\(\int \frac {x^3 \sqrt {c+d x}}{a-b x^2} \, dx\) [563]

Optimal result

Integrand size = 23, antiderivative size = 179 \[ \int \frac {x^3 \sqrt {c+d x}}{a-b x^2} \, dx=-\frac {2 a \sqrt {c+d x}}{b^2}+\frac {2 c (c+d x)^{3/2}}{3 b d^2}-\frac {2 (c+d x)^{5/2}}{5 b d^2}+\frac {a \sqrt {\sqrt {b} c-\sqrt {a} d} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{b^{9/4}}+\frac {a \sqrt {\sqrt {b} c+\sqrt {a} d} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c+\sqrt {a} d}}\right )}{b^{9/4}} \] 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+a 
*(b^(1/2)*c-a^(1/2)*d)^(1/2)*arctanh(b^(1/4)*(d*x+c)^(1/2)/(b^(1/2)*c-a^(1 
/2)*d)^(1/2))/b^(9/4)+a*(b^(1/2)*c+a^(1/2)*d)^(1/2)*arctanh(b^(1/4)*(d*x+c 
)^(1/2)/(b^(1/2)*c+a^(1/2)*d)^(1/2))/b^(9/4)
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.15 \[ \int \frac {x^3 \sqrt {c+d x}}{a-b x^2} \, dx=-\frac {2 \sqrt {c+d x} \left (-2 b c^2+15 a d^2+b c d x+3 b d^2 x^2\right )}{15 b^2 d^2}-\frac {a \sqrt {-b c-\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c-\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+\sqrt {a} d}\right )}{b^{5/2}}-\frac {a \sqrt {-b c+\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c+\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-\sqrt {a} d}\right )}{b^{5/2}} \] 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) - Sqrt[a]*Sqrt[b]*d]*ArcTan[(Sqrt[-(b*c) - Sqrt[a]*Sq 
rt[b]*d]*Sqrt[c + d*x])/(Sqrt[b]*c + Sqrt[a]*d)])/b^(5/2) - (a*Sqrt[-(b*c) 
 + Sqrt[a]*Sqrt[b]*d]*ArcTan[(Sqrt[-(b*c) + Sqrt[a]*Sqrt[b]*d]*Sqrt[c + d* 
x])/(Sqrt[b]*c - Sqrt[a]*d)])/b^(5/2)
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, 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

\(\displaystyle -\frac {2 \left (-\frac {a d^4 \sqrt {\sqrt {b} c-\sqrt {a} d} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{2 b^{9/4}}-\frac {a d^4 \sqrt {\sqrt {a} d+\sqrt {b} c} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {a} d+\sqrt {b} c}}\right )}{2 b^{9/4}}+\frac {a d^4 \sqrt {c+d x}}{b^2}+\frac {d^2 (c+d x)^{5/2}}{5 b}-\frac {c d^2 (c+d x)^{3/2}}{3 b}\right )}{d^4}\)

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*Sqrt[Sqrt[b]*c - Sqrt[a]*d]*ArcTanh[(b^(1/4)*S 
qrt[c + d*x])/Sqrt[Sqrt[b]*c - Sqrt[a]*d]])/(2*b^(9/4)) - (a*d^4*Sqrt[Sqrt 
[b]*c + Sqrt[a]*d]*ArcTanh[(b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c + Sqrt[a 
]*d]])/(2*b^(9/4))))/d^4
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 561

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]
 

rule 1610

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.06

method	result	size
risch	\(-\frac {2 \left (3 b \,x^{2} d^{2}+b c d x +15 a \,d^{2}-2 b \,c^{2}\right ) \sqrt {d x +c}}{15 d^{2} b^{2}}-\frac {2 a \left (\frac {\left (-a \,d^{2}+\sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}-\frac {\left (a \,d^{2}+\sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{b}\)	\(190\)
derivativedivides	\(-\frac {2 \left (\frac {\frac {b \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {b c \left (d x +c \right )^{\frac {3}{2}}}{3}+a \,d^{2} \sqrt {d x +c}}{b^{2}}+\frac {a \,d^{2} \left (\frac {\left (-a \,d^{2}+\sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}-\frac {\left (a \,d^{2}+\sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{b}\right )}{d^{2}}\)	\(193\)
default	\(\frac {-\frac {2 \left (\frac {b \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {b c \left (d x +c \right )^{\frac {3}{2}}}{3}+a \,d^{2} \sqrt {d x +c}\right )}{b^{2}}-\frac {2 a \,d^{2} \left (\frac {\left (-a \,d^{2}+\sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}-\frac {\left (a \,d^{2}+\sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{b}}{d^{2}}\)	\(195\)
pseudoelliptic	\(-\frac {2 \left (-\frac {d^{2} b \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (a \,d^{2}-\sqrt {a b \,d^{2}}\, c \right ) a \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2}+\left (-\frac {a b \,d^{2} \left (a \,d^{2}+\sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2}+\left (-\frac {2 \left (-\frac {3 d x}{2}+c \right ) \left (d x +c \right ) b}{15}+a \,d^{2}\right ) \sqrt {d x +c}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {a b \,d^{2}}\right ) \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\right )}{\sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, b^{2} d^{2}}\)	\(238\)

Input:

int(x^3*(d*x+c)^(1/2)/(-b*x^2+a),x,method=_RETURNVERBOSE)
 

Output:

-2/15*(3*b*d^2*x^2+b*c*d*x+15*a*d^2-2*b*c^2)/d^2*(d*x+c)^(1/2)/b^2-2*a/b*( 
1/2*(-a*d^2+(a*b*d^2)^(1/2)*c)/(a*b*d^2)^(1/2)/((-b*c+(a*b*d^2)^(1/2))*b)^ 
(1/2)*arctan(b*(d*x+c)^(1/2)/((-b*c+(a*b*d^2)^(1/2))*b)^(1/2))-1/2*(a*d^2+ 
(a*b*d^2)^(1/2)*c)/(a*b*d^2)^(1/2)/((b*c+(a*b*d^2)^(1/2))*b)^(1/2)*arctanh 
(b*(d*x+c)^(1/2)/((b*c+(a*b*d^2)^(1/2))*b)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (133) = 266\).

Time = 0.12 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.07 \[ \int \frac {x^3 \sqrt {c+d x}}{a-b x^2} \, dx=\frac {15 \, b^{2} d^{2} \sqrt {\frac {b^{4} \sqrt {\frac {a^{5} d^{2}}{b^{9}}} + a^{2} c}{b^{4}}} \log \left (b^{2} \sqrt {\frac {b^{4} \sqrt {\frac {a^{5} d^{2}}{b^{9}}} + a^{2} c}{b^{4}}} + \sqrt {d x + c} a\right ) - 15 \, b^{2} d^{2} \sqrt {\frac {b^{4} \sqrt {\frac {a^{5} d^{2}}{b^{9}}} + a^{2} c}{b^{4}}} \log \left (-b^{2} \sqrt {\frac {b^{4} \sqrt {\frac {a^{5} d^{2}}{b^{9}}} + a^{2} c}{b^{4}}} + \sqrt {d x + c} a\right ) + 15 \, b^{2} d^{2} \sqrt {-\frac {b^{4} \sqrt {\frac {a^{5} d^{2}}{b^{9}}} - a^{2} c}{b^{4}}} \log \left (b^{2} \sqrt {-\frac {b^{4} \sqrt {\frac {a^{5} d^{2}}{b^{9}}} - a^{2} c}{b^{4}}} + \sqrt {d x + c} a\right ) - 15 \, b^{2} d^{2} \sqrt {-\frac {b^{4} \sqrt {\frac {a^{5} d^{2}}{b^{9}}} - a^{2} c}{b^{4}}} \log \left (-b^{2} \sqrt {-\frac {b^{4} \sqrt {\frac {a^{5} d^{2}}{b^{9}}} - a^{2} c}{b^{4}}} + \sqrt {d x + c} a\right ) - 4 \, {\left (3 \, b d^{2} x^{2} + b c d x - 2 \, b c^{2} + 15 \, a d^{2}\right )} \sqrt {d x + c}}{30 \, b^{2} d^{2}} \] 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) + 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) - 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)
 

Sympy [F]

\[ \int \frac {x^3 \sqrt {c+d x}}{a-b x^2} \, dx=- \int \frac {x^{3} \sqrt {c + d x}}{- a + b x^{2}}\, dx \] 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)
 

Maxima [F]

\[ \int \frac {x^3 \sqrt {c+d x}}{a-b x^2} \, dx=\int { -\frac {\sqrt {d x + c} x^{3}}{b x^{2} - a} \,d x } \] 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)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (133) = 266\).

Time = 0.17 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.75 \[ \int \frac {x^3 \sqrt {c+d x}}{a-b x^2} \, dx=-\frac {{\left (a b^{2} c^{2} - a^{2} b d^{2}\right )} {\left | b \right |} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {b^{6} c d^{12} + \sqrt {b^{12} c^{2} d^{24} - {\left (b^{6} c^{2} d^{12} - a b^{5} d^{14}\right )} b^{6} d^{12}}}{b^{6} d^{12}}}}\right )}{{\left (b^{4} c - \sqrt {a b} b^{3} d\right )} \sqrt {-b^{2} c - \sqrt {a b} b d}} - \frac {{\left (a b^{2} c^{2} - a^{2} b d^{2}\right )} {\left | b \right |} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {b^{6} c d^{12} - \sqrt {b^{12} c^{2} d^{24} - {\left (b^{6} c^{2} d^{12} - a b^{5} d^{14}\right )} b^{6} d^{12}}}{b^{6} d^{12}}}}\right )}{{\left (b^{4} c + \sqrt {a b} b^{3} d\right )} \sqrt {-b^{2} c + \sqrt {a b} b d}} - \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} d^{8} - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c d^{8} + 15 \, \sqrt {d x + c} a b^{3} d^{10}\right )}}{15 \, b^{5} d^{10}} \] 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^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)) - 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)
 

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 515, normalized size of antiderivative = 2.88 \[ \int \frac {x^3 \sqrt {c+d x}}{a-b x^2} \, dx=\frac {2\,c\,{\left (c+d\,x\right )}^{3/2}}{3\,b\,d^2}-\frac {2\,{\left (c+d\,x\right )}^{5/2}}{5\,b\,d^2}-\left (\frac {2\,\left (a\,d^4-b\,c^2\,d^2\right )}{b^2\,d^4}+\frac {2\,c^2}{b\,d^2}\right )\,\sqrt {c+d\,x}-\mathrm {atan}\left (\frac {a^4\,d^4\,\sqrt {\frac {a^2\,c}{4\,b^4}+\frac {d\,\sqrt {a^5\,b^9}}{4\,b^9}}\,\sqrt {c+d\,x}\,32{}\mathrm {i}}{\frac {16\,a^3\,d^5\,\sqrt {a^5\,b^9}}{b^7}-\frac {16\,a^2\,c^2\,d^3\,\sqrt {a^5\,b^9}}{b^6}}-\frac {a\,c\,d^3\,\sqrt {\frac {a^2\,c}{4\,b^4}+\frac {d\,\sqrt {a^5\,b^9}}{4\,b^9}}\,\sqrt {a^5\,b^9}\,\sqrt {c+d\,x}\,32{}\mathrm {i}}{\frac {16\,a^3\,d^5\,\sqrt {a^5\,b^9}}{b^3}-\frac {16\,a^2\,c^2\,d^3\,\sqrt {a^5\,b^9}}{b^2}}\right )\,\sqrt {\frac {d\,\sqrt {a^5\,b^9}+a^2\,b^5\,c}{4\,b^9}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {a^4\,d^4\,\sqrt {\frac {a^2\,c}{4\,b^4}-\frac {d\,\sqrt {a^5\,b^9}}{4\,b^9}}\,\sqrt {c+d\,x}\,32{}\mathrm {i}}{\frac {16\,a^3\,d^5\,\sqrt {a^5\,b^9}}{b^7}-\frac {16\,a^2\,c^2\,d^3\,\sqrt {a^5\,b^9}}{b^6}}+\frac {a\,c\,d^3\,\sqrt {\frac {a^2\,c}{4\,b^4}-\frac {d\,\sqrt {a^5\,b^9}}{4\,b^9}}\,\sqrt {a^5\,b^9}\,\sqrt {c+d\,x}\,32{}\mathrm {i}}{\frac {16\,a^3\,d^5\,\sqrt {a^5\,b^9}}{b^3}-\frac {16\,a^2\,c^2\,d^3\,\sqrt {a^5\,b^9}}{b^2}}\right )\,\sqrt {-\frac {d\,\sqrt {a^5\,b^9}-a^2\,b^5\,c}{4\,b^9}}\,2{}\mathrm {i} \] 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)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.11 \[ \int \frac {x^3 \sqrt {c+d x}}{a-b x^2} \, dx=\frac {30 \sqrt {b}\, \sqrt {\sqrt {b}\, \sqrt {a}\, d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {\sqrt {b}\, \sqrt {a}\, d -b c}}\right ) a \,d^{2}-15 \sqrt {b}\, \sqrt {\sqrt {b}\, \sqrt {a}\, d +b c}\, \mathrm {log}\left (-\sqrt {\sqrt {b}\, \sqrt {a}\, d +b c}+\sqrt {b}\, \sqrt {d x +c}\right ) a \,d^{2}+15 \sqrt {b}\, \sqrt {\sqrt {b}\, \sqrt {a}\, d +b c}\, \mathrm {log}\left (\sqrt {\sqrt {b}\, \sqrt {a}\, d +b c}+\sqrt {b}\, \sqrt {d x +c}\right ) a \,d^{2}-60 \sqrt {d x +c}\, a b \,d^{2}+8 \sqrt {d x +c}\, b^{2} c^{2}-4 \sqrt {d x +c}\, b^{2} c d x -12 \sqrt {d x +c}\, b^{2} d^{2} x^{2}}{30 b^{3} d^{2}} \] Input:

int(x^3*(d*x+c)^(1/2)/(-b*x^2+a),x)
 

Output:

(30*sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)* 
sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a*d**2 - 15*sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d 
 + b*c)*log( - sqrt(sqrt(b)*sqrt(a)*d + b*c) + sqrt(b)*sqrt(c + d*x))*a*d* 
*2 + 15*sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d + b*c)*log(sqrt(sqrt(b)*sqrt(a)*d + 
 b*c) + sqrt(b)*sqrt(c + d*x))*a*d**2 - 60*sqrt(c + d*x)*a*b*d**2 + 8*sqrt 
(c + d*x)*b**2*c**2 - 4*sqrt(c + d*x)*b**2*c*d*x - 12*sqrt(c + d*x)*b**2*d 
**2*x**2)/(30*b**3*d**2)