)))/((a^4 + a^3*b)*d*cos(d*x + c)^2 - (a^4 + 2*a^3*b + a^2*b^2)*d), -1/
4*(4*a*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*cos(d*x + c)*sin(d*x + c) - ((a^2 + a*b)*cos(d*
x + c)^2 - a^2 - 2*a*b - b^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(d*x + c)^4 - 8*(a^2 + a*b)*cos(d*x + c)^2 + a^2 +
2*a*b + b^2)*sqrt(a)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c)/(2*a^3*cos(d*x + c)^5
- 3*(a^3 + a^2*b)*cos(d*x + c)^3 + (a^3 + 2*a^2*b + a*b^2)*cos(d*x + c))))/((a^4 + a^3*b)*d*cos(d*x + c)^2 - (
a^4 + 2*a^3*b + a^2*b^2)*d)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*csc(d*x+c)**2)**(3/2),x)
[Out]
Integral((a + b*csc(c + d*x)**2)**(-3/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \csc \left (d x + c\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*csc(d*x+c)^2)^(3/2),x, algorithm="giac")
[Out]
integrate((b*csc(d*x + c)^2 + a)^(-3/2), x)