\(\int \frac {(c+d x) (a x^2+b x^3)^{3/2}}{(e x)^{11/2}} \, dx\) [316]

Optimal result

Integrand size = 28, antiderivative size = 147 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{11/2}} \, dx=-\frac {2 (b c+a d) \sqrt {a x^2+b x^3}}{e^4 (e x)^{3/2}}+\frac {b d \sqrt {a x^2+b x^3}}{e^5 \sqrt {e x}}-\frac {2 c \left (a x^2+b x^3\right )^{3/2}}{3 e (e x)^{9/2}}+\frac {\sqrt {b} (2 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a x^2+b x^3}}\right )}{e^{11/2}} \] Output:

-2*(a*d+b*c)*(b*x^3+a*x^2)^(1/2)/e^4/(e*x)^(3/2)+b*d*(b*x^3+a*x^2)^(1/2)/e 
^5/(e*x)^(1/2)-2/3*c*(b*x^3+a*x^2)^(3/2)/e/(e*x)^(9/2)+b^(1/2)*(3*a*d+2*b* 
c)*arctanh(b^(1/2)*(e*x)^(3/2)/e^(3/2)/(b*x^3+a*x^2)^(1/2))/e^(11/2)
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{11/2}} \, dx=\frac {\sqrt {x^2 (a+b x)} \left (-\sqrt {a+b x} (b x (8 c-3 d x)+2 a (c+3 d x))+6 \sqrt {b} (2 b c+3 a d) x^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )\right )}{3 e^3 (e x)^{5/2} \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/(e*x)^(11/2),x]
 

Output:

(Sqrt[x^2*(a + b*x)]*(-(Sqrt[a + b*x]*(b*x*(8*c - 3*d*x) + 2*a*(c + 3*d*x) 
)) + 6*Sqrt[b]*(2*b*c + 3*a*d)*x^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] 
 + Sqrt[a + b*x])]))/(3*e^3*(e*x)^(5/2)*Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1944, 1926, 1927, 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.

\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{3/2} (c+d x)}{(e x)^{11/2}} \, dx\)

\(\Big \downarrow \) 1944

\(\displaystyle \frac {(3 a d+2 b c) \int \frac {\left (b x^3+a x^2\right )^{3/2}}{(e x)^{9/2}}dx}{3 a e}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{3 a (e x)^{13/2}}\)

\(\Big \downarrow \) 1926

\(\displaystyle \frac {(3 a d+2 b c) \left (\frac {3 b \int \frac {\sqrt {b x^3+a x^2}}{(e x)^{3/2}}dx}{e^3}-\frac {2 \left (a x^2+b x^3\right )^{3/2}}{e (e x)^{7/2}}\right )}{3 a e}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{3 a (e x)^{13/2}}\)

\(\Big \downarrow \) 1927

\(\displaystyle \frac {(3 a d+2 b c) \left (\frac {3 b \left (\frac {a \int \frac {\sqrt {e x}}{\sqrt {b x^3+a x^2}}dx}{2 e^2}+\frac {\sqrt {a x^2+b x^3}}{e \sqrt {e x}}\right )}{e^3}-\frac {2 \left (a x^2+b x^3\right )^{3/2}}{e (e x)^{7/2}}\right )}{3 a e}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{3 a (e x)^{13/2}}\)

\(\Big \downarrow \) 1937

\(\displaystyle \frac {(3 a d+2 b c) \left (\frac {3 b \left (\frac {a \sqrt {e x} \int \frac {\sqrt {x}}{\sqrt {b x^3+a x^2}}dx}{2 e^2 \sqrt {x}}+\frac {\sqrt {a x^2+b x^3}}{e \sqrt {e x}}\right )}{e^3}-\frac {2 \left (a x^2+b x^3\right )^{3/2}}{e (e x)^{7/2}}\right )}{3 a e}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{3 a (e x)^{13/2}}\)

\(\Big \downarrow \) 1935

\(\displaystyle \frac {(3 a d+2 b c) \left (\frac {3 b \left (\frac {a \sqrt {e x} \int \frac {1}{1-\frac {b x^3}{b x^3+a x^2}}d\frac {x^{3/2}}{\sqrt {b x^3+a x^2}}}{e^2 \sqrt {x}}+\frac {\sqrt {a x^2+b x^3}}{e \sqrt {e x}}\right )}{e^3}-\frac {2 \left (a x^2+b x^3\right )^{3/2}}{e (e x)^{7/2}}\right )}{3 a e}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{3 a (e x)^{13/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(3 a d+2 b c) \left (\frac {3 b \left (\frac {a \sqrt {e x} \text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{\sqrt {b} e^2 \sqrt {x}}+\frac {\sqrt {a x^2+b x^3}}{e \sqrt {e x}}\right )}{e^3}-\frac {2 \left (a x^2+b x^3\right )^{3/2}}{e (e x)^{7/2}}\right )}{3 a e}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{3 a (e x)^{13/2}}\)

Input:

Int[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/(e*x)^(11/2),x]
 

Output:

(-2*c*e*(a*x^2 + b*x^3)^(5/2))/(3*a*(e*x)^(13/2)) + ((2*b*c + 3*a*d)*((-2* 
(a*x^2 + b*x^3)^(3/2))/(e*(e*x)^(7/2)) + (3*b*(Sqrt[a*x^2 + b*x^3]/(e*Sqrt 
[e*x]) + (a*Sqrt[e*x]*ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a*x^2 + b*x^3]])/(Sqr 
t[b]*e^2*Sqrt[x])))/e^3))/(3*a*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 1927

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 + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) 
, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*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 = 0.45 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {\left (-3 b d \,x^{2}+6 a d x +8 c b x +2 a c \right ) \sqrt {x^{2} \left (b x +a \right )}}{3 x^{2} e^{5} \sqrt {e x}}+\frac {b \left (3 a d +2 b c \right ) \ln \left (\frac {\frac {1}{2} a e +b e x}{\sqrt {b e}}+\sqrt {b e \,x^{2}+a e x}\right ) \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x \left (b x +a \right )}}{2 \sqrt {b e}\, e^{5} x \left (b x +a \right ) \sqrt {e x}}\)	\(134\)
default	\(\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (9 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a b d e \,x^{2}+6 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) b^{2} c e \,x^{2}+6 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, b d \,x^{2}-12 a d x \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}-16 b c x \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}-4 a c \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\right )}{6 x^{4} \left (b x +a \right ) e^{5} \sqrt {e x}\, \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}}\)	\(218\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(11/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3*(-3*b*d*x^2+6*a*d*x+8*b*c*x+2*a*c)/x^2/e^5*(x^2*(b*x+a))^(1/2)/(e*x)^ 
(1/2)+1/2*b*(3*a*d+2*b*c)*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^5*(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 = 0.11 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.59 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{11/2}} \, dx=\left [\frac {3 \, {\left (2 \, b c + 3 \, a d\right )} e x^{3} \sqrt {\frac {b}{e}} \log \left (\frac {2 \, b x^{2} + a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {e x} \sqrt {\frac {b}{e}}}{x}\right ) + 2 \, {\left (3 \, b d x^{2} - 2 \, a c - 2 \, {\left (4 \, b c + 3 \, a d\right )} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{6 \, e^{6} x^{3}}, -\frac {3 \, {\left (2 \, b c + 3 \, a d\right )} e x^{3} \sqrt {-\frac {b}{e}} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {e x} \sqrt {-\frac {b}{e}}}{b x^{2} + a x}\right ) - {\left (3 \, b d x^{2} - 2 \, a c - 2 \, {\left (4 \, b c + 3 \, a d\right )} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{3 \, e^{6} x^{3}}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(11/2),x, algorithm="fricas")
 

Output:

[1/6*(3*(2*b*c + 3*a*d)*e*x^3*sqrt(b/e)*log((2*b*x^2 + a*x + 2*sqrt(b*x^3 
+ a*x^2)*sqrt(e*x)*sqrt(b/e))/x) + 2*(3*b*d*x^2 - 2*a*c - 2*(4*b*c + 3*a*d 
)*x)*sqrt(b*x^3 + a*x^2)*sqrt(e*x))/(e^6*x^3), -1/3*(3*(2*b*c + 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 
)) - (3*b*d*x^2 - 2*a*c - 2*(4*b*c + 3*a*d)*x)*sqrt(b*x^3 + a*x^2)*sqrt(e* 
x))/(e^6*x^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{11/2}} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)*(b*x**3+a*x**2)**(3/2)/(e*x)**(11/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{11/2}} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} {\left (d x + c\right )}}{\left (e x\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(11/2),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(3/2)*(d*x + c)/(e*x)^(11/2), x)
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.27 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{11/2}} \, dx=-\frac {b^{3} {\left (\frac {3 \, {\left (2 \, b c \mathrm {sgn}\left (x\right ) + 3 \, a d \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {{\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} b} - \frac {{\left ({\left (3 \, {\left (b x + a\right )} d e \mathrm {sgn}\left (x\right ) - \frac {4 \, {\left (2 \, a b^{2} c e^{2} \mathrm {sgn}\left (x\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (x\right )\right )}}{a b e}\right )} {\left (b x + a\right )} + \frac {3 \, {\left (2 \, a^{2} b^{2} c e^{2} \mathrm {sgn}\left (x\right ) + 3 \, a^{3} b d e^{2} \mathrm {sgn}\left (x\right )\right )}}{a b e}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b e - a b e\right )}^{\frac {3}{2}}}\right )}}{3 \, e^{5} {\left | b \right |}} \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(11/2),x, algorithm="giac")
 

Output:

-1/3*b^3*(3*(2*b*c*sgn(x) + 3*a*d*sgn(x))*log(abs(-sqrt(b*e)*sqrt(b*x + a) 
 + sqrt((b*x + a)*b*e - a*b*e)))/(sqrt(b*e)*b) - ((3*(b*x + a)*d*e*sgn(x) 
- 4*(2*a*b^2*c*e^2*sgn(x) + 3*a^2*b*d*e^2*sgn(x))/(a*b*e))*(b*x + a) + 3*( 
2*a^2*b^2*c*e^2*sgn(x) + 3*a^3*b*d*e^2*sgn(x))/(a*b*e))*sqrt(b*x + a)/((b* 
x + a)*b*e - a*b*e)^(3/2))/(e^5*abs(b))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{11/2}} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}\,\left (c+d\,x\right )}{{\left (e\,x\right )}^{11/2}} \,d x \] Input:

int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/(e*x)^(11/2),x)
 

Output:

int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/(e*x)^(11/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{11/2}} \, dx=\frac {\sqrt {e}\, \left (-4 \sqrt {x}\, \sqrt {b x +a}\, a c -12 \sqrt {x}\, \sqrt {b x +a}\, a d x -16 \sqrt {x}\, \sqrt {b x +a}\, b c x +6 \sqrt {x}\, \sqrt {b x +a}\, b d \,x^{2}+18 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a d \,x^{2}+12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) b c \,x^{2}+5 \sqrt {b}\, a d \,x^{2}\right )}{6 e^{6} x^{2}} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(11/2),x)
 

Output:

(sqrt(e)*( - 4*sqrt(x)*sqrt(a + b*x)*a*c - 12*sqrt(x)*sqrt(a + b*x)*a*d*x 
- 16*sqrt(x)*sqrt(a + b*x)*b*c*x + 6*sqrt(x)*sqrt(a + b*x)*b*d*x**2 + 18*s 
qrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a*d*x**2 + 12*sqrt(b 
)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*b*c*x**2 + 5*sqrt(b)*a*d* 
x**2))/(6*e**6*x**2)