3.10 \(\int (a+b \csc ^2(c+d x))^{3/2} \, dx\)
Optimal. Leaf size=119 \[ -\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)+b}}\right )}{d}-\frac{b \cot (c+d x) \sqrt{a+b \cot ^2(c+d x)+b}}{2 d}-\frac{\sqrt{b} (3 a+b) \tanh ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)+b}}\right )}{2 d} \]
[Out]
-((a^(3/2)*ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/d) - (Sqrt[b]*(3*a + b)*ArcTanh[(Sqr
t[b]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/(2*d) - (b*Cot[c + d*x]*Sqrt[a + b + b*Cot[c + d*x]^2])/(2
*d)
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Rubi [A] time = 0.0963157, antiderivative size = 119, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used =
{4128, 416, 523, 217, 206, 377, 203} \[ -\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)+b}}\right )}{d}-\frac{b \cot (c+d x) \sqrt{a+b \cot ^2(c+d x)+b}}{2 d}-\frac{\sqrt{b} (3 a+b) \tanh ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)+b}}\right )}{2 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Csc[c + d*x]^2)^(3/2),x]
[Out]
-((a^(3/2)*ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/d) - (Sqrt[b]*(3*a + b)*ArcTanh[(Sqr
t[b]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/(2*d) - (b*Cot[c + d*x]*Sqrt[a + b + b*Cot[c + d*x]^2])/(2
*d)
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 416
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] && !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]
Rule 523
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]
Rule 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 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 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 \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{b \cot (c+d x) \sqrt{a+b+b \cot ^2(c+d x)}}{2 d}-\frac{\operatorname{Subst}\left (\int \frac{(a+b) (2 a+b)+b (3 a+b) x^2}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=-\frac{b \cot (c+d x) \sqrt{a+b+b \cot ^2(c+d x)}}{2 d}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\cot (c+d x)\right )}{d}-\frac{(b (3 a+b)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b+b x^2}} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=-\frac{b \cot (c+d x) \sqrt{a+b+b \cot ^2(c+d x)}}{2 d}-\frac{a^2 \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 )}{d}-\frac{(b (3 a+b)) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\cot (c+d x)}{\sqrt{a+b+b \cot ^2(c+d x)}}\right )}{2 d}\\ &=-\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \cot (c+d x)}{\sqrt{a+b+b \cot ^2(c+d x)}}\right )}{d}-\frac{\sqrt{b} (3 a+b) \tanh ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a+b+b \cot ^2(c+d x)}}\right )}{2 d}-\frac{b \cot (c+d x) \sqrt{a+b+b \cot ^2(c+d x)}}{2 d}\\ \end{align*}
Mathematica [A] time = 0.914206, size = 208, normalized size = 1.75 \[ \frac{b \sin ^3(c+d x) \left (a+b \csc ^2(c+d x)\right )^{3/2} \left (\sqrt{-b} \left (2 \sqrt{2} a^{3/2} \log \left (\sqrt{a \cos (2 (c+d x))-a-2 b}+\sqrt{2} \sqrt{a} \cos (c+d x)\right )-b \cot (c+d x) \csc (c+d x) \sqrt{a \cos (2 (c+d x))-a-2 b}\right )+\sqrt{2} b (3 a+b) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{-b} \cos (c+d x)}{\sqrt{a \cos (2 (c+d x))-a-2 b}}\right )\right )}{(-b)^{3/2} d (a \cos (2 (c+d x))-a-2 b)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Csc[c + d*x]^2)^(3/2),x]
[Out]
(b*(a + b*Csc[c + d*x]^2)^(3/2)*(Sqrt[2]*b*(3*a + b)*ArcTanh[(Sqrt[2]*Sqrt[-b]*Cos[c + d*x])/Sqrt[-a - 2*b + a
*Cos[2*(c + d*x)]]] + Sqrt[-b]*(-(b*Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]*Cot[c + d*x]*Csc[c + d*x]) + 2*Sqrt[2]
*a^(3/2)*Log[Sqrt[2]*Sqrt[a]*Cos[c + d*x] + Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]]))*Sin[c + d*x]^3)/((-b)^(3/2)
*d*(-a - 2*b + a*Cos[2*(c + d*x)])^(3/2))
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Maple [B] time = 0.427, size = 1286, normalized size = 10.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*csc(d*x+c)^2)^(3/2),x)
[Out]
-1/4/d/(-a)^(1/2)*((a*cos(d*x+c)^2-a-b)/(cos(d*x+c)^2-1))^(3/2)*(-1+cos(d*x+c))^2*(cos(d*x+c)*ln(-2/b^(1/2)*(-
1+cos(d*x+c))*(cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*b^(1/2)+a*cos(d*x+c)+b^(1/2)*(-(a*cos
(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)+a+b)/sin(d*x+c)^2)*b^(3/2)*(-a)^(1/2)-cos(d*x+c)*ln(-4*(cos(d*x+c)*(-(a
*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*b^(1/2)-a*cos(d*x+c)+b^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^
2)^(1/2)+a+b)/(-1+cos(d*x+c)))*b^(3/2)*(-a)^(1/2)+3*cos(d*x+c)*ln(-2/b^(1/2)*(-1+cos(d*x+c))*(cos(d*x+c)*(-(a*
cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*b^(1/2)+a*cos(d*x+c)+b^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2
)^(1/2)+a+b)/sin(d*x+c)^2)*b^(1/2)*(-a)^(1/2)*a-3*cos(d*x+c)*ln(-4*(cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x
+c)+1)^2)^(1/2)*b^(1/2)-a*cos(d*x+c)+b^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)+a+b)/(-1+cos(d*x+c
)))*b^(1/2)*(-a)^(1/2)*a-b^(3/2)*ln(-2/b^(1/2)*(-1+cos(d*x+c))*(cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+
1)^2)^(1/2)*b^(1/2)+a*cos(d*x+c)+b^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)+a+b)/sin(d*x+c)^2)*(-a
)^(1/2)+b^(3/2)*ln(-4*(cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*b^(1/2)-a*cos(d*x+c)+b^(1/2)*
(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)+a+b)/(-1+cos(d*x+c)))*(-a)^(1/2)+2*cos(d*x+c)*(-(a*cos(d*x+c)^2
-a-b)/(cos(d*x+c)+1)^2)^(1/2)*(-a)^(1/2)*b-4*cos(d*x+c)*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^2-3*b^(1/2)*
ln(-2/b^(1/2)*(-1+cos(d*x+c))*(cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*b^(1/2)+a*cos(d*x+c)+
b^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)+a+b)/sin(d*x+c)^2)*a*(-a)^(1/2)+3*b^(1/2)*ln(-4*(cos(d*
x+c)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*b^(1/2)-a*cos(d*x+c)+b^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d
*x+c)+1)^2)^(1/2)+a+b)/(-1+cos(d*x+c)))*a*(-a)^(1/2)+4*a^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)))/sin(d*x+c)
^3/(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(3/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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((a+b*csc(d*x+c)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate((b*csc(d*x + c)^2 + a)^(3/2), x)
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Fricas [B] time = 1.62462, size = 3914, normalized size = 32.89 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*csc(d*x+c)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/8*(sqrt(-a)*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))*sin(d*x + c) + (3*a + b)*sqrt(b)*log(2*((a^2 - 6*a*b + b^2)*cos(d*x + c)^4 - 2*(a^2 - 2*a*b -
3*b^2)*cos(d*x + c)^2 + 4*((a - b)*cos(d*x + c)^3 - (a + b)*cos(d*x + c))*sqrt(b)*sqrt((a*cos(d*x + c)^2 - a -
b)/(cos(d*x + c)^2 - 1))*sin(d*x + c) + a^2 + 2*a*b + b^2)/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1))*sin(d*x +
c) - 4*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*cos(d*x + c))/(d*sin(d*x + c)), -1/8*(2*(3*a +
b)*sqrt(-b)*arctan(-1/2*((a - b)*cos(d*x + c)^2 - a - b)*sqrt(-b)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x +
c)^2 - 1))*sin(d*x + c)/(a*b*cos(d*x + c)^3 - (a*b + b^2)*cos(d*x + c)))*sin(d*x + c) - sqrt(-a)*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))*sin(d*x + c)
+ 4*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*cos(d*x + c))/(d*sin(d*x + c)), 1/8*(2*a^(3/2)*ar
ctan(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)))*sin(d*x + c) + (3*a + b)*sqrt(b)*log(2*((a^2 - 6*a*b + b^2)*cos(d*x + c)^4 -
2*(a^2 - 2*a*b - 3*b^2)*cos(d*x + c)^2 + 4*((a - b)*cos(d*x + c)^3 - (a + b)*cos(d*x + c))*sqrt(b)*sqrt((a*co
s(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c) + a^2 + 2*a*b + b^2)/(cos(d*x + c)^4 - 2*cos(d*x + c)
^2 + 1))*sin(d*x + c) - 4*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*cos(d*x + c))/(d*sin(d*x + c
)), 1/4*(a^(3/2)*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)))*sin(d*x + c) - (3*a + b)*sqrt(-b)*arctan(-1/2*((a - b)*co
s(d*x + c)^2 - a - b)*sqrt(-b)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c)/(a*b*cos(d*x
+ c)^3 - (a*b + b^2)*cos(d*x + c)))*sin(d*x + c) - 2*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*
cos(d*x + c))/(d*sin(d*x + c))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \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((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{\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((a+b*csc(d*x+c)^2)^(3/2),x, algorithm="giac")
[Out]
integrate((b*csc(d*x + c)^2 + a)^(3/2), x)