\(\int \frac {(c+d x)^{5/6}}{(a+b x)^{3/2}} \, dx\) [496]

Optimal result

Integrand size = 19, antiderivative size = 798 \[ \int \frac {(c+d x)^{5/6}}{(a+b x)^{3/2}} \, dx=-\frac {2 (c+d x)^{5/6}}{b \sqrt {a+b x}}-\frac {5 \left (1+\sqrt {3}\right ) d \sqrt {a+b x} \sqrt [6]{c+d x}}{b^{5/3} \left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )}-\frac {5 \sqrt [4]{3} \sqrt [3]{b c-a d} \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} E\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{b^{5/3} \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}-\frac {5 \left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d} \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} b^{5/3} \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \] Output:

-2*(d*x+c)^(5/6)/b/(b*x+a)^(1/2)-5*(1+3^(1/2))*d*(b*x+a)^(1/2)*(d*x+c)^(1/ 
6)/b^(5/3)/((-a*d+b*c)^(1/3)-(1+3^(1/2))*b^(1/3)*(d*x+c)^(1/3))-5*3^(1/4)* 
(-a*d+b*c)^(1/3)*(d*x+c)^(1/6)*((-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))/((-a*d+b*c)^(1/3)-(1+3^(1/2))*b^(1/3)*(d*x+c)^(1/3))^2)^(1/2)*Ellipti 
cE((1-((-a*d+b*c)^(1/3)-(1-3^(1/2))*b^(1/3)*(d*x+c)^(1/3))^2/((-a*d+b*c)^( 
1/3)-(1+3^(1/2))*b^(1/3)*(d*x+c)^(1/3))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))/ 
b^(5/3)/(b*x+a)^(1/2)/(-b^(1/3)*(d*x+c)^(1/3)*((-a*d+b*c)^(1/3)-b^(1/3)*(d 
*x+c)^(1/3))/((-a*d+b*c)^(1/3)-(1+3^(1/2))*b^(1/3)*(d*x+c)^(1/3))^2)^(1/2) 
-5/6*(1-3^(1/2))*(-a*d+b*c)^(1/3)*(d*x+c)^(1/6)*((-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))/((-a*d+b*c)^(1/3)-(1+3^(1/2))*b^(1/3)*(d*x+c)^(1/3)) 
^2)^(1/2)*InverseJacobiAM(arccos(((-a*d+b*c)^(1/3)-(1-3^(1/2))*b^(1/3)*(d* 
x+c)^(1/3))/((-a*d+b*c)^(1/3)-(1+3^(1/2))*b^(1/3)*(d*x+c)^(1/3))),1/4*6^(1 
/2)+1/4*2^(1/2))*3^(3/4)/b^(5/3)/(b*x+a)^(1/2)/(-b^(1/3)*(d*x+c)^(1/3)*((- 
a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))/((-a*d+b*c)^(1/3)-(1+3^(1/2))*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 = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.09 \[ \int \frac {(c+d x)^{5/6}}{(a+b x)^{3/2}} \, dx=-\frac {2 (c+d x)^{5/6} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},-\frac {1}{2},\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{b \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/6}} \] Input:

Integrate[(c + d*x)^(5/6)/(a + b*x)^(3/2),x]
 

Output:

(-2*(c + d*x)^(5/6)*Hypergeometric2F1[-5/6, -1/2, 1/2, (d*(a + b*x))/(-(b* 
c) + a*d)])/(b*Sqrt[a + b*x]*((b*(c + d*x))/(b*c - a*d))^(5/6))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 846, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {57, 73, 837, 25, 766, 2420}

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.

\(\displaystyle \int \frac {(c+d x)^{5/6}}{(a+b x)^{3/2}} \, dx\)

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {10 \int \frac {(c+d x)^{2/3}}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [6]{c+d x}}{b}-\frac {2 (c+d x)^{5/6}}{b \sqrt {a+b x}}\)

\(\Big \downarrow \) 837

\(\displaystyle \frac {10 \left (-\frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3} \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [6]{c+d x}}{2 b^{2/3}}-\frac {\int -\frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3}+2 b^{2/3} (c+d x)^{2/3}}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [6]{c+d x}}{2 b^{2/3}}\right )}{b}-\frac {2 (c+d x)^{5/6}}{b \sqrt {a+b x}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {10 \left (\frac {\int \frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3}+2 b^{2/3} (c+d x)^{2/3}}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [6]{c+d x}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3} \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [6]{c+d x}}{2 b^{2/3}}\right )}{b}-\frac {2 (c+d x)^{5/6}}{b \sqrt {a+b x}}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {10 \left (\frac {\int \frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3}+2 b^{2/3} (c+d x)^{2/3}}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [6]{c+d x}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [6]{c+d x} \sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} b^{2/3} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{b}-\frac {2 (c+d x)^{5/6}}{b \sqrt {a+b x}}\)

\(\Big \downarrow \) 2420

\(\displaystyle \frac {10 \left (\frac {-\frac {\left (1+\sqrt {3}\right ) \sqrt [6]{c+d x} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}} d}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}-\frac {\sqrt [4]{3} \sqrt [3]{b c-a d} \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} E\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d} \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} b^{2/3} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{b}-\frac {2 (c+d x)^{5/6}}{b \sqrt {a+b x}}\)

Input:

Int[(c + d*x)^(5/6)/(a + b*x)^(3/2),x]
 

Output:

(-2*(c + d*x)^(5/6))/(b*Sqrt[a + b*x]) + (10*((-(((1 + Sqrt[3])*d*(c + d*x 
)^(1/6)*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])/((b*c - a*d)^(1/3) - (1 + Sqr 
t[3])*b^(1/3)*(c + d*x)^(1/3))) - (3^(1/4)*(b*c - a*d)^(1/3)*(c + d*x)^(1/ 
6)*((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))/((b* 
c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))^2]*EllipticE[ArcCo 
s[((b*c - a*d)^(1/3) - (1 - Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))/((b*c - a*d) 
^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))], (2 + Sqrt[3])/4])/(Sqrt[ 
-((b^(1/3)*(c + d*x)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)))/ 
((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))^2)]*Sqrt[a - ( 
b*c)/d + (b*(c + d*x))/d]))/(2*b^(2/3)) - ((1 - Sqrt[3])*(b*c - a*d)^(1/3) 
*(c + d*x)^(1/6)*((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))/((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))^2]*E 
llipticF[ArcCos[((b*c - a*d)^(1/3) - (1 - Sqrt[3])*b^(1/3)*(c + d*x)^(1/3) 
)/((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))], (2 + Sqrt[ 
3])/4])/(4*3^(1/4)*b^(2/3)*Sqrt[-((b^(1/3)*(c + d*x)^(1/3)*((b*c - a*d)^(1 
/3) - b^(1/3)*(c + d*x)^(1/3)))/((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/3) 
*(c + d*x)^(1/3))^2)]*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])))/b
 

Defintions of rubi rules used

rule 25

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

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[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 766

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ 
(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x 
]
 

rule 837

Int[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 
3]], s = Denom[Rt[b/a, 3]]}, Simp[(Sqrt[3] - 1)*(s^2/(2*r^2))   Int[1/Sqrt[ 
a + b*x^6], x], x] - Simp[1/(2*r^2)   Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4)/S 
qrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, x]
 

rule 2420

Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = 
 Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(1 + Sqrt[3])*d*s^3*x*(Sqr 
t[a + b*x^6]/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2))), x] - Simp[3^(1/4)*d*s*x* 
(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2 
*r^2*Sqrt[(r*x^2*(s + r*x^2))/(s + (1 + Sqrt[3])*r*x^2)^2]*Sqrt[a + b*x^6]) 
)*EllipticE[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 
 + Sqrt[3])/4], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 
- Sqrt[3])*d, 0]
 

Maple [F]

\[\int \frac {\left (x d +c \right )^{\frac {5}{6}}}{\left (b x +a \right )^{\frac {3}{2}}}d x\]

Input:

int((d*x+c)^(5/6)/(b*x+a)^(3/2),x)
 

Output:

int((d*x+c)^(5/6)/(b*x+a)^(3/2),x)
 

Fricas [F]

\[ \int \frac {(c+d x)^{5/6}}{(a+b x)^{3/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{6}}}{{\left (b x + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((d*x+c)^(5/6)/(b*x+a)^(3/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*x + a)*(d*x + c)^(5/6)/(b^2*x^2 + 2*a*b*x + a^2), x)
 

Sympy [F]

\[ \int \frac {(c+d x)^{5/6}}{(a+b x)^{3/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{6}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((d*x+c)**(5/6)/(b*x+a)**(3/2),x)
 

Output:

Integral((c + d*x)**(5/6)/(a + b*x)**(3/2), x)
 

Maxima [F]

\[ \int \frac {(c+d x)^{5/6}}{(a+b x)^{3/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{6}}}{{\left (b x + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((d*x+c)^(5/6)/(b*x+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((d*x + c)^(5/6)/(b*x + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {(c+d x)^{5/6}}{(a+b x)^{3/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{6}}}{{\left (b x + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((d*x+c)^(5/6)/(b*x+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((d*x + c)^(5/6)/(b*x + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/6}}{(a+b x)^{3/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/6}}{{\left (a+b\,x\right )}^{3/2}} \,d x \] Input:

int((c + d*x)^(5/6)/(a + b*x)^(3/2),x)
 

Output:

int((c + d*x)^(5/6)/(a + b*x)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.06 \[ \int \frac {(c+d x)^{5/6}}{(a+b x)^{3/2}} \, dx=\frac {2 \sqrt {d x +c}\, \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{\left (d x +c \right )^{\frac {2}{3}} \sqrt {b x +a}\, \left (a d -b c \right )} \] Input:

int((d*x+c)^(5/6)/(b*x+a)^(3/2),x)
 

Output:

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