Integrand size = 22, antiderivative size = 409 \[ \int \frac {x^3 \sqrt {c+d x}}{a+b x^2} \, dx=-\frac {2 a \sqrt {c+d x}}{b^2}-\frac {2 c (c+d x)^{3/2}}{3 b d^2}+\frac {2 (c+d x)^{5/2}}{5 b d^2}-\frac {a \sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}} \arctan \left (\frac {\sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}}-\sqrt {2} \sqrt [4]{b} \sqrt {c+d x}}{\sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}}}\right )}{\sqrt {2} b^{9/4}}+\frac {a \sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}} \arctan \left (\frac {\sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}}+\sqrt {2} \sqrt [4]{b} \sqrt {c+d x}}{\sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}}}\right )}{\sqrt {2} b^{9/4}}+\frac {a \sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}} \sqrt {c+d x}}{\sqrt {b c^2+a d^2}+\sqrt {b} (c+d x)}\right )}{\sqrt {2} b^{9/4}} \] Output:
-2*a*(d*x+c)^(1/2)/b^2-2/3*c*(d*x+c)^(3/2)/b/d^2+2/5*(d*x+c)^(5/2)/b/d^2-1 /2*a*(-b^(1/2)*c+(a*d^2+b*c^2)^(1/2))^(1/2)*arctan(((b^(1/2)*c+(a*d^2+b*c^ 2)^(1/2))^(1/2)-2^(1/2)*b^(1/4)*(d*x+c)^(1/2))/(-b^(1/2)*c+(a*d^2+b*c^2)^( 1/2))^(1/2))*2^(1/2)/b^(9/4)+1/2*a*(-b^(1/2)*c+(a*d^2+b*c^2)^(1/2))^(1/2)* arctan(((b^(1/2)*c+(a*d^2+b*c^2)^(1/2))^(1/2)+2^(1/2)*b^(1/4)*(d*x+c)^(1/2 ))/(-b^(1/2)*c+(a*d^2+b*c^2)^(1/2))^(1/2))*2^(1/2)/b^(9/4)+1/2*a*(b^(1/2)* c+(a*d^2+b*c^2)^(1/2))^(1/2)*arctanh(2^(1/2)*b^(1/4)*(b^(1/2)*c+(a*d^2+b*c ^2)^(1/2))^(1/2)*(d*x+c)^(1/2)/((a*d^2+b*c^2)^(1/2)+b^(1/2)*(d*x+c)))*2^(1 /2)/b^(9/4)
Result contains complex when optimal does not.
Time = 0.63 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.54 \[ \int \frac {x^3 \sqrt {c+d x}}{a+b x^2} \, dx=\frac {2 \sqrt {c+d x} \left (-2 b c^2-15 a d^2+b c d x+3 b d^2 x^2\right )}{15 b^2 d^2}-\frac {a \sqrt {-b c-i \sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c-i \sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+i \sqrt {a} d}\right )}{b^{5/2}}-\frac {a \sqrt {-b c+i \sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c+i \sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-i \sqrt {a} d}\right )}{b^{5/2}} \] Input:
Integrate[(x^3*Sqrt[c + d*x])/(a + b*x^2),x]
Output:
(2*Sqrt[c + d*x]*(-2*b*c^2 - 15*a*d^2 + b*c*d*x + 3*b*d^2*x^2))/(15*b^2*d^ 2) - (a*Sqrt[-(b*c) - I*Sqrt[a]*Sqrt[b]*d]*ArcTan[(Sqrt[-(b*c) - I*Sqrt[a] *Sqrt[b]*d]*Sqrt[c + d*x])/(Sqrt[b]*c + I*Sqrt[a]*d)])/b^(5/2) - (a*Sqrt[- (b*c) + I*Sqrt[a]*Sqrt[b]*d]*ArcTan[(Sqrt[-(b*c) + I*Sqrt[a]*Sqrt[b]*d]*Sq rt[c + d*x])/(Sqrt[b]*c - I*Sqrt[a]*d)])/b^(5/2)
Time = 2.46 (sec) , antiderivative size = 654, normalized size of antiderivative = 1.60, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {561, 25, 27, 1610, 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.
| \(\displaystyle \int \frac {x^3 \sqrt {c+d x}}{a+b x^2} \, dx\) |
| \(\Big \downarrow \) 561 |
| \(\displaystyle \frac {2 \int \frac {x^3 (c+d x)}{\frac {b c^2}{d^2}-\frac {2 b (c+d x) c}{d^2}+\frac {b (c+d x)^2}{d^2}+a}d\sqrt {c+d x}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 \int -\frac {x^3 (c+d x)}{\frac {b c^2}{d^2}-\frac {2 b (c+d x) c}{d^2}+\frac {b (c+d x)^2}{d^2}+a}d\sqrt {c+d x}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 \int -\frac {d^3 x^3 (c+d x)}{\frac {b c^2}{d^2}-\frac {2 b (c+d x) c}{d^2}+\frac {b (c+d x)^2}{d^2}+a}d\sqrt {c+d x}}{d^4}\) |
| \(\Big \downarrow \) 1610 |
| \(\displaystyle -\frac {2 \int \left (\frac {a d^4}{b^2}-\frac {(c+d x)^2 d^2}{b}+\frac {c (c+d x) d^2}{b}-\frac {a d^2 \left (b c^2+a d^2\right )-a b c d^2 (c+d x)}{b^2 \left (\frac {b c^2}{d^2}-\frac {2 b (c+d x) c}{d^2}+\frac {b (c+d x)^2}{d^2}+a\right )}\right )d\sqrt {c+d x}}{d^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \left (-\frac {a d^4 \left (-\sqrt {b} c \sqrt {a d^2+b c^2}+a d^2+b c^2\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c}-\sqrt {2} \sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a d^2+b c^2}}}\right )}{2 \sqrt {2} b^{9/4} \sqrt {a d^2+b c^2} \sqrt {\sqrt {b} c-\sqrt {a d^2+b c^2}}}+\frac {a d^4 \left (-\sqrt {b} c \sqrt {a d^2+b c^2}+a d^2+b c^2\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c}+\sqrt {2} \sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a d^2+b c^2}}}\right )}{2 \sqrt {2} b^{9/4} \sqrt {a d^2+b c^2} \sqrt {\sqrt {b} c-\sqrt {a d^2+b c^2}}}+\frac {a d^4 \sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt {c+d x} \sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c}+\sqrt {a d^2+b c^2}+\sqrt {b} (c+d x)\right )}{4 \sqrt {2} b^{9/4}}-\frac {a d^4 \sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt {c+d x} \sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c}+\sqrt {a d^2+b c^2}+\sqrt {b} (c+d x)\right )}{4 \sqrt {2} b^{9/4}}+\frac {a d^4 \sqrt {c+d x}}{b^2}-\frac {d^2 (c+d x)^{5/2}}{5 b}+\frac {c d^2 (c+d x)^{3/2}}{3 b}\right )}{d^4}\) |
Input:
Int[(x^3*Sqrt[c + d*x])/(a + b*x^2),x]
Output:
(-2*((a*d^4*Sqrt[c + d*x])/b^2 + (c*d^2*(c + d*x)^(3/2))/(3*b) - (d^2*(c + d*x)^(5/2))/(5*b) - (a*d^4*(b*c^2 + a*d^2 - Sqrt[b]*c*Sqrt[b*c^2 + a*d^2] )*ArcTanh[(Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]] - Sqrt[2]*b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c - Sqrt[b*c^2 + a*d^2]]])/(2*Sqrt[2]*b^(9/4)*Sqrt[b* c^2 + a*d^2]*Sqrt[Sqrt[b]*c - Sqrt[b*c^2 + a*d^2]]) + (a*d^4*(b*c^2 + a*d^ 2 - Sqrt[b]*c*Sqrt[b*c^2 + a*d^2])*ArcTanh[(Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]] + Sqrt[2]*b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c - Sqrt[b*c^2 + a*d ^2]]])/(2*Sqrt[2]*b^(9/4)*Sqrt[b*c^2 + a*d^2]*Sqrt[Sqrt[b]*c - Sqrt[b*c^2 + a*d^2]]) + (a*d^4*Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]]*Log[Sqrt[b*c^2 + a*d^2] - Sqrt[2]*b^(1/4)*Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]]*Sqrt[c + d *x] + Sqrt[b]*(c + d*x)])/(4*Sqrt[2]*b^(9/4)) - (a*d^4*Sqrt[Sqrt[b]*c + Sq rt[b*c^2 + a*d^2]]*Log[Sqrt[b*c^2 + a*d^2] + Sqrt[2]*b^(1/4)*Sqrt[Sqrt[b]* c + Sqrt[b*c^2 + a*d^2]]*Sqrt[c + d*x] + Sqrt[b]*(c + d*x)])/(4*Sqrt[2]*b^ (9/4))))/d^4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{k = Denominator[n]}, Simp[k/d Subst[Int[x^(k*(n + 1) - 1)*(-c /d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac tionQ[n] && IntegerQ[p] && IntegerQ[m]
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*((d + e*x^2)^q/(a + b*x^2 + c*x^4)), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4 *a*c, 0] && IntegerQ[q] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(699\) vs. \(2(319)=638\).
Time = 2.91 (sec) , antiderivative size = 700, normalized size of antiderivative = 1.71
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {\left (\left (\sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}+b c \right ) \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-b^{2} c^{2}-b^{\frac {3}{2}} \sqrt {a \,d^{2}+b \,c^{2}}\, c \right ) \sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}\, \sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}\, \ln \left (\sqrt {b}\, \left (d x +c \right )-\sqrt {d x +c}\, \sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}+\sqrt {a \,d^{2}+b \,c^{2}}\right )}{4}+\frac {\left (\left (\sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}+b c \right ) \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-b^{2} c^{2}-b^{\frac {3}{2}} \sqrt {a \,d^{2}+b \,c^{2}}\, c \right ) \sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}\, \sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}\, \ln \left (\sqrt {b}\, \left (d x +c \right )+\sqrt {d x +c}\, \sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}+\sqrt {a \,d^{2}+b \,c^{2}}\right )}{4}-2 \left (b^{\frac {3}{2}} a \,d^{2}+\frac {2 \left (d x +c \right ) \left (-\frac {3 d x}{2}+c \right ) b^{\frac {5}{2}}}{15}\right ) \sqrt {d x +c}\, \sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}+\left (-b^{2} c +\sqrt {a \,d^{2}+b \,c^{2}}\, b^{\frac {3}{2}}\right ) d^{2} \left (\arctan \left (\frac {2 \sqrt {b}\, \sqrt {d x +c}+\sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}}{\sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}}\right )-\arctan \left (\frac {-2 \sqrt {b}\, \sqrt {d x +c}+\sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}}{\sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}}\right )\right ) a}{b^{\frac {7}{2}} \sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}\, d^{2}}\) | \(700\) |
| risch | \(\text {Expression too large to display}\) | \(1271\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1274\) |
| default | \(\text {Expression too large to display}\) | \(1274\) |
Input:
int(x^3*(d*x+c)^(1/2)/(b*x^2+a),x,method=_RETURNVERBOSE)
Output:
1/b^(7/2)/(4*(a*d^2+b*c^2)^(1/2)*b^(1/2)-2*((a*d^2+b*c^2)*b)^(1/2)-2*b*c)^ (1/2)*(-1/4*(((a*d^2+b*c^2)^(1/2)*b^(1/2)+b*c)*((a*d^2+b*c^2)*b)^(1/2)-b^2 *c^2-b^(3/2)*(a*d^2+b*c^2)^(1/2)*c)*(2*((a*d^2+b*c^2)*b)^(1/2)+2*b*c)^(1/2 )*(4*(a*d^2+b*c^2)^(1/2)*b^(1/2)-2*((a*d^2+b*c^2)*b)^(1/2)-2*b*c)^(1/2)*ln (b^(1/2)*(d*x+c)-(d*x+c)^(1/2)*(2*((a*d^2+b*c^2)*b)^(1/2)+2*b*c)^(1/2)+(a* d^2+b*c^2)^(1/2))+1/4*(((a*d^2+b*c^2)^(1/2)*b^(1/2)+b*c)*((a*d^2+b*c^2)*b) ^(1/2)-b^2*c^2-b^(3/2)*(a*d^2+b*c^2)^(1/2)*c)*(2*((a*d^2+b*c^2)*b)^(1/2)+2 *b*c)^(1/2)*(4*(a*d^2+b*c^2)^(1/2)*b^(1/2)-2*((a*d^2+b*c^2)*b)^(1/2)-2*b*c )^(1/2)*ln(b^(1/2)*(d*x+c)+(d*x+c)^(1/2)*(2*((a*d^2+b*c^2)*b)^(1/2)+2*b*c) ^(1/2)+(a*d^2+b*c^2)^(1/2))-2*(b^(3/2)*a*d^2+2/15*(d*x+c)*(-3/2*d*x+c)*b^( 5/2))*(d*x+c)^(1/2)*(4*(a*d^2+b*c^2)^(1/2)*b^(1/2)-2*((a*d^2+b*c^2)*b)^(1/ 2)-2*b*c)^(1/2)+(-b^2*c+(a*d^2+b*c^2)^(1/2)*b^(3/2))*d^2*(arctan((2*b^(1/2 )*(d*x+c)^(1/2)+(2*((a*d^2+b*c^2)*b)^(1/2)+2*b*c)^(1/2))/(4*(a*d^2+b*c^2)^ (1/2)*b^(1/2)-2*((a*d^2+b*c^2)*b)^(1/2)-2*b*c)^(1/2))-arctan((-2*b^(1/2)*( d*x+c)^(1/2)+(2*((a*d^2+b*c^2)*b)^(1/2)+2*b*c)^(1/2))/(4*(a*d^2+b*c^2)^(1/ 2)*b^(1/2)-2*((a*d^2+b*c^2)*b)^(1/2)-2*b*c)^(1/2)))*a)/d^2
Time = 0.14 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.93 \[ \int \frac {x^3 \sqrt {c+d x}}{a+b x^2} \, dx=\frac {15 \, b^{2} d^{2} \sqrt {\frac {b^{4} \sqrt {-\frac {a^{5} d^{2}}{b^{9}}} + a^{2} c}{b^{4}}} \log \left (b^{2} \sqrt {\frac {b^{4} \sqrt {-\frac {a^{5} d^{2}}{b^{9}}} + a^{2} c}{b^{4}}} + \sqrt {d x + c} a\right ) - 15 \, b^{2} d^{2} \sqrt {\frac {b^{4} \sqrt {-\frac {a^{5} d^{2}}{b^{9}}} + a^{2} c}{b^{4}}} \log \left (-b^{2} \sqrt {\frac {b^{4} \sqrt {-\frac {a^{5} d^{2}}{b^{9}}} + a^{2} c}{b^{4}}} + \sqrt {d x + c} a\right ) + 15 \, b^{2} d^{2} \sqrt {-\frac {b^{4} \sqrt {-\frac {a^{5} d^{2}}{b^{9}}} - a^{2} c}{b^{4}}} \log \left (b^{2} \sqrt {-\frac {b^{4} \sqrt {-\frac {a^{5} d^{2}}{b^{9}}} - a^{2} c}{b^{4}}} + \sqrt {d x + c} a\right ) - 15 \, b^{2} d^{2} \sqrt {-\frac {b^{4} \sqrt {-\frac {a^{5} d^{2}}{b^{9}}} - a^{2} c}{b^{4}}} \log \left (-b^{2} \sqrt {-\frac {b^{4} \sqrt {-\frac {a^{5} d^{2}}{b^{9}}} - a^{2} c}{b^{4}}} + \sqrt {d x + c} a\right ) + 4 \, {\left (3 \, b d^{2} x^{2} + b c d x - 2 \, b c^{2} - 15 \, a d^{2}\right )} \sqrt {d x + c}}{30 \, b^{2} d^{2}} \] Input:
integrate(x^3*(d*x+c)^(1/2)/(b*x^2+a),x, algorithm="fricas")
Output:
1/30*(15*b^2*d^2*sqrt((b^4*sqrt(-a^5*d^2/b^9) + a^2*c)/b^4)*log(b^2*sqrt(( b^4*sqrt(-a^5*d^2/b^9) + a^2*c)/b^4) + sqrt(d*x + c)*a) - 15*b^2*d^2*sqrt( (b^4*sqrt(-a^5*d^2/b^9) + a^2*c)/b^4)*log(-b^2*sqrt((b^4*sqrt(-a^5*d^2/b^9 ) + a^2*c)/b^4) + sqrt(d*x + c)*a) + 15*b^2*d^2*sqrt(-(b^4*sqrt(-a^5*d^2/b ^9) - a^2*c)/b^4)*log(b^2*sqrt(-(b^4*sqrt(-a^5*d^2/b^9) - a^2*c)/b^4) + sq rt(d*x + c)*a) - 15*b^2*d^2*sqrt(-(b^4*sqrt(-a^5*d^2/b^9) - a^2*c)/b^4)*lo g(-b^2*sqrt(-(b^4*sqrt(-a^5*d^2/b^9) - a^2*c)/b^4) + sqrt(d*x + c)*a) + 4* (3*b*d^2*x^2 + b*c*d*x - 2*b*c^2 - 15*a*d^2)*sqrt(d*x + c))/(b^2*d^2)
\[ \int \frac {x^3 \sqrt {c+d x}}{a+b x^2} \, dx=\int \frac {x^{3} \sqrt {c + d x}}{a + b x^{2}}\, dx \] Input:
integrate(x**3*(d*x+c)**(1/2)/(b*x**2+a),x)
Output:
Integral(x**3*sqrt(c + d*x)/(a + b*x**2), x)
\[ \int \frac {x^3 \sqrt {c+d x}}{a+b x^2} \, dx=\int { \frac {\sqrt {d x + c} x^{3}}{b x^{2} + a} \,d x } \] Input:
integrate(x^3*(d*x+c)^(1/2)/(b*x^2+a),x, algorithm="maxima")
Output:
integrate(sqrt(d*x + c)*x^3/(b*x^2 + a), x)
Time = 0.18 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.77 \[ \int \frac {x^3 \sqrt {c+d x}}{a+b x^2} \, dx=-\frac {{\left (a b^{2} c^{2} + a^{2} b d^{2}\right )} {\left | b \right |} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {b^{6} c d^{12} + \sqrt {b^{12} c^{2} d^{24} - {\left (b^{6} c^{2} d^{12} + a b^{5} d^{14}\right )} b^{6} d^{12}}}{b^{6} d^{12}}}}\right )}{{\left (b^{4} c - \sqrt {-a b} b^{3} d\right )} \sqrt {-b^{2} c - \sqrt {-a b} b d}} - \frac {{\left (a b^{2} c^{2} + a^{2} b d^{2}\right )} {\left | b \right |} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {b^{6} c d^{12} - \sqrt {b^{12} c^{2} d^{24} - {\left (b^{6} c^{2} d^{12} + a b^{5} d^{14}\right )} b^{6} d^{12}}}{b^{6} d^{12}}}}\right )}{{\left (b^{4} c + \sqrt {-a b} b^{3} d\right )} \sqrt {-b^{2} c + \sqrt {-a b} b d}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} d^{8} - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c d^{8} - 15 \, \sqrt {d x + c} a b^{3} d^{10}\right )}}{15 \, b^{5} d^{10}} \] Input:
integrate(x^3*(d*x+c)^(1/2)/(b*x^2+a),x, algorithm="giac")
Output:
-(a*b^2*c^2 + a^2*b*d^2)*abs(b)*arctan(sqrt(d*x + c)/sqrt(-(b^6*c*d^12 + s qrt(b^12*c^2*d^24 - (b^6*c^2*d^12 + a*b^5*d^14)*b^6*d^12))/(b^6*d^12)))/(( b^4*c - sqrt(-a*b)*b^3*d)*sqrt(-b^2*c - sqrt(-a*b)*b*d)) - (a*b^2*c^2 + a^ 2*b*d^2)*abs(b)*arctan(sqrt(d*x + c)/sqrt(-(b^6*c*d^12 - sqrt(b^12*c^2*d^2 4 - (b^6*c^2*d^12 + a*b^5*d^14)*b^6*d^12))/(b^6*d^12)))/((b^4*c + sqrt(-a* b)*b^3*d)*sqrt(-b^2*c + sqrt(-a*b)*b*d)) + 2/15*(3*(d*x + c)^(5/2)*b^4*d^8 - 5*(d*x + c)^(3/2)*b^4*c*d^8 - 15*sqrt(d*x + c)*a*b^3*d^10)/(b^5*d^10)
Time = 0.41 (sec) , antiderivative size = 530, normalized size of antiderivative = 1.30 \[ \int \frac {x^3 \sqrt {c+d x}}{a+b x^2} \, dx=\frac {2\,{\left (c+d\,x\right )}^{5/2}}{5\,b\,d^2}-\left (\frac {2\,\left (b\,c^2\,d^2+a\,d^4\right )}{b^2\,d^4}-\frac {2\,c^2}{b\,d^2}\right )\,\sqrt {c+d\,x}-\frac {2\,c\,{\left (c+d\,x\right )}^{3/2}}{3\,b\,d^2}-\mathrm {atan}\left (\frac {a^4\,d^4\,\sqrt {\frac {a^2\,c}{4\,b^4}+\frac {d\,\sqrt {-a^5\,b^9}}{4\,b^9}}\,\sqrt {c+d\,x}\,32{}\mathrm {i}}{\frac {16\,a^3\,d^5\,\sqrt {-a^5\,b^9}}{b^7}+\frac {16\,a^2\,c^2\,d^3\,\sqrt {-a^5\,b^9}}{b^6}}+\frac {a\,c\,d^3\,\sqrt {\frac {a^2\,c}{4\,b^4}+\frac {d\,\sqrt {-a^5\,b^9}}{4\,b^9}}\,\sqrt {-a^5\,b^9}\,\sqrt {c+d\,x}\,32{}\mathrm {i}}{\frac {16\,a^3\,d^5\,\sqrt {-a^5\,b^9}}{b^3}+\frac {16\,a^2\,c^2\,d^3\,\sqrt {-a^5\,b^9}}{b^2}}\right )\,\sqrt {\frac {d\,\sqrt {-a^5\,b^9}+a^2\,b^5\,c}{4\,b^9}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {a^4\,d^4\,\sqrt {\frac {a^2\,c}{4\,b^4}-\frac {d\,\sqrt {-a^5\,b^9}}{4\,b^9}}\,\sqrt {c+d\,x}\,32{}\mathrm {i}}{\frac {16\,a^3\,d^5\,\sqrt {-a^5\,b^9}}{b^7}+\frac {16\,a^2\,c^2\,d^3\,\sqrt {-a^5\,b^9}}{b^6}}-\frac {a\,c\,d^3\,\sqrt {\frac {a^2\,c}{4\,b^4}-\frac {d\,\sqrt {-a^5\,b^9}}{4\,b^9}}\,\sqrt {-a^5\,b^9}\,\sqrt {c+d\,x}\,32{}\mathrm {i}}{\frac {16\,a^3\,d^5\,\sqrt {-a^5\,b^9}}{b^3}+\frac {16\,a^2\,c^2\,d^3\,\sqrt {-a^5\,b^9}}{b^2}}\right )\,\sqrt {-\frac {d\,\sqrt {-a^5\,b^9}-a^2\,b^5\,c}{4\,b^9}}\,2{}\mathrm {i} \] Input:
int((x^3*(c + d*x)^(1/2))/(a + b*x^2),x)
Output:
atan((a^4*d^4*((a^2*c)/(4*b^4) - (d*(-a^5*b^9)^(1/2))/(4*b^9))^(1/2)*(c + d*x)^(1/2)*32i)/((16*a^3*d^5*(-a^5*b^9)^(1/2))/b^7 + (16*a^2*c^2*d^3*(-a^5 *b^9)^(1/2))/b^6) - (a*c*d^3*((a^2*c)/(4*b^4) - (d*(-a^5*b^9)^(1/2))/(4*b^ 9))^(1/2)*(-a^5*b^9)^(1/2)*(c + d*x)^(1/2)*32i)/((16*a^3*d^5*(-a^5*b^9)^(1 /2))/b^3 + (16*a^2*c^2*d^3*(-a^5*b^9)^(1/2))/b^2))*(-(d*(-a^5*b^9)^(1/2) - a^2*b^5*c)/(4*b^9))^(1/2)*2i - atan((a^4*d^4*((a^2*c)/(4*b^4) + (d*(-a^5* b^9)^(1/2))/(4*b^9))^(1/2)*(c + d*x)^(1/2)*32i)/((16*a^3*d^5*(-a^5*b^9)^(1 /2))/b^7 + (16*a^2*c^2*d^3*(-a^5*b^9)^(1/2))/b^6) + (a*c*d^3*((a^2*c)/(4*b ^4) + (d*(-a^5*b^9)^(1/2))/(4*b^9))^(1/2)*(-a^5*b^9)^(1/2)*(c + d*x)^(1/2) *32i)/((16*a^3*d^5*(-a^5*b^9)^(1/2))/b^3 + (16*a^2*c^2*d^3*(-a^5*b^9)^(1/2 ))/b^2))*((d*(-a^5*b^9)^(1/2) + a^2*b^5*c)/(4*b^9))^(1/2)*2i - ((2*(a*d^4 + b*c^2*d^2))/(b^2*d^4) - (2*c^2)/(b*d^2))*(c + d*x)^(1/2) + (2*(c + d*x)^ (5/2))/(5*b*d^2) - (2*c*(c + d*x)^(3/2))/(3*b*d^2)
Time = 0.20 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.03 \[ \int \frac {x^3 \sqrt {c+d x}}{a+b x^2} \, dx=\frac {-30 \sqrt {b}\, \sqrt {\sqrt {b}\, \sqrt {a \,d^{2}+b \,c^{2}}-b c}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {b}\, \sqrt {a \,d^{2}+b \,c^{2}}+b c}\, \sqrt {2}-2 \sqrt {b}\, \sqrt {d x +c}}{\sqrt {\sqrt {b}\, \sqrt {a \,d^{2}+b \,c^{2}}-b c}\, \sqrt {2}}\right ) a \,d^{2}+30 \sqrt {b}\, \sqrt {\sqrt {b}\, \sqrt {a \,d^{2}+b \,c^{2}}-b c}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {b}\, \sqrt {a \,d^{2}+b \,c^{2}}+b c}\, \sqrt {2}+2 \sqrt {b}\, \sqrt {d x +c}}{\sqrt {\sqrt {b}\, \sqrt {a \,d^{2}+b \,c^{2}}-b c}\, \sqrt {2}}\right ) a \,d^{2}-15 \sqrt {b}\, \sqrt {\sqrt {b}\, \sqrt {a \,d^{2}+b \,c^{2}}+b c}\, \sqrt {2}\, \mathrm {log}\left (-\sqrt {d x +c}\, \sqrt {\sqrt {b}\, \sqrt {a \,d^{2}+b \,c^{2}}+b c}\, \sqrt {2}+\sqrt {a \,d^{2}+b \,c^{2}}+\sqrt {b}\, c +\sqrt {b}\, d x \right ) a \,d^{2}+15 \sqrt {b}\, \sqrt {\sqrt {b}\, \sqrt {a \,d^{2}+b \,c^{2}}+b c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {d x +c}\, \sqrt {\sqrt {b}\, \sqrt {a \,d^{2}+b \,c^{2}}+b c}\, \sqrt {2}+\sqrt {a \,d^{2}+b \,c^{2}}+\sqrt {b}\, c +\sqrt {b}\, d x \right ) a \,d^{2}-120 \sqrt {d x +c}\, a b \,d^{2}-16 \sqrt {d x +c}\, b^{2} c^{2}+8 \sqrt {d x +c}\, b^{2} c d x +24 \sqrt {d x +c}\, b^{2} d^{2} x^{2}}{60 b^{3} d^{2}} \] Input:
int(x^3*(d*x+c)^(1/2)/(b*x^2+a),x)
Output:
( - 30*sqrt(b)*sqrt(sqrt(b)*sqrt(a*d**2 + b*c**2) - b*c)*sqrt(2)*atan((sqr t(sqrt(b)*sqrt(a*d**2 + b*c**2) + b*c)*sqrt(2) - 2*sqrt(b)*sqrt(c + d*x))/ (sqrt(sqrt(b)*sqrt(a*d**2 + b*c**2) - b*c)*sqrt(2)))*a*d**2 + 30*sqrt(b)*s qrt(sqrt(b)*sqrt(a*d**2 + b*c**2) - b*c)*sqrt(2)*atan((sqrt(sqrt(b)*sqrt(a *d**2 + b*c**2) + b*c)*sqrt(2) + 2*sqrt(b)*sqrt(c + d*x))/(sqrt(sqrt(b)*sq rt(a*d**2 + b*c**2) - b*c)*sqrt(2)))*a*d**2 - 15*sqrt(b)*sqrt(sqrt(b)*sqrt (a*d**2 + b*c**2) + b*c)*sqrt(2)*log( - sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a* d**2 + b*c**2) + b*c)*sqrt(2) + sqrt(a*d**2 + b*c**2) + sqrt(b)*c + sqrt(b )*d*x)*a*d**2 + 15*sqrt(b)*sqrt(sqrt(b)*sqrt(a*d**2 + b*c**2) + b*c)*sqrt( 2)*log(sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a*d**2 + b*c**2) + b*c)*sqrt(2) + s qrt(a*d**2 + b*c**2) + sqrt(b)*c + sqrt(b)*d*x)*a*d**2 - 120*sqrt(c + d*x) *a*b*d**2 - 16*sqrt(c + d*x)*b**2*c**2 + 8*sqrt(c + d*x)*b**2*c*d*x + 24*s qrt(c + d*x)*b**2*d**2*x**2)/(60*b**3*d**2)