\(\int \frac {(a+b x)^{3/2}}{(c+d x)^{2/3}} \, dx\) [426]

Optimal result

Integrand size = 19, antiderivative size = 416 \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{2/3}} \, dx=-\frac {54 (b c-a d) \sqrt {a+b x} \sqrt [3]{c+d x}}{55 d^2}+\frac {6 (a+b x)^{3/2} \sqrt [3]{c+d x}}{11 d}-\frac {54\ 3^{3/4} \sqrt {2-\sqrt {3}} (b c-a d)^2 \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{55 \sqrt [3]{b} d^3 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \] Output:

-54/55*(-a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/3)/d^2+6/11*(b*x+a)^(3/2)*(d*x+ 
c)^(1/3)/d-54/55*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*(-a*d+b*c)^2*((-a*d+b*c 
)^(1/3)-b^(1/3)*(d*x+c)^(1/3))*(((-a*d+b*c)^(2/3)+b^(1/3)*(-a*d+b*c)^(1/3) 
*(d*x+c)^(1/3)+b^(2/3)*(d*x+c)^(2/3))/((1-3^(1/2))*(-a*d+b*c)^(1/3)-b^(1/3 
)*(d*x+c)^(1/3))^2)^(1/2)*EllipticF(((1+3^(1/2))*(-a*d+b*c)^(1/3)-b^(1/3)* 
(d*x+c)^(1/3))/((1-3^(1/2))*(-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)),2*I-I* 
3^(1/2))/b^(1/3)/d^3/(b*x+a)^(1/2)/(-(-a*d+b*c)^(1/3)*((-a*d+b*c)^(1/3)-b^ 
(1/3)*(d*x+c)^(1/3))/((1-3^(1/2))*(-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))^ 
2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.18 \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{2/3}} \, dx=\frac {2 (a+b x)^{5/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{2},\frac {7}{2},\frac {d (a+b x)}{-b c+a d}\right )}{5 b (c+d x)^{2/3}} \] Input:

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

Output:

(2*(a + b*x)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(2/3)*Hypergeometric2F1[2/3 
, 5/2, 7/2, (d*(a + b*x))/(-(b*c) + a*d)])/(5*b*(c + d*x)^(2/3))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 436, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {60, 60, 73, 760}

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.

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {6 (a+b x)^{3/2} \sqrt [3]{c+d x}}{11 d}-\frac {9 (b c-a d) \int \frac {\sqrt {a+b x}}{(c+d x)^{2/3}}dx}{11 d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {6 (a+b x)^{3/2} \sqrt [3]{c+d x}}{11 d}-\frac {9 (b c-a d) \left (\frac {6 \sqrt {a+b x} \sqrt [3]{c+d x}}{5 d}-\frac {3 (b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{2/3}}dx}{5 d}\right )}{11 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {6 (a+b x)^{3/2} \sqrt [3]{c+d x}}{11 d}-\frac {9 (b c-a d) \left (\frac {6 \sqrt {a+b x} \sqrt [3]{c+d x}}{5 d}-\frac {9 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [3]{c+d x}}{5 d^2}\right )}{11 d}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {6 (a+b x)^{3/2} \sqrt [3]{c+d x}}{11 d}-\frac {9 (b c-a d) \left (\frac {6\ 3^{3/4} \sqrt {2-\sqrt {3}} (b c-a d) \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{5 \sqrt [3]{b} d^2 \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}+\frac {6 \sqrt {a+b x} \sqrt [3]{c+d x}}{5 d}\right )}{11 d}\)

Input:

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

Output:

(6*(a + b*x)^(3/2)*(c + d*x)^(1/3))/(11*d) - (9*(b*c - a*d)*((6*Sqrt[a + b 
*x]*(c + d*x)^(1/3))/(5*d) + (6*3^(3/4)*Sqrt[2 - Sqrt[3]]*(b*c - a*d)*((b* 
c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))*Sqrt[((b*c - a*d)^(2/3) + b^(1/3 
)*(b*c - a*d)^(1/3)*(c + d*x)^(1/3) + b^(2/3)*(c + d*x)^(2/3))/((1 - Sqrt[ 
3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2]*EllipticF[ArcSin[((1 + 
 Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))/((1 - Sqrt[3])*(b*c 
 - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))], -7 + 4*Sqrt[3]])/(5*b^(1/3)*d^2 
*Sqrt[-(((b*c - a*d)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)))/ 
((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2)]*Sqrt[a - ( 
b*c)/d + (b*(c + d*x))/d])))/(11*d)
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 760

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- 
s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) 
*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x 
] && NegQ[a]
 

Maple [F]

\[\int \frac {\left (b x +a \right )^{\frac {3}{2}}}{\left (x d +c \right )^{\frac {2}{3}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*x + a)^(3/2)/(d*x + c)^(2/3), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(a + b*x)/(c + d*x)**(2/3),x)*a + int((sqrt(a + b*x)*x)/(c + d*x)* 
*(2/3),x)*b