Integrand size = 22, antiderivative size = 255 \[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=-\frac {2 i b^2 x}{d}+\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {8 a b \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 a b \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d} \]
-2*I*b^2*x/d+2/3*a^2*x^(3/2)-8*I*a*b*x*arctan(exp(I*(c+d*x^(1/2))))/d-2*I* b^2*polylog(2,-exp(2*I*(c+d*x^(1/2))))/d^3-8*a*b*polylog(3,-I*exp(I*(c+d*x ^(1/2))))/d^3+8*a*b*polylog(3,I*exp(I*(c+d*x^(1/2))))/d^3+4*b^2*ln(1+exp(2 *I*(c+d*x^(1/2))))*x^(1/2)/d^2+8*I*a*b*polylog(2,-I*exp(I*(c+d*x^(1/2))))* x^(1/2)/d^2-8*I*a*b*polylog(2,I*exp(I*(c+d*x^(1/2))))*x^(1/2)/d^2+2*b^2*x* tan(c+d*x^(1/2))/d
Time = 0.77 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.97 \[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {2 \left (-3 i b^2 d^2 x+a^2 d^3 x^{3/2}-12 i a b d^2 x \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )+6 b^2 d \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )+12 i a b d \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )-12 i a b d \sqrt {x} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )-3 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-12 a b \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )+12 a b \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )+3 b^2 d^2 x \tan \left (c+d \sqrt {x}\right )\right )}{3 d^3} \]
(2*((-3*I)*b^2*d^2*x + a^2*d^3*x^(3/2) - (12*I)*a*b*d^2*x*ArcTan[E^(I*(c + d*Sqrt[x]))] + 6*b^2*d*Sqrt[x]*Log[1 + E^((2*I)*(c + d*Sqrt[x]))] + (12*I )*a*b*d*Sqrt[x]*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))] - (12*I)*a*b*d*Sqrt [x]*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))] - (3*I)*b^2*PolyLog[2, -E^((2*I)*( c + d*Sqrt[x]))] - 12*a*b*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))] + 12*a*b* PolyLog[3, I*E^(I*(c + d*Sqrt[x]))] + 3*b^2*d^2*x*Tan[c + d*Sqrt[x]]))/(3* d^3)
Time = 0.55 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4692, 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.
\(\displaystyle \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 4692 |
\(\displaystyle 2 \int x \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2d\sqrt {x}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \int x \left (a+b \csc \left (c+d \sqrt {x}+\frac {\pi }{2}\right )\right )^2d\sqrt {x}\) |
\(\Big \downarrow \) 4678 |
\(\displaystyle 2 \int \left (x a^2+2 b x \sec \left (c+d \sqrt {x}\right ) a+b^2 x \sec ^2\left (c+d \sqrt {x}\right )\right )d\sqrt {x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{3} a^2 x^{3/2}-\frac {4 i a b x \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {4 a b \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 a b \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 i a b \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i a b \sqrt {x} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {b^2 x \tan \left (c+d \sqrt {x}\right )}{d}-\frac {i b^2 x}{d}\right )\) |
2*(((-I)*b^2*x)/d + (a^2*x^(3/2))/3 - ((4*I)*a*b*x*ArcTan[E^(I*(c + d*Sqrt [x]))])/d + (2*b^2*Sqrt[x]*Log[1 + E^((2*I)*(c + d*Sqrt[x]))])/d^2 + ((4*I )*a*b*Sqrt[x]*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((4*I)*a*b*Sqr t[x]*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - (I*b^2*PolyLog[2, -E^((2*I )*(c + d*Sqrt[x]))])/d^3 - (4*a*b*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))])/ d^3 + (4*a*b*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 + (b^2*x*Tan[c + d*S qrt[x]])/d)
3.1.57.3.1 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[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sec[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]
\[\int \left (a +b \sec \left (c +d \sqrt {x}\right )\right )^{2} \sqrt {x}d x\]
\[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2} \sqrt {x} \,d x } \]
integral(b^2*sqrt(x)*sec(d*sqrt(x) + c)^2 + 2*a*b*sqrt(x)*sec(d*sqrt(x) + c) + a^2*sqrt(x), x)
\[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int \sqrt {x} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1272 vs. \(2 (194) = 388\).
Time = 0.45 (sec) , antiderivative size = 1272, normalized size of antiderivative = 4.99 \[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\text {Too large to display} \]
2/3*((d*sqrt(x) + c)^3*a^2 - 3*(d*sqrt(x) + c)^2*a^2*c + 3*(d*sqrt(x) + c) *a^2*c^2 + 6*a*b*c^2*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c)) + 3*(2*b ^2*c^2 - 2*((d*sqrt(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c + ((d*sqrt(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c)*cos(2*d*sqrt(x) + 2*c) + (I*(d*sqrt( x) + c)^2*a*b - 2*I*(d*sqrt(x) + c)*a*b*c)*sin(2*d*sqrt(x) + 2*c))*arctan2 (cos(d*sqrt(x) + c), sin(d*sqrt(x) + c) + 1) - 2*((d*sqrt(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c + ((d*sqrt(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c )*cos(2*d*sqrt(x) + 2*c) + (I*(d*sqrt(x) + c)^2*a*b - 2*I*(d*sqrt(x) + c)* a*b*c)*sin(2*d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) + c), -sin(d*sqrt(x) + c) + 1) + 2*((d*sqrt(x) + c)*b^2 - b^2*c + ((d*sqrt(x) + c)*b^2 - b^2*c) *cos(2*d*sqrt(x) + 2*c) - (-I*(d*sqrt(x) + c)*b^2 + I*b^2*c)*sin(2*d*sqrt( x) + 2*c))*arctan2(sin(2*d*sqrt(x) + 2*c), cos(2*d*sqrt(x) + 2*c) + 1) - 2 *((d*sqrt(x) + c)^2*b^2 - 2*(d*sqrt(x) + c)*b^2*c)*cos(2*d*sqrt(x) + 2*c) - (b^2*cos(2*d*sqrt(x) + 2*c) + I*b^2*sin(2*d*sqrt(x) + 2*c) + b^2)*dilog( -e^(2*I*d*sqrt(x) + 2*I*c)) - 4*((d*sqrt(x) + c)*a*b - a*b*c + ((d*sqrt(x) + c)*a*b - a*b*c)*cos(2*d*sqrt(x) + 2*c) + (I*(d*sqrt(x) + c)*a*b - I*a*b *c)*sin(2*d*sqrt(x) + 2*c))*dilog(I*e^(I*d*sqrt(x) + I*c)) + 4*((d*sqrt(x) + c)*a*b - a*b*c + ((d*sqrt(x) + c)*a*b - a*b*c)*cos(2*d*sqrt(x) + 2*c) - (-I*(d*sqrt(x) + c)*a*b + I*a*b*c)*sin(2*d*sqrt(x) + 2*c))*dilog(-I*e^(I* d*sqrt(x) + I*c)) + (-I*(d*sqrt(x) + c)*b^2 + I*b^2*c + (-I*(d*sqrt(x) ...
\[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2} \sqrt {x} \,d x } \]
Timed out. \[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int \sqrt {x}\,{\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]