\(\int \frac {(b c^2-b d^2 x^2)^{3/2}}{(e x)^{3/2} (c+d x)^{3/2}} \, dx\) [1101]

Optimal result

Integrand size = 36, antiderivative size = 155 \[ \int \frac {\left (b c^2-b d^2 x^2\right )^{3/2}}{(e x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {2 b c \sqrt {b c^2-b d^2 x^2}}{e \sqrt {e x} \sqrt {c+d x}}-\frac {b d \sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{e^2 \sqrt {c+d x}}+\frac {3 b^{3/2} c \sqrt {d} \arctan \left (\frac {\sqrt {e} \sqrt {b c^2-b d^2 x^2}}{\sqrt {b} \sqrt {d} \sqrt {e x} \sqrt {c+d x}}\right )}{e^{3/2}} \] Output:

-2*b*c*(-b*d^2*x^2+b*c^2)^(1/2)/e/(e*x)^(1/2)/(d*x+c)^(1/2)-b*d*(e*x)^(1/2 
)*(-b*d^2*x^2+b*c^2)^(1/2)/e^2/(d*x+c)^(1/2)+3*b^(3/2)*c*d^(1/2)*arctan(e^ 
(1/2)*(-b*d^2*x^2+b*c^2)^(1/2)/b^(1/2)/d^(1/2)/(e*x)^(1/2)/(d*x+c)^(1/2))/ 
e^(3/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.44 \[ \int \frac {\left (b c^2-b d^2 x^2\right )^{3/2}}{(e x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {2 b c x \sqrt {b \left (c^2-d^2 x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},\frac {d x}{c}\right )}{(e x)^{3/2} \sqrt {c+d x} \sqrt {1-\frac {d x}{c}}} \] Input:

Integrate[(b*c^2 - b*d^2*x^2)^(3/2)/((e*x)^(3/2)*(c + d*x)^(3/2)),x]
 

Output:

(-2*b*c*x*Sqrt[b*(c^2 - d^2*x^2)]*Hypergeometric2F1[-3/2, -1/2, 1/2, (d*x) 
/c])/((e*x)^(3/2)*Sqrt[c + d*x]*Sqrt[1 - (d*x)/c])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {575, 576, 586, 65, 218}

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.

\(\displaystyle \int \frac {\left (b c^2-b d^2 x^2\right )^{3/2}}{(e x)^{3/2} (c+d x)^{3/2}} \, dx\)

\(\Big \downarrow \) 575

\(\displaystyle -\frac {3 b d \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} \sqrt {c+d x}}dx}{e}-\frac {2 \left (b c^2-b d^2 x^2\right )^{3/2}}{e \sqrt {e x} (c+d x)^{3/2}}\)

\(\Big \downarrow \) 576

\(\displaystyle -\frac {3 b d \left (\frac {1}{2} b c \int \frac {\sqrt {c+d x}}{\sqrt {e x} \sqrt {b c^2-b d^2 x^2}}dx+\frac {\sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{e \sqrt {c+d x}}\right )}{e}-\frac {2 \left (b c^2-b d^2 x^2\right )^{3/2}}{e \sqrt {e x} (c+d x)^{3/2}}\)

\(\Big \downarrow \) 586

\(\displaystyle -\frac {3 b d \left (\frac {b c \sqrt {c+d x} \sqrt {b c-b d x} \int \frac {1}{\sqrt {e x} \sqrt {b c-b d x}}dx}{2 \sqrt {b c^2-b d^2 x^2}}+\frac {\sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{e \sqrt {c+d x}}\right )}{e}-\frac {2 \left (b c^2-b d^2 x^2\right )^{3/2}}{e \sqrt {e x} (c+d x)^{3/2}}\)

\(\Big \downarrow \) 65

\(\displaystyle -\frac {3 b d \left (\frac {b c \sqrt {c+d x} \sqrt {b c-b d x} \int \frac {1}{\frac {b d x e}{b c-b d x}+e}d\frac {\sqrt {e x}}{\sqrt {b c-b d x}}}{\sqrt {b c^2-b d^2 x^2}}+\frac {\sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{e \sqrt {c+d x}}\right )}{e}-\frac {2 \left (b c^2-b d^2 x^2\right )^{3/2}}{e \sqrt {e x} (c+d x)^{3/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {3 b d \left (\frac {\sqrt {b} c \sqrt {c+d x} \sqrt {b c-b d x} \arctan \left (\frac {\sqrt {b} \sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {b c-b d x}}\right )}{\sqrt {d} \sqrt {e} \sqrt {b c^2-b d^2 x^2}}+\frac {\sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{e \sqrt {c+d x}}\right )}{e}-\frac {2 \left (b c^2-b d^2 x^2\right )^{3/2}}{e \sqrt {e x} (c+d x)^{3/2}}\)

Input:

Int[(b*c^2 - b*d^2*x^2)^(3/2)/((e*x)^(3/2)*(c + d*x)^(3/2)),x]
 

Output:

(-2*(b*c^2 - b*d^2*x^2)^(3/2))/(e*Sqrt[e*x]*(c + d*x)^(3/2)) - (3*b*d*((Sq 
rt[e*x]*Sqrt[b*c^2 - b*d^2*x^2])/(e*Sqrt[c + d*x]) + (Sqrt[b]*c*Sqrt[c + d 
*x]*Sqrt[b*c - b*d*x]*ArcTan[(Sqrt[b]*Sqrt[d]*Sqrt[e*x])/(Sqrt[e]*Sqrt[b*c 
 - b*d*x])])/(Sqrt[d]*Sqrt[e]*Sqrt[b*c^2 - b*d^2*x^2])))/e
 

Defintions of rubi rules used

rule 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 575

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[(e*x)^(m + 1)*(c + d*x)^n*((a + b*x^2)^p/(e*(m + 1))), x] 
 + Simp[b*(n/(d*e*(m + 1)))   Int[(e*x)^(m + 1)*(c + d*x)^(n + 1)*(a + b*x^ 
2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[b*c^2 + a*d^2, 0] && 
 EqQ[n + p, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m + p] && LeQ[m + 
p + 2, 0])
 

rule 576

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[(-(e*x)^(m + 1))*(c + d*x)^n*((a + b*x^2)^p/(e*(n - m - 1 
))), x] - Simp[b*c*(n/(d^2*(n - m - 1)))   Int[(e*x)^m*(c + d*x)^(n + 1)*(a 
 + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c^2 + a* 
d^2, 0] && EqQ[n + p, 0] && GtQ[p, 0] && NeQ[m - n + 1, 0] &&  !IGtQ[m, 0] 
&&  !(IntegerQ[m + p] && LtQ[m + p + 2, 0]) && RationalQ[m]
 

rule 586

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_) 
, x_Symbol] :> Simp[(a + b*x^2)^FracPart[p]/((c + d*x)^FracPart[p]*(a/c + ( 
b*x)/d)^FracPart[p])   Int[(e*x)^m*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0]
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {\sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, b \left (3 \arctan \left (\frac {\sqrt {b d e}\, \left (-2 d x +c \right )}{2 d \sqrt {\left (-d x +c \right ) b e x}}\right ) b c d e x -2 \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}\, d x -4 \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}\, c \right )}{2 e \sqrt {e x}\, \sqrt {d x +c}\, \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}}\)	\(134\)

Input:

int((-b*d^2*x^2+b*c^2)^(3/2)/(e*x)^(3/2)/(d*x+c)^(3/2),x,method=_RETURNVER 
BOSE)
 

Output:

1/2*(b*(-d^2*x^2+c^2))^(1/2)*b/e*(3*arctan(1/2*(b*d*e)^(1/2)*(-2*d*x+c)/d/ 
((-d*x+c)*b*e*x)^(1/2))*b*c*d*e*x-2*((-d*x+c)*b*e*x)^(1/2)*(b*d*e)^(1/2)*d 
*x-4*((-d*x+c)*b*e*x)^(1/2)*(b*d*e)^(1/2)*c)/(e*x)^(1/2)/(d*x+c)^(1/2)/((- 
d*x+c)*b*e*x)^(1/2)/(b*d*e)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.08 \[ \int \frac {\left (b c^2-b d^2 x^2\right )^{3/2}}{(e x)^{3/2} (c+d x)^{3/2}} \, dx=\left [-\frac {4 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (b d x + 2 \, b c\right )} \sqrt {d x + c} \sqrt {e x} - 3 \, {\left (b c d e x^{2} + b c^{2} e x\right )} \sqrt {-\frac {b d}{e}} \log \left (-\frac {8 \, b d^{3} x^{3} - 7 \, b c^{2} d x + b c^{3} - 4 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (2 \, d x - c\right )} \sqrt {d x + c} \sqrt {e x} \sqrt {-\frac {b d}{e}}}{d x + c}\right )}{4 \, {\left (d e^{2} x^{2} + c e^{2} x\right )}}, -\frac {2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (b d x + 2 \, b c\right )} \sqrt {d x + c} \sqrt {e x} - 3 \, {\left (b c d e x^{2} + b c^{2} e x\right )} \sqrt {\frac {b d}{e}} \arctan \left (\frac {\sqrt {-b d^{2} x^{2} + b c^{2}} {\left (2 \, d x - c\right )} \sqrt {d x + c} \sqrt {e x} \sqrt {\frac {b d}{e}}}{2 \, {\left (b d^{3} x^{3} - b c^{2} d x\right )}}\right )}{2 \, {\left (d e^{2} x^{2} + c e^{2} x\right )}}\right ] \] Input:

integrate((-b*d^2*x^2+b*c^2)^(3/2)/(e*x)^(3/2)/(d*x+c)^(3/2),x, algorithm= 
"fricas")
 

Output:

[-1/4*(4*sqrt(-b*d^2*x^2 + b*c^2)*(b*d*x + 2*b*c)*sqrt(d*x + c)*sqrt(e*x) 
- 3*(b*c*d*e*x^2 + b*c^2*e*x)*sqrt(-b*d/e)*log(-(8*b*d^3*x^3 - 7*b*c^2*d*x 
 + b*c^3 - 4*sqrt(-b*d^2*x^2 + b*c^2)*(2*d*x - c)*sqrt(d*x + c)*sqrt(e*x)* 
sqrt(-b*d/e))/(d*x + c)))/(d*e^2*x^2 + c*e^2*x), -1/2*(2*sqrt(-b*d^2*x^2 + 
 b*c^2)*(b*d*x + 2*b*c)*sqrt(d*x + c)*sqrt(e*x) - 3*(b*c*d*e*x^2 + b*c^2*e 
*x)*sqrt(b*d/e)*arctan(1/2*sqrt(-b*d^2*x^2 + b*c^2)*(2*d*x - c)*sqrt(d*x + 
 c)*sqrt(e*x)*sqrt(b*d/e)/(b*d^3*x^3 - b*c^2*d*x)))/(d*e^2*x^2 + c*e^2*x)]
 

Sympy [F]

\[ \int \frac {\left (b c^2-b d^2 x^2\right )^{3/2}}{(e x)^{3/2} (c+d x)^{3/2}} \, dx=\int \frac {\left (- b \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {3}{2}}}{\left (e x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((-b*d**2*x**2+b*c**2)**(3/2)/(e*x)**(3/2)/(d*x+c)**(3/2),x)
 

Output:

Integral((-b*(-c + d*x)*(c + d*x))**(3/2)/((e*x)**(3/2)*(c + d*x)**(3/2)), 
 x)
 

Maxima [F]

\[ \int \frac {\left (b c^2-b d^2 x^2\right )^{3/2}}{(e x)^{3/2} (c+d x)^{3/2}} \, dx=\int { \frac {{\left (-b d^{2} x^{2} + b c^{2}\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((-b*d^2*x^2+b*c^2)^(3/2)/(e*x)^(3/2)/(d*x+c)^(3/2),x, algorithm= 
"maxima")
 

Output:

integrate((-b*d^2*x^2 + b*c^2)^(3/2)/((d*x + c)^(3/2)*(e*x)^(3/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.01 \[ \int \frac {\left (b c^2-b d^2 x^2\right )^{3/2}}{(e x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {{\left (\frac {3 \, b^{3} c \log \left ({\left | -\sqrt {-b d e} \sqrt {-{\left (d x + c\right )} b + 2 \, b c} + \sqrt {b^{2} c d e + {\left ({\left (d x + c\right )} b - 2 \, b c\right )} b d e} \right |}\right )}{\sqrt {-b d e} {\left | b \right |}} + \frac {{\left (\frac {3 \, b^{3} c}{{\left | b \right |}} + \frac {{\left ({\left (d x + c\right )} b - 2 \, b c\right )} b^{2}}{{\left | b \right |}}\right )} \sqrt {-{\left (d x + c\right )} b + 2 \, b c}}{\sqrt {b^{2} c d e + {\left ({\left (d x + c\right )} b - 2 \, b c\right )} b d e}}\right )} d^{2}}{e {\left | d \right |}} \] Input:

integrate((-b*d^2*x^2+b*c^2)^(3/2)/(e*x)^(3/2)/(d*x+c)^(3/2),x, algorithm= 
"giac")
 

Output:

-(3*b^3*c*log(abs(-sqrt(-b*d*e)*sqrt(-(d*x + c)*b + 2*b*c) + sqrt(b^2*c*d* 
e + ((d*x + c)*b - 2*b*c)*b*d*e)))/(sqrt(-b*d*e)*abs(b)) + (3*b^3*c/abs(b) 
 + ((d*x + c)*b - 2*b*c)*b^2/abs(b))*sqrt(-(d*x + c)*b + 2*b*c)/sqrt(b^2*c 
*d*e + ((d*x + c)*b - 2*b*c)*b*d*e))*d^2/(e*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b c^2-b d^2 x^2\right )^{3/2}}{(e x)^{3/2} (c+d x)^{3/2}} \, dx=\int \frac {{\left (b\,c^2-b\,d^2\,x^2\right )}^{3/2}}{{\left (e\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.52 \[ \int \frac {\left (b c^2-b d^2 x^2\right )^{3/2}}{(e x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {\sqrt {e}\, \sqrt {b}\, b \left (-6 \sqrt {x}\, \sqrt {d}\, \mathit {asin} \left (\frac {\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) c +3 \sqrt {x}\, \sqrt {d}\, \mathit {atan} \left (\frac {\sqrt {x}\, \sqrt {d}\, \sqrt {-d x +c}}{-d x +c}\right ) c -2 \sqrt {-d x +c}\, c -\sqrt {-d x +c}\, d x \right )}{\sqrt {x}\, e^{2}} \] Input:

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

Output:

(sqrt(e)*sqrt(b)*b*( - 6*sqrt(x)*sqrt(d)*asin((sqrt(x)*sqrt(d))/sqrt(c))*c 
 + 3*sqrt(x)*sqrt(d)*atan((sqrt(x)*sqrt(d)*sqrt(c - d*x))/(c - d*x))*c - 2 
*sqrt(c - d*x)*c - sqrt(c - d*x)*d*x))/(sqrt(x)*e**2)