\(\int \frac {(c+d x) \sqrt {a x^2+b x^3}}{(e x)^{7/2}} \, dx\) [306]

Optimal result

Integrand size = 28, antiderivative size = 107 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{(e x)^{7/2}} \, dx=-\frac {2 d \sqrt {a x^2+b x^3}}{e^2 (e x)^{3/2}}-\frac {2 c e \left (a x^2+b x^3\right )^{3/2}}{3 a (e x)^{9/2}}+\frac {2 \sqrt {b} d \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a x^2+b x^3}}\right )}{e^{7/2}} \] Output:

-2*d*(b*x^3+a*x^2)^(1/2)/e^2/(e*x)^(3/2)-2/3*c*e*(b*x^3+a*x^2)^(3/2)/a/(e* 
x)^(9/2)+2*b^(1/2)*d*arctanh(b^(1/2)*(e*x)^(3/2)/e^(3/2)/(b*x^3+a*x^2)^(1/ 
2))/e^(7/2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.92 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{(e x)^{7/2}} \, dx=-\frac {2 x \sqrt {x^2 (a+b x)} \left (\sqrt {a+b x} (b c x+a (c+3 d x))+3 a \sqrt {b} d x^{3/2} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )\right )}{3 a (e x)^{7/2} \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)*Sqrt[a*x^2 + b*x^3])/(e*x)^(7/2),x]
 

Output:

(-2*x*Sqrt[x^2*(a + b*x)]*(Sqrt[a + b*x]*(b*c*x + a*(c + 3*d*x)) + 3*a*Sqr 
t[b]*d*x^(3/2)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]]))/(3*a*(e*x)^(7/2)* 
Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1944, 1926, 1937, 1935, 219}

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.

\(\displaystyle \int \frac {\sqrt {a x^2+b x^3} (c+d x)}{(e x)^{7/2}} \, dx\)

\(\Big \downarrow \) 1944

\(\displaystyle \frac {d \int \frac {\sqrt {b x^3+a x^2}}{(e x)^{5/2}}dx}{e}-\frac {2 c e \left (a x^2+b x^3\right )^{3/2}}{3 a (e x)^{9/2}}\)

\(\Big \downarrow \) 1926

\(\displaystyle \frac {d \left (\frac {b \int \frac {\sqrt {e x}}{\sqrt {b x^3+a x^2}}dx}{e^3}-\frac {2 \sqrt {a x^2+b x^3}}{e (e x)^{3/2}}\right )}{e}-\frac {2 c e \left (a x^2+b x^3\right )^{3/2}}{3 a (e x)^{9/2}}\)

\(\Big \downarrow \) 1937

\(\displaystyle \frac {d \left (\frac {b \sqrt {e x} \int \frac {\sqrt {x}}{\sqrt {b x^3+a x^2}}dx}{e^3 \sqrt {x}}-\frac {2 \sqrt {a x^2+b x^3}}{e (e x)^{3/2}}\right )}{e}-\frac {2 c e \left (a x^2+b x^3\right )^{3/2}}{3 a (e x)^{9/2}}\)

\(\Big \downarrow \) 1935

\(\displaystyle \frac {d \left (\frac {2 b \sqrt {e x} \int \frac {1}{1-\frac {b x^3}{b x^3+a x^2}}d\frac {x^{3/2}}{\sqrt {b x^3+a x^2}}}{e^3 \sqrt {x}}-\frac {2 \sqrt {a x^2+b x^3}}{e (e x)^{3/2}}\right )}{e}-\frac {2 c e \left (a x^2+b x^3\right )^{3/2}}{3 a (e x)^{9/2}}\)

\(\Big \downarrow \) 219

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

Input:

Int[((c + d*x)*Sqrt[a*x^2 + b*x^3])/(e*x)^(7/2),x]
 

Output:

(-2*c*e*(a*x^2 + b*x^3)^(3/2))/(3*a*(e*x)^(9/2)) + (d*((-2*Sqrt[a*x^2 + b* 
x^3])/(e*(e*x)^(3/2)) + (2*Sqrt[b]*Sqrt[e*x]*ArcTanh[(Sqrt[b]*x^(3/2))/Sqr 
t[a*x^2 + b*x^3]])/(e^3*Sqrt[x])))/e
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1926

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] 
 :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + j*p + 1))), x] - Simp[b*p 
*((n - j)/(c^n*(m + j*p + 1)))   Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p - 1), 
 x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Integer 
sQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && LtQ[m + j*p + 1, 0]
 

rule 1935

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp 
[-2/(n - j)   Subst[Int[1/(1 - a*x^2), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], 
 x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]
 

rule 1937

Int[((c_)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] 
:> Simp[c^IntPart[m]*((c*x)^FracPart[m]/x^FracPart[m])   Int[x^m*(a*x^j + b 
*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && IntegerQ[p + 1/2] && 
NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0]
 

rule 1944

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

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.11

method	result	size
risch	\(-\frac {2 \left (3 a d x +c b x +a c \right ) \sqrt {x^{2} \left (b x +a \right )}}{3 x^{2} a \,e^{3} \sqrt {e x}}+\frac {b d \ln \left (\frac {\frac {1}{2} a e +b e x}{\sqrt {b e}}+\sqrt {b e \,x^{2}+a e x}\right ) \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x \left (b x +a \right )}}{\sqrt {b e}\, e^{3} x \left (b x +a \right ) \sqrt {e x}}\)	\(119\)
default	\(-\frac {\sqrt {b \,x^{3}+a \,x^{2}}\, \left (-3 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a b d e \,x^{2}+6 a d x \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+2 b c x \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+2 a c \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\right )}{3 x^{2} e^{3} \sqrt {e x}\, \sqrt {e x \left (b x +a \right )}\, a \sqrt {b e}}\)	\(148\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(1/2)/(e*x)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.91 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{(e x)^{7/2}} \, dx=\left [\frac {3 \, a d e x^{3} \sqrt {\frac {b}{e}} \log \left (\frac {2 \, b x^{2} + a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {e x} \sqrt {\frac {b}{e}}}{x}\right ) - 2 \, \sqrt {b x^{3} + a x^{2}} {\left (a c + {\left (b c + 3 \, a d\right )} x\right )} \sqrt {e x}}{3 \, a e^{4} x^{3}}, -\frac {2 \, {\left (3 \, a d e x^{3} \sqrt {-\frac {b}{e}} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {e x} \sqrt {-\frac {b}{e}}}{b x^{2} + a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (a c + {\left (b c + 3 \, a d\right )} x\right )} \sqrt {e x}\right )}}{3 \, a e^{4} x^{3}}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/(e*x)^(7/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{(e x)^{7/2}} \, dx=\int \frac {\sqrt {x^{2} \left (a + b x\right )} \left (c + d x\right )}{\left (e x\right )^{\frac {7}{2}}}\, dx \] Input:

integrate((d*x+c)*(b*x**3+a*x**2)**(1/2)/(e*x)**(7/2),x)
 

Output:

Integral(sqrt(x**2*(a + b*x))*(c + d*x)/(e*x)**(7/2), x)
 

Maxima [F]

\[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{(e x)^{7/2}} \, dx=\int { \frac {\sqrt {b x^{3} + a x^{2}} {\left (d x + c\right )}}{\left (e x\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/(e*x)^(7/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*x^3 + a*x^2)*(d*x + c)/(e*x)^(7/2), x)
 

Giac [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.14 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{(e x)^{7/2}} \, dx=-\frac {2 \, {\left (\frac {3 \, d \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {{\left (b x + a\right )} b e - a b e} \right |}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {b e} b} - \frac {{\left (3 \, a d e \mathrm {sgn}\left (x\right ) - \frac {{\left (b^{2} c e \mathrm {sgn}\left (x\right ) + 3 \, a b d e \mathrm {sgn}\left (x\right )\right )} {\left (b x + a\right )}}{a b}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b e - a b e\right )}^{\frac {3}{2}}}\right )} b^{3}}{3 \, e^{3} {\left | b \right |}} \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/(e*x)^(7/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{(e x)^{7/2}} \, dx=\int \frac {\sqrt {b\,x^3+a\,x^2}\,\left (c+d\,x\right )}{{\left (e\,x\right )}^{7/2}} \,d x \] Input:

int(((a*x^2 + b*x^3)^(1/2)*(c + d*x))/(e*x)^(7/2),x)
 

Output:

int(((a*x^2 + b*x^3)^(1/2)*(c + d*x))/(e*x)^(7/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{(e x)^{7/2}} \, dx=\frac {2 \sqrt {e}\, \left (-\sqrt {x}\, \sqrt {b x +a}\, a c -3 \sqrt {x}\, \sqrt {b x +a}\, a d x -\sqrt {x}\, \sqrt {b x +a}\, b c x +3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a d \,x^{2}+\sqrt {b}\, a d \,x^{2}-\sqrt {b}\, b c \,x^{2}\right )}{3 a \,e^{4} x^{2}} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(1/2)/(e*x)^(7/2),x)
 

Output:

(2*sqrt(e)*( - sqrt(x)*sqrt(a + b*x)*a*c - 3*sqrt(x)*sqrt(a + b*x)*a*d*x - 
 sqrt(x)*sqrt(a + b*x)*b*c*x + 3*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt 
(b))/sqrt(a))*a*d*x**2 + sqrt(b)*a*d*x**2 - sqrt(b)*b*c*x**2))/(3*a*e**4*x 
**2)