\(\int (a+b \csc ^2(c+d x))^{3/2} \, dx\) [10]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (warning: unable to verify)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F(-2)]
   Mupad [F(-1)]

Optimal result

Integrand size = 16, antiderivative size = 119 \[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=-\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{d}-\frac {\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{2 d}-\frac {b \cot (c+d x) \sqrt {a+b+b \cot ^2(c+d x)}}{2 d} \]

[Out]

-a^(3/2)*arctan(cot(d*x+c)*a^(1/2)/(a+b+b*cot(d*x+c)^2)^(1/2))/d-1/2*(3*a+b)*arctanh(cot(d*x+c)*b^(1/2)/(a+b+b
*cot(d*x+c)^2)^(1/2))*b^(1/2)/d-1/2*b*cot(d*x+c)*(a+b+b*cot(d*x+c)^2)^(1/2)/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4213, 427, 537, 223, 212, 385, 209} \[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=-\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )}{d}-\frac {\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )}{2 d}-\frac {b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}}{2 d} \]

[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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 385

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 427

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 537

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 4213

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {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 {\text {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 \text {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)) \text {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 \text {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)) \text {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} \arctan \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) \text {arctanh}\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] (verified)

Time = 1.21 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.75 \[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\frac {b \left (a+b \csc ^2(c+d x)\right )^{3/2} \left (\sqrt {2} b (3 a+b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {-b} \cos (c+d x)}{\sqrt {-a-2 b+a \cos (2 (c+d x))}}\right )+\sqrt {-b} \left (-b \sqrt {-a-2 b+a \cos (2 (c+d x))} \cot (c+d x) \csc (c+d x)+2 \sqrt {2} a^{3/2} \log \left (\sqrt {2} \sqrt {a} \cos (c+d x)+\sqrt {-a-2 b+a \cos (2 (c+d x))}\right )\right )\right ) \sin ^3(c+d x)}{(-b)^{3/2} d (-a-2 b+a \cos (2 (c+d x)))^{3/2}} \]

[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))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1132\) vs. \(2(101)=202\).

Time = 2.51 (sec) , antiderivative size = 1133, normalized size of antiderivative = 9.52

method result size
default \(\text {Expression too large to display}\) \(1133\)

[In]

int((a+b*csc(d*x+c)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/16/d*csc(d*x+c)*(1/(1-cos(d*x+c))^2*(csc(d*x+c)^2*b*(1-cos(d*x+c))^4+4*a*(1-cos(d*x+c))^2+2*b*(1-cos(d*x+c))
^2+b*sin(d*x+c)^2))^(3/2)*(1-cos(d*x+c))*(2*csc(d*x+c)^2*b^(3/2)*ln((b*(1-cos(d*x+c))^2*csc(d*x+c)^2+b^(1/2)*(
b*(1-cos(d*x+c))^4*csc(d*x+c)^4+4*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+2*b*(1-cos(d*x+c))^2*csc(d*x+c)^2+b)^(1/2)+2
*a+b)/b^(1/2))*(1-cos(d*x+c))^2*(-a)^(1/2)-2*csc(d*x+c)^2*b^(3/2)*ln(2/(1-cos(d*x+c))^2*(2*a*(1-cos(d*x+c))^2+
b*(1-cos(d*x+c))^2+sin(d*x+c)^2*b^(1/2)*(b*(1-cos(d*x+c))^4*csc(d*x+c)^4+4*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+2*b
*(1-cos(d*x+c))^2*csc(d*x+c)^2+b)^(1/2)+b*sin(d*x+c)^2))*(1-cos(d*x+c))^2*(-a)^(1/2)+6*csc(d*x+c)^2*a*b^(1/2)*
ln((b*(1-cos(d*x+c))^2*csc(d*x+c)^2+b^(1/2)*(b*(1-cos(d*x+c))^4*csc(d*x+c)^4+4*a*(1-cos(d*x+c))^2*csc(d*x+c)^2
+2*b*(1-cos(d*x+c))^2*csc(d*x+c)^2+b)^(1/2)+2*a+b)/b^(1/2))*(1-cos(d*x+c))^2*(-a)^(1/2)-6*csc(d*x+c)^2*b^(1/2)
*ln(2/(1-cos(d*x+c))^2*(2*a*(1-cos(d*x+c))^2+b*(1-cos(d*x+c))^2+sin(d*x+c)^2*b^(1/2)*(b*(1-cos(d*x+c))^4*csc(d
*x+c)^4+4*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+2*b*(1-cos(d*x+c))^2*csc(d*x+c)^2+b)^(1/2)+b*sin(d*x+c)^2))*a*(1-cos
(d*x+c))^2*(-a)^(1/2)+csc(d*x+c)^2*b*(b*(1-cos(d*x+c))^4*csc(d*x+c)^4+4*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+2*b*(1
-cos(d*x+c))^2*csc(d*x+c)^2+b)^(1/2)*(1-cos(d*x+c))^2*(-a)^(1/2)-8*csc(d*x+c)^2*a^2*ln(4*(a*(1-cos(d*x+c))^2*c
sc(d*x+c)^2+(-a)^(1/2)*(b*(1-cos(d*x+c))^4*csc(d*x+c)^4+4*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+2*b*(1-cos(d*x+c))^2
*csc(d*x+c)^2+b)^(1/2)-a)/((1-cos(d*x+c))^2*csc(d*x+c)^2+1))*(1-cos(d*x+c))^2-b*(b*(1-cos(d*x+c))^4*csc(d*x+c)
^4+4*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+2*b*(1-cos(d*x+c))^2*csc(d*x+c)^2+b)^(1/2)*(-a)^(1/2))/(b*(1-cos(d*x+c))^
4*csc(d*x+c)^4+4*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+2*b*(1-cos(d*x+c))^2*csc(d*x+c)^2+b)^(3/2)*4^(1/2)/(-a)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (101) = 202\).

Time = 0.55 (sec) , antiderivative size = 1607, normalized size of antiderivative = 13.50 \[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\text {Too large to display} \]

[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))]

Sympy [F]

\[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\int \left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\int { {\left (b \csc \left (d x + c\right )^{2} + a\right )}^{\frac {3}{2}} \,d x } \]

[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)

Giac [F(-2)]

Exception generated. \[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx=\int {\left (a+\frac {b}{{\sin \left (c+d\,x\right )}^2}\right )}^{3/2} \,d x \]

[In]

int((a + b/sin(c + d*x)^2)^(3/2),x)

[Out]

int((a + b/sin(c + d*x)^2)^(3/2), x)