3.9 \(\int (a+b \csc ^2(c+d x))^{5/2} \, dx\)

Optimal. Leaf size=167 \[ -\frac{\sqrt{b} \left (15 a^2+10 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)+b}}\right )}{8 d}-\frac{a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)+b}}\right )}{d}-\frac{b \cot (c+d x) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}{4 d}-\frac{b (7 a+3 b) \cot (c+d x) \sqrt{a+b \cot ^2(c+d x)+b}}{8 d} \]

[Out]

-((a^(5/2)*ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/d) - (Sqrt[b]*(15*a^2 + 10*a*b + 3*b
^2)*ArcTanh[(Sqrt[b]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/(8*d) - (b*(7*a + 3*b)*Cot[c + d*x]*Sqrt[a
+ b + b*Cot[c + d*x]^2])/(8*d) - (b*Cot[c + d*x]*(a + b + b*Cot[c + d*x]^2)^(3/2))/(4*d)

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Rubi [A]  time = 0.181948, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4128, 416, 528, 523, 217, 206, 377, 203} \[ -\frac{\sqrt{b} \left (15 a^2+10 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)+b}}\right )}{8 d}-\frac{a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)+b}}\right )}{d}-\frac{b \cot (c+d x) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}{4 d}-\frac{b (7 a+3 b) \cot (c+d x) \sqrt{a+b \cot ^2(c+d x)+b}}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x]^2)^(5/2),x]

[Out]

-((a^(5/2)*ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/d) - (Sqrt[b]*(15*a^2 + 10*a*b + 3*b
^2)*ArcTanh[(Sqrt[b]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/(8*d) - (b*(7*a + 3*b)*Cot[c + d*x]*Sqrt[a
+ b + b*Cot[c + d*x]^2])/(8*d) - (b*Cot[c + d*x]*(a + b + b*Cot[c + d*x]^2)^(3/2))/(4*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 528

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
(f*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(n*(p + q + 1) + 1)), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]

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 )^{5/2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^{5/2}}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{b \cot (c+d x) \left (a+b+b \cot ^2(c+d x)\right )^{3/2}}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b+b x^2} \left ((a+b) (4 a+3 b)+b (7 a+3 b) x^2\right )}{1+x^2} \, dx,x,\cot (c+d x)\right )}{4 d}\\ &=-\frac{b (7 a+3 b) \cot (c+d x) \sqrt{a+b+b \cot ^2(c+d x)}}{8 d}-\frac{b \cot (c+d x) \left (a+b+b \cot ^2(c+d x)\right )^{3/2}}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{(a+b) \left (8 a^2+7 a b+3 b^2\right )+b \left (15 a^2+10 a b+3 b^2\right ) x^2}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\cot (c+d x)\right )}{8 d}\\ &=-\frac{b (7 a+3 b) \cot (c+d x) \sqrt{a+b+b \cot ^2(c+d x)}}{8 d}-\frac{b \cot (c+d x) \left (a+b+b \cot ^2(c+d x)\right )^{3/2}}{4 d}-\frac{a^3 \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{\left (b \left (15 a^2+10 a b+3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b+b x^2}} \, dx,x,\cot (c+d x)\right )}{8 d}\\ &=-\frac{b (7 a+3 b) \cot (c+d x) \sqrt{a+b+b \cot ^2(c+d x)}}{8 d}-\frac{b \cot (c+d x) \left (a+b+b \cot ^2(c+d x)\right )^{3/2}}{4 d}-\frac{a^3 \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{\left (b \left (15 a^2+10 a b+3 b^2\right )\right ) \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 )}{8 d}\\ &=-\frac{a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \cot (c+d x)}{\sqrt{a+b+b \cot ^2(c+d x)}}\right )}{d}-\frac{\sqrt{b} \left (15 a^2+10 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a+b+b \cot ^2(c+d x)}}\right )}{8 d}-\frac{b (7 a+3 b) \cot (c+d x) \sqrt{a+b+b \cot ^2(c+d x)}}{8 d}-\frac{b \cot (c+d x) \left (a+b+b \cot ^2(c+d x)\right )^{3/2}}{4 d}\\ \end{align*}

Mathematica [A]  time = 3.56811, size = 246, normalized size = 1.47 \[ \frac{\sin ^5(c+d x) \left (a+b \csc ^2(c+d x)\right )^{5/2} \left (\sqrt{2} b \left (15 a^2+10 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{-b} \cos (c+d x)}{\sqrt{a \cos (2 (c+d x))-a-2 b}}\right )+\frac{1}{2} \sqrt{-b} \left (16 \sqrt{2} a^{5/2} \log \left (\sqrt{a \cos (2 (c+d x))-a-2 b}+\sqrt{2} \sqrt{a} \cos (c+d x)\right )+b \cot (c+d x) \csc ^3(c+d x) \sqrt{a \cos (2 (c+d x))-a-2 b} (3 (3 a+b) \cos (2 (c+d x))-9 a-7 b)\right )\right )}{2 \sqrt{-b} d (a \cos (2 (c+d x))-a-2 b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Csc[c + d*x]^2)^(5/2)*(Sqrt[2]*b*(15*a^2 + 10*a*b + 3*b^2)*ArcTanh[(Sqrt[2]*Sqrt[-b]*Cos[c + d*x])/Sqr
t[-a - 2*b + a*Cos[2*(c + d*x)]]] + (Sqrt[-b]*(b*Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]*(-9*a - 7*b + 3*(3*a + b)
*Cos[2*(c + d*x)])*Cot[c + d*x]*Csc[c + d*x]^3 + 16*Sqrt[2]*a^(5/2)*Log[Sqrt[2]*Sqrt[a]*Cos[c + d*x] + Sqrt[-a
- 2*b + a*Cos[2*(c + d*x)]]]))/2)*Sin[c + d*x]^5)/(2*Sqrt[-b]*d*(-a - 2*b + a*Cos[2*(c + d*x)])^(5/2))

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Maple [B]  time = 0.452, size = 3621, normalized size = 21.7 \begin{align*} \text{output 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)^(5/2),x)

[Out]

1/16/d/(-a)^(1/2)*((a*cos(d*x+c)^2-a-b)/(cos(d*x+c)^2-1))^(5/2)*(-1+cos(d*x+c))^4*(15*cos(d*x+c)^3*(-a)^(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)*b^(1/2)*a^2-15*cos(d*x+c)^3*(-a)^(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)))*b^(1/2)*a^2-10*cos(d*x+c)^2*(-a)^(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*co
s(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+10*cos(d*x+c)^2*(-a)^(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)))*b^(3/2)*a+18*cos(d*x+c)^3*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)
+1)^2)^(1/2)*a*b-15*cos(d*x+c)^2*(-a)^(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)*b^(1/2)*a^2+15*cos(d*x+c)^2*(-a)^(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)))*b^(1/2)*a^
2-10*cos(d*x+c)*(-a)^(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)*b^(3/2)*a
+10*cos(d*x+c)*(-a)^(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)))*b^(3/2)*a-15*cos(d*x+c)*(-a)^(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)*b^(1/2)*a^2+15*cos(d*x+c)*(-a)^(
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)))*b^(1/2)*a^2-18*cos(d*x+c)*(-a)^(1/2)*(-(a*cos(d*x+c)
^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*a*b+10*cos(d*x+c)^3*(-a)^(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)*b^(3/2)*a-10*cos(d*x+c)^3*(-a)^(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)))*b^(3/2)*a+6*cos(d*x+c)^3*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*b^2-10*cos(d*x+c)*(-
a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*b^2+3*cos(d*x+c)^3*(-a)^(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)*b^(5/2)-3*cos(d*x+c)^3*(-a)^(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)))*b^(5/2)-3*cos(d*x+c)^2*(-a)^(1/2)*ln(-2/b^(1/2)*(-1+cos(d*x+c))*(cos(d*x+c)*(-(a*co
s(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^(5/2)+3*cos(d*x+c)^2*(-a)^(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))
)*b^(5/2)-3*cos(d*x+c)*(-a)^(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)*b^
(5/2)+3*cos(d*x+c)*(-a)^(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)))*b^(5/2)+10*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*(-a)^(1/2)-10*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*(-a)^(1/2)+15*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^2*(-a)^(1/2)-15*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^2*(-a)^(
1/2)+16*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^3-16*cos(d*x+c)^3*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^3+3*b^(5/2)*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)-3*
b^(5/2)*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)+16*cos(d*x+c)^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^3-16*a^3*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)^7/(-(a*cos(d*x+c)^2-a-b)/(cos
(d*x+c)+1)^2)^(5/2)

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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{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csc(d*x+c)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*csc(d*x + c)^2 + a)^(5/2), x)

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Fricas [B]  time = 4.11659, size = 4761, normalized size = 28.51 \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)^(5/2),x, algorithm="fricas")

[Out]

[1/32*(4*(a^2*cos(d*x + c)^2 - a^2)*sqrt(-a)*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 + a^3*b)*cos(d*x + c)^6 + 1
60*(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) + ((15*a^2 + 10*a*b + 3*b^2)*cos(d*x + c)^2 - 15*a^2
- 10*a*b - 3*b^2)*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*(3*(3*a*b + b^2
)*cos(d*x + c)^3 - (9*a*b + 5*b^2)*cos(d*x + c))*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1)))/((d*co
s(d*x + c)^2 - d)*sin(d*x + c)), -1/16*(((15*a^2 + 10*a*b + 3*b^2)*cos(d*x + c)^2 - 15*a^2 - 10*a*b - 3*b^2)*s
qrt(-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) - 2*(a^2*cos(d*x + c)^2 - a^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))*sin(d*x + c) + 2*(3*(3*a*b + b^2)*cos(d*x + c)^3 - (9*a*b + 5*b^2)*cos(d*x + c))*sqrt((a*cos(d*x + c
)^2 - a - b)/(cos(d*x + c)^2 - 1)))/((d*cos(d*x + c)^2 - d)*sin(d*x + c)), 1/32*(8*(a^2*cos(d*x + c)^2 - a^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*c
os(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) + ((15*a^2 + 10*a*b + 3*b^2)*cos(d*x + c)^2 - 15*a^2
- 10*a*b - 3*b^2)*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*(3*(3*a*b + b^2
)*cos(d*x + c)^3 - (9*a*b + 5*b^2)*cos(d*x + c))*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1)))/((d*co
s(d*x + c)^2 - d)*sin(d*x + c)), 1/16*(4*(a^2*cos(d*x + c)^2 - a^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))
)*sin(d*x + c) - ((15*a^2 + 10*a*b + 3*b^2)*cos(d*x + c)^2 - 15*a^2 - 10*a*b - 3*b^2)*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) - 2*(3*(3*a*b + b^2)*cos(d*x + c)^3 - (9*a*b + 5*b^2
)*cos(d*x + c))*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1)))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))]

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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((a+b*csc(d*x+c)**2)**(5/2),x)

[Out]

Timed out

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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{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csc(d*x+c)^2)^(5/2),x, algorithm="giac")

[Out]

integrate((b*csc(d*x + c)^2 + a)^(5/2), x)