Integrand size = 16, antiderivative size = 180 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{a^{7/2} d}+\frac {b \cot (c+d x)}{5 a (a+b) d \left (a+b+b \cot ^2(c+d x)\right )^{5/2}}+\frac {b (9 a+5 b) \cot (c+d x)}{15 a^2 (a+b)^2 d \left (a+b+b \cot ^2(c+d x)\right )^{3/2}}+\frac {b \left (33 a^2+40 a b+15 b^2\right ) \cot (c+d x)}{15 a^3 (a+b)^3 d \sqrt {a+b+b \cot ^2(c+d x)}} \]
-arctan(cot(d*x+c)*a^(1/2)/(a+b+b*cot(d*x+c)^2)^(1/2))/a^(7/2)/d+1/5*b*cot (d*x+c)/a/(a+b)/d/(a+b+b*cot(d*x+c)^2)^(5/2)+1/15*b*(9*a+5*b)*cot(d*x+c)/a ^2/(a+b)^2/d/(a+b+b*cot(d*x+c)^2)^(3/2)+1/15*b*(33*a^2+40*a*b+15*b^2)*cot( d*x+c)/a^3/(a+b)^3/d/(a+b+b*cot(d*x+c)^2)^(1/2)
Time = 1.69 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\frac {\csc ^7(c+d x) \left (\frac {b \cos (c+d x) (a+2 b-a \cos (2 (c+d x))) \left (135 a^4+480 a^3 b+709 a^2 b^2+460 a b^3+120 b^4-4 a \left (45 a^3+135 a^2 b+117 a b^2+35 b^3\right ) \cos (2 (c+d x))+a^2 \left (45 a^2+60 a b+23 b^2\right ) \cos (4 (c+d x))\right )}{15 a^3 (a+b)^3}+\frac {\sqrt {2} (-a-2 b+a \cos (2 (c+d x)))^{7/2} \log \left (\sqrt {2} \sqrt {a} \cos (c+d x)+\sqrt {-a-2 b+a \cos (2 (c+d x))}\right )}{a^{7/2}}\right )}{16 d \left (a+b \csc ^2(c+d x)\right )^{7/2}} \]
(Csc[c + d*x]^7*((b*Cos[c + d*x]*(a + 2*b - a*Cos[2*(c + d*x)])*(135*a^4 + 480*a^3*b + 709*a^2*b^2 + 460*a*b^3 + 120*b^4 - 4*a*(45*a^3 + 135*a^2*b + 117*a*b^2 + 35*b^3)*Cos[2*(c + d*x)] + a^2*(45*a^2 + 60*a*b + 23*b^2)*Cos [4*(c + d*x)]))/(15*a^3*(a + b)^3) + (Sqrt[2]*(-a - 2*b + a*Cos[2*(c + d*x )])^(7/2)*Log[Sqrt[2]*Sqrt[a]*Cos[c + d*x] + Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]])/a^(7/2)))/(16*d*(a + b*Csc[c + d*x]^2)^(7/2))
Time = 0.34 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4616, 316, 402, 402, 27, 291, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a+b \sec \left (c+d x+\frac {\pi }{2}\right )^2\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 4616 |
| \(\displaystyle -\frac {\int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \left (b \cot ^2(c+d x)+a+b\right )^{7/2}}d\cot (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle -\frac {\frac {\int \frac {-4 b \cot ^2(c+d x)+5 a+b}{\left (\cot ^2(c+d x)+1\right ) \left (b \cot ^2(c+d x)+a+b\right )^{5/2}}d\cot (c+d x)}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle -\frac {\frac {\frac {\int \frac {15 a^2+12 b a+5 b^2-2 b (9 a+5 b) \cot ^2(c+d x)}{\left (\cot ^2(c+d x)+1\right ) \left (b \cot ^2(c+d x)+a+b\right )^{3/2}}d\cot (c+d x)}{3 a (a+b)}-\frac {b (9 a+5 b) \cot (c+d x)}{3 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle -\frac {\frac {\frac {\frac {\int \frac {15 (a+b)^3}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)}{a (a+b)}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \cot (c+d x)}{a (a+b) \sqrt {a+b \cot ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \cot (c+d x)}{3 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {\frac {\frac {15 (a+b)^2 \int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)}{a}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \cot (c+d x)}{a (a+b) \sqrt {a+b \cot ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \cot (c+d x)}{3 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle -\frac {\frac {\frac {\frac {15 (a+b)^2 \int \frac {1}{\frac {a \cot ^2(c+d x)}{b \cot ^2(c+d x)+a+b}+1}d\frac {\cot (c+d x)}{\sqrt {b \cot ^2(c+d x)+a+b}}}{a}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \cot (c+d x)}{a (a+b) \sqrt {a+b \cot ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \cot (c+d x)}{3 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {\frac {\frac {\frac {15 (a+b)^2 \arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )}{a^{3/2}}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \cot (c+d x)}{a (a+b) \sqrt {a+b \cot ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \cot (c+d x)}{3 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\) |
-((-1/5*(b*Cot[c + d*x])/(a*(a + b)*(a + b + b*Cot[c + d*x]^2)^(5/2)) + (- 1/3*(b*(9*a + 5*b)*Cot[c + d*x])/(a*(a + b)*(a + b + b*Cot[c + d*x]^2)^(3/ 2)) + ((15*(a + b)^2*ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/a^(3/2) - (b*(33*a^2 + 40*a*b + 15*b^2)*Cot[c + d*x])/(a*(a + b) *Sqrt[a + b + b*Cot[c + d*x]^2]))/(3*a*(a + b)))/(5*a*(a + b)))/d)
3.1.15.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[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]
Leaf count of result is larger than twice the leaf count of optimal. \(4845\) vs. \(2(162)=324\).
Time = 7.53 (sec) , antiderivative size = 4846, normalized size of antiderivative = 26.92
1/30/d*csc(d*x+c)^3*(-45*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*si n(d*x+c)^2*cos(d*x+c)^5*ln(4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2 )*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1) ^2)^(1/2)-4*cos(d*x+c)*a)*a^4*b^2+45*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1) ^2)^(1/2)*sin(d*x+c)^4*cos(d*x+c)^2*ln(4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c )+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(c os(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a^5*b-45*(-(a*cos(d*x+c)^2-a-b)/(cos (d*x+c)+1)^2)^(1/2)*sin(d*x+c)^2*cos(d*x+c)^4*ln(4*(-(a*cos(d*x+c)^2-a-b)/ (cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c) ^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a^4*b^2-90*(-(a*cos(d*x+c) ^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)^4*cos(d*x+c)*ln(4*(-(a*cos(d*x+ c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a* cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a^5*b+225*(-(a*c os(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)^2*cos(d*x+c)^3*ln(4*(- (a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^ (1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a^4*b ^2+225*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)^2*cos(d*x +c)^2*ln(4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^ (1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d* x+c)*a)*a^4*b^2-225*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*sin(...
Leaf count of result is larger than twice the leaf count of optimal. 667 vs. \(2 (162) = 324\).
Time = 1.99 (sec) , antiderivative size = 1445, normalized size of antiderivative = 8.03 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\text {Too large to display} \]
[-1/120*(15*((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*cos(d*x + c)^6 - a^6 - 6*a^5*b - 15*a^4*b^2 - 20*a^3*b^3 - 15*a^2*b^4 - 6*a*b^5 - b^6 - 3*(a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*cos(d*x + c)^4 + 3*(a^6 + 5*a^5 *b + 10*a^4*b^2 + 10*a^3*b^3 + 5*a^2*b^4 + a*b^5)*cos(d*x + c)^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)) + 8* ((45*a^5*b + 60*a^4*b^2 + 23*a^3*b^3)*cos(d*x + c)^5 - 5*(18*a^5*b + 39*a^ 4*b^2 + 28*a^3*b^3 + 7*a^2*b^4)*cos(d*x + c)^3 + 15*(3*a^5*b + 9*a^4*b^2 + 10*a^3*b^3 + 5*a^2*b^4 + a*b^5)*cos(d*x + c))*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c))/((a^10 + 3*a^9*b + 3*a^8*b^2 + a^ 7*b^3)*d*cos(d*x + c)^6 - 3*(a^10 + 4*a^9*b + 6*a^8*b^2 + 4*a^7*b^3 + a^6* b^4)*d*cos(d*x + c)^4 + 3*(a^10 + 5*a^9*b + 10*a^8*b^2 + 10*a^7*b^3 + 5*a^ 6*b^4 + a^5*b^5)*d*cos(d*x + c)^2 - (a^10 + 6*a^9*b + 15*a^8*b^2 + 20*a^7* b^3 + 15*a^6*b^4 + 6*a^5*b^5 + a^4*b^6)*d), 1/60*(15*((a^6 + 3*a^5*b + 3*a ^4*b^2 + a^3*b^3)*cos(d*x + c)^6 - a^6 - 6*a^5*b - 15*a^4*b^2 - 20*a^3*b^3 - 15*a^2*b^4 - 6*a*b^5 - b^6 - 3*(a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^...
\[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\int \frac {1}{\left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]
Timed out. \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 737 vs. \(2 (162) = 324\).
Time = 0.85 (sec) , antiderivative size = 737, normalized size of antiderivative = 4.09 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\text {Too large to display} \]
-1/15*(((((((33*a^20*b^3*sgn(sin(d*x + c)) + 40*a^19*b^4*sgn(sin(d*x + c)) + 15*a^18*b^5*sgn(sin(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/(a^24 + 3*a^23*b + 3*a^22*b^2 + a^21*b^3) + 5*(60*a^21*b^2*sgn(sin(d*x + c)) + 95*a^20*b^3* sgn(sin(d*x + c)) + 52*a^19*b^4*sgn(sin(d*x + c)) + 9*a^18*b^5*sgn(sin(d*x + c)))/(a^24 + 3*a^23*b + 3*a^22*b^2 + a^21*b^3))*tan(1/2*d*x + 1/2*c)^2 + 10*(72*a^22*b*sgn(sin(d*x + c)) + 126*a^21*b^2*sgn(sin(d*x + c)) + 81*a^ 20*b^3*sgn(sin(d*x + c)) + 22*a^19*b^4*sgn(sin(d*x + c)) + 3*a^18*b^5*sgn( sin(d*x + c)))/(a^24 + 3*a^23*b + 3*a^22*b^2 + a^21*b^3))*tan(1/2*d*x + 1/ 2*c)^2 - 10*(72*a^22*b*sgn(sin(d*x + c)) + 126*a^21*b^2*sgn(sin(d*x + c)) + 81*a^20*b^3*sgn(sin(d*x + c)) + 22*a^19*b^4*sgn(sin(d*x + c)) + 3*a^18*b ^5*sgn(sin(d*x + c)))/(a^24 + 3*a^23*b + 3*a^22*b^2 + a^21*b^3))*tan(1/2*d *x + 1/2*c)^2 - 5*(60*a^21*b^2*sgn(sin(d*x + c)) + 95*a^20*b^3*sgn(sin(d*x + c)) + 52*a^19*b^4*sgn(sin(d*x + c)) + 9*a^18*b^5*sgn(sin(d*x + c)))/(a^ 24 + 3*a^23*b + 3*a^22*b^2 + a^21*b^3))*tan(1/2*d*x + 1/2*c)^2 - (33*a^20* b^3*sgn(sin(d*x + c)) + 40*a^19*b^4*sgn(sin(d*x + c)) + 15*a^18*b^5*sgn(si n(d*x + c)))/(a^24 + 3*a^23*b + 3*a^22*b^2 + a^21*b^3))/(b*tan(1/2*d*x + 1 /2*c)^4 + 4*a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c)^2 + b)^(5/ 2) - 30*arctan(-1/2*(sqrt(b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(b*tan(1/2*d*x + 1/2*c)^4 + 4*a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c)^2 + b) + sqrt(b))/sqrt(a))/(a^(7/2)*sgn(sin(d*x + c))))/d
Timed out. \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\sin \left (c+d\,x\right )}^2}\right )}^{7/2}} \,d x \]