Integrand size = 20, antiderivative size = 695 \[ \int x^3 \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \, dx=-\frac {2 i b^2 x^{7/2}}{d}+\frac {a^2 x^4}{4}-\frac {8 a b x^{7/2} \text {arctanh}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^{7/2} \cot \left (c+d \sqrt {x}\right )}{d}+\frac {14 b^2 x^3 \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {28 i a b x^3 \operatorname {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {28 i a b x^3 \operatorname {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {42 i b^2 x^{5/2} \operatorname {PolyLog}\left (2,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {168 a b x^{5/2} \operatorname {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {168 a b x^{5/2} \operatorname {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {105 b^2 x^2 \operatorname {PolyLog}\left (3,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {840 i a b x^2 \operatorname {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {840 i a b x^2 \operatorname {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {210 i b^2 x^{3/2} \operatorname {PolyLog}\left (4,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {3360 a b x^{3/2} \operatorname {PolyLog}\left (5,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {3360 a b x^{3/2} \operatorname {PolyLog}\left (5,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {315 b^2 x \operatorname {PolyLog}\left (5,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {10080 i a b x \operatorname {PolyLog}\left (6,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 i a b x \operatorname {PolyLog}\left (6,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {315 i b^2 \sqrt {x} \operatorname {PolyLog}\left (6,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {20160 a b \sqrt {x} \operatorname {PolyLog}\left (7,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {20160 a b \sqrt {x} \operatorname {PolyLog}\left (7,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {315 b^2 \operatorname {PolyLog}\left (7,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^8}-\frac {20160 i a b \operatorname {PolyLog}\left (8,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {20160 i a b \operatorname {PolyLog}\left (8,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8} \]
-28*I*a*b*x^3*polylog(2,exp(I*(c+d*x^(1/2))))/d^2-840*I*a*b*x^2*polylog(4, -exp(I*(c+d*x^(1/2))))/d^4-10080*I*a*b*x*polylog(6,exp(I*(c+d*x^(1/2))))/d ^6-2*b^2*x^(7/2)*cot(c+d*x^(1/2))/d+14*b^2*x^3*ln(1-exp(2*I*(c+d*x^(1/2))) )/d^2+105*b^2*x^2*polylog(3,exp(2*I*(c+d*x^(1/2))))/d^4-315*b^2*x*polylog( 5,exp(2*I*(c+d*x^(1/2))))/d^6-2*I*b^2*x^(7/2)/d-20160*a*b*polylog(7,-exp(I *(c+d*x^(1/2))))*x^(1/2)/d^7+20160*a*b*polylog(7,exp(I*(c+d*x^(1/2))))*x^( 1/2)/d^7-42*I*b^2*x^(5/2)*polylog(2,exp(2*I*(c+d*x^(1/2))))/d^3-20160*I*a* b*polylog(8,-exp(I*(c+d*x^(1/2))))/d^8-315*I*b^2*polylog(6,exp(2*I*(c+d*x^ (1/2))))*x^(1/2)/d^7-8*a*b*x^(7/2)*arctanh(exp(I*(c+d*x^(1/2))))/d-168*a*b *x^(5/2)*polylog(3,-exp(I*(c+d*x^(1/2))))/d^3+168*a*b*x^(5/2)*polylog(3,ex p(I*(c+d*x^(1/2))))/d^3+3360*a*b*x^(3/2)*polylog(5,-exp(I*(c+d*x^(1/2))))/ d^5-3360*a*b*x^(3/2)*polylog(5,exp(I*(c+d*x^(1/2))))/d^5+1/4*a^2*x^4+28*I* a*b*x^3*polylog(2,-exp(I*(c+d*x^(1/2))))/d^2+840*I*a*b*x^2*polylog(4,exp(I *(c+d*x^(1/2))))/d^4+10080*I*a*b*x*polylog(6,-exp(I*(c+d*x^(1/2))))/d^6+21 0*I*b^2*x^(3/2)*polylog(4,exp(2*I*(c+d*x^(1/2))))/d^5+20160*I*a*b*polylog( 8,exp(I*(c+d*x^(1/2))))/d^8+315/2*b^2*polylog(7,exp(2*I*(c+d*x^(1/2))))/d^ 8
Time = 21.47 (sec) , antiderivative size = 1323, normalized size of antiderivative = 1.90 \[ \int x^3 \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \, dx =\text {Too large to display} \]
(a^2*x^4*(a + b*Csc[c + d*Sqrt[x]])^2*Sin[c + d*Sqrt[x]]^2)/(4*(b + a*Sin[ c + d*Sqrt[x]])^2) - (b^2*E^(I*c)*Csc[c]*(a + b*Csc[c + d*Sqrt[x]])^2*((2* d^7*x^(7/2))/E^((2*I)*c) + (7*I)*d^6*(1 - E^((-2*I)*c))*x^3*Log[1 - E^((-I )*(c + d*Sqrt[x]))] + (7*I)*d^6*(1 - E^((-2*I)*c))*x^3*Log[1 + E^((-I)*(c + d*Sqrt[x]))] - 42*d^5*(1 - E^((-2*I)*c))*x^(5/2)*PolyLog[2, -E^((-I)*(c + d*Sqrt[x]))] - 42*d^5*(1 - E^((-2*I)*c))*x^(5/2)*PolyLog[2, E^((-I)*(c + d*Sqrt[x]))] + (210*I)*d^4*(1 - E^((-2*I)*c))*x^2*PolyLog[3, -E^((-I)*(c + d*Sqrt[x]))] + (210*I)*d^4*(1 - E^((-2*I)*c))*x^2*PolyLog[3, E^((-I)*(c + d*Sqrt[x]))] + 840*d^3*(1 - E^((-2*I)*c))*x^(3/2)*PolyLog[4, -E^((-I)*(c + d*Sqrt[x]))] + 840*d^3*(1 - E^((-2*I)*c))*x^(3/2)*PolyLog[4, E^((-I)*(c + d*Sqrt[x]))] - (2520*I)*d^2*(1 - E^((-2*I)*c))*x*PolyLog[5, -E^((-I)*(c + d*Sqrt[x]))] - (2520*I)*d^2*(1 - E^((-2*I)*c))*x*PolyLog[5, E^((-I)*(c + d*Sqrt[x]))] - 5040*d*(1 - E^((-2*I)*c))*Sqrt[x]*PolyLog[6, -E^((-I)*(c + d*Sqrt[x]))] - 5040*d*(1 - E^((-2*I)*c))*Sqrt[x]*PolyLog[6, E^((-I)*(c + d*Sqrt[x]))] + (5040*I)*(1 - E^((-2*I)*c))*PolyLog[7, -E^((-I)*(c + d*Sqr t[x]))] + (5040*I)*(1 - E^((-2*I)*c))*PolyLog[7, E^((-I)*(c + d*Sqrt[x]))] )*Sin[c + d*Sqrt[x]]^2)/(d^8*(b + a*Sin[c + d*Sqrt[x]])^2) + (4*a*b*(a + b *Csc[c + d*Sqrt[x]])^2*(d^7*x^(7/2)*Log[1 - E^(I*(c + d*Sqrt[x]))] - d^7*x ^(7/2)*Log[1 + E^(I*(c + d*Sqrt[x]))] + (7*I)*d^6*x^3*PolyLog[2, -E^(I*(c + d*Sqrt[x]))] - (7*I)*d^6*x^3*PolyLog[2, E^(I*(c + d*Sqrt[x]))] - 42*d...
Time = 1.01 (sec) , antiderivative size = 703, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4693, 3042, 4678, 2009}
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.
| \(\displaystyle \int x^3 \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \, dx\) |
| \(\Big \downarrow \) 4693 |
| \(\displaystyle 2 \int x^{7/2} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2d\sqrt {x}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \int x^{7/2} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2d\sqrt {x}\) |
| \(\Big \downarrow \) 4678 |
| \(\displaystyle 2 \int \left (a^2 x^{7/2}+b^2 \csc ^2\left (c+d \sqrt {x}\right ) x^{7/2}+2 a b \csc \left (c+d \sqrt {x}\right ) x^{7/2}\right )d\sqrt {x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (\frac {a^2 x^4}{8}-\frac {4 a b x^{7/2} \text {arctanh}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {10080 i a b \operatorname {PolyLog}\left (8,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {10080 i a b \operatorname {PolyLog}\left (8,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}-\frac {10080 a b \sqrt {x} \operatorname {PolyLog}\left (7,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 a b \sqrt {x} \operatorname {PolyLog}\left (7,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {5040 i a b x \operatorname {PolyLog}\left (6,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i a b x \operatorname {PolyLog}\left (6,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {1680 a b x^{3/2} \operatorname {PolyLog}\left (5,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 a b x^{3/2} \operatorname {PolyLog}\left (5,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {420 i a b x^2 \operatorname {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i a b x^2 \operatorname {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {84 a b x^{5/2} \operatorname {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 a b x^{5/2} \operatorname {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {14 i a b x^3 \operatorname {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i a b x^3 \operatorname {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {315 b^2 \operatorname {PolyLog}\left (7,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8}-\frac {315 i b^2 \sqrt {x} \operatorname {PolyLog}\left (6,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {315 b^2 x \operatorname {PolyLog}\left (5,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}+\frac {105 i b^2 x^{3/2} \operatorname {PolyLog}\left (4,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {105 b^2 x^2 \operatorname {PolyLog}\left (3,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}-\frac {21 i b^2 x^{5/2} \operatorname {PolyLog}\left (2,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {7 b^2 x^3 \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {b^2 x^{7/2} \cot \left (c+d \sqrt {x}\right )}{d}-\frac {i b^2 x^{7/2}}{d}\right )\) |
2*(((-I)*b^2*x^(7/2))/d + (a^2*x^4)/8 - (4*a*b*x^(7/2)*ArcTanh[E^(I*(c + d *Sqrt[x]))])/d - (b^2*x^(7/2)*Cot[c + d*Sqrt[x]])/d + (7*b^2*x^3*Log[1 - E ^((2*I)*(c + d*Sqrt[x]))])/d^2 + ((14*I)*a*b*x^3*PolyLog[2, -E^(I*(c + d*S qrt[x]))])/d^2 - ((14*I)*a*b*x^3*PolyLog[2, E^(I*(c + d*Sqrt[x]))])/d^2 - ((21*I)*b^2*x^(5/2)*PolyLog[2, E^((2*I)*(c + d*Sqrt[x]))])/d^3 - (84*a*b*x ^(5/2)*PolyLog[3, -E^(I*(c + d*Sqrt[x]))])/d^3 + (84*a*b*x^(5/2)*PolyLog[3 , E^(I*(c + d*Sqrt[x]))])/d^3 + (105*b^2*x^2*PolyLog[3, E^((2*I)*(c + d*Sq rt[x]))])/(2*d^4) - ((420*I)*a*b*x^2*PolyLog[4, -E^(I*(c + d*Sqrt[x]))])/d ^4 + ((420*I)*a*b*x^2*PolyLog[4, E^(I*(c + d*Sqrt[x]))])/d^4 + ((105*I)*b^ 2*x^(3/2)*PolyLog[4, E^((2*I)*(c + d*Sqrt[x]))])/d^5 + (1680*a*b*x^(3/2)*P olyLog[5, -E^(I*(c + d*Sqrt[x]))])/d^5 - (1680*a*b*x^(3/2)*PolyLog[5, E^(I *(c + d*Sqrt[x]))])/d^5 - (315*b^2*x*PolyLog[5, E^((2*I)*(c + d*Sqrt[x]))] )/(2*d^6) + ((5040*I)*a*b*x*PolyLog[6, -E^(I*(c + d*Sqrt[x]))])/d^6 - ((50 40*I)*a*b*x*PolyLog[6, E^(I*(c + d*Sqrt[x]))])/d^6 - (((315*I)/2)*b^2*Sqrt [x]*PolyLog[6, E^((2*I)*(c + d*Sqrt[x]))])/d^7 - (10080*a*b*Sqrt[x]*PolyLo g[7, -E^(I*(c + d*Sqrt[x]))])/d^7 + (10080*a*b*Sqrt[x]*PolyLog[7, E^(I*(c + d*Sqrt[x]))])/d^7 + (315*b^2*PolyLog[7, E^((2*I)*(c + d*Sqrt[x]))])/(4*d ^8) - ((10080*I)*a*b*PolyLog[8, -E^(I*(c + d*Sqrt[x]))])/d^8 + ((10080*I)* a*b*PolyLog[8, E^(I*(c + d*Sqrt[x]))])/d^8)
3.1.36.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Csc[c + d*x])^ p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 1)/n], 0] && IntegerQ[p]
\[\int x^{3} \left (a +b \csc \left (c +d \sqrt {x}\right )\right )^{2}d x\]
\[ \int x^3 \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \csc \left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{3} \,d x } \]
\[ \int x^3 \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x^{3} \left (a + b \csc {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 6462 vs. \(2 (550) = 1100\).
Time = 0.67 (sec) , antiderivative size = 6462, normalized size of antiderivative = 9.30 \[ \int x^3 \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \, dx=\text {Too large to display} \]
1/4*((d*sqrt(x) + c)^8*a^2 - 8*(d*sqrt(x) + c)^7*a^2*c + 28*(d*sqrt(x) + c )^6*a^2*c^2 - 56*(d*sqrt(x) + c)^5*a^2*c^3 + 70*(d*sqrt(x) + c)^4*a^2*c^4 - 56*(d*sqrt(x) + c)^3*a^2*c^5 + 28*(d*sqrt(x) + c)^2*a^2*c^6 - 8*(d*sqrt( x) + c)*a^2*c^7 + 16*a*b*c^7*log(cot(d*sqrt(x) + c) + csc(d*sqrt(x) + c)) + 8*(4*b^2*c^7 + 2*(2*(d*sqrt(x) + c)^7*a*b - 7*b^2*c^6 - 7*(2*a*b*c + b^2 )*(d*sqrt(x) + c)^6 + 42*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^5 - 35*(2*a*b*c ^3 + 3*b^2*c^2)*(d*sqrt(x) + c)^4 + 70*(a*b*c^4 + 2*b^2*c^3)*(d*sqrt(x) + c)^3 - 21*(2*a*b*c^5 + 5*b^2*c^4)*(d*sqrt(x) + c)^2 + 14*(a*b*c^6 + 3*b^2* c^5)*(d*sqrt(x) + c) - (2*(d*sqrt(x) + c)^7*a*b - 7*b^2*c^6 - 7*(2*a*b*c + b^2)*(d*sqrt(x) + c)^6 + 42*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^5 - 35*(2*a *b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c)^4 + 70*(a*b*c^4 + 2*b^2*c^3)*(d*sqrt(x ) + c)^3 - 21*(2*a*b*c^5 + 5*b^2*c^4)*(d*sqrt(x) + c)^2 + 14*(a*b*c^6 + 3* b^2*c^5)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + (-2*I*(d*sqrt(x) + c)^7 *a*b + 7*I*b^2*c^6 + 7*(2*I*a*b*c + I*b^2)*(d*sqrt(x) + c)^6 + 42*(-I*a*b* c^2 - I*b^2*c)*(d*sqrt(x) + c)^5 + 35*(2*I*a*b*c^3 + 3*I*b^2*c^2)*(d*sqrt( x) + c)^4 + 70*(-I*a*b*c^4 - 2*I*b^2*c^3)*(d*sqrt(x) + c)^3 + 21*(2*I*a*b* c^5 + 5*I*b^2*c^4)*(d*sqrt(x) + c)^2 + 14*(-I*a*b*c^6 - 3*I*b^2*c^5)*(d*sq rt(x) + c))*sin(2*d*sqrt(x) + 2*c))*arctan2(sin(d*sqrt(x) + c), cos(d*sqrt (x) + c) + 1) + 14*(b^2*c^6*cos(2*d*sqrt(x) + 2*c) + I*b^2*c^6*sin(2*d*sqr t(x) + 2*c) - b^2*c^6)*arctan2(sin(d*sqrt(x) + c), cos(d*sqrt(x) + c) -...
\[ \int x^3 \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \csc \left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{3} \,d x } \]
Timed out. \[ \int x^3 \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x^3\,{\left (a+\frac {b}{\sin \left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]