\(\int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx\) [458]

Optimal result

Integrand size = 19, antiderivative size = 144 \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=-\frac {8 (b c-a d) \sqrt {a+b x} \sqrt [4]{c+d x}}{7 d^2}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}+\frac {16 (b c-a d)^{9/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{7 \sqrt [4]{b} d^3 \sqrt {a+b x}} \] Output:

-8/7*(-a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/4)/d^2+4/7*(b*x+a)^(3/2)*(d*x+c)^ 
(1/4)/d+16/7*(-a*d+b*c)^(9/4)*(-d*(b*x+a)/(-a*d+b*c))^(1/2)*EllipticF(b^(1 
/4)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4),I)/b^(1/4)/d^3/(b*x+a)^(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 = 73, normalized size of antiderivative = 0.51 \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\frac {2 (a+b x)^{5/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{2},\frac {d (a+b x)}{-b c+a d}\right )}{5 b (c+d x)^{3/4}} \] Input:

Integrate[(a + b*x)^(3/2)/(c + d*x)^(3/4),x]
 

Output:

(2*(a + b*x)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[3/4 
, 5/2, 7/2, (d*(a + b*x))/(-(b*c) + a*d)])/(5*b*(c + d*x)^(3/4))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {60, 60, 73, 765, 762}

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.

\(\displaystyle \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}}dx}{7 d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {2 (b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{3 d}\right )}{7 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{3 d^2}\right )}{7 d}\)

\(\Big \downarrow \) 765

\(\displaystyle \frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{3 d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{7 d}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d)^{5/4} \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 \sqrt [4]{b} d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{7 d}\)

Input:

Int[(a + b*x)^(3/2)/(c + d*x)^(3/4),x]
 

Output:

(4*(a + b*x)^(3/2)*(c + d*x)^(1/4))/(7*d) - (6*(b*c - a*d)*((4*Sqrt[a + b* 
x]*(c + d*x)^(1/4))/(3*d) - (8*(b*c - a*d)^(5/4)*Sqrt[1 - (b*(c + d*x))/(b 
*c - a*d)]*EllipticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], 
-1])/(3*b^(1/4)*d^2*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])))/(7*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 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 765

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt 
[a + b*x^4]   Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ 
[b/a] &&  !GtQ[a, 0]
 

Maple [F]

\[\int \frac {\left (b x +a \right )^{\frac {3}{2}}}{\left (x d +c \right )^{\frac {3}{4}}}d x\]

Input:

int((b*x+a)^(3/2)/(d*x+c)^(3/4),x)
 

Output:

int((b*x+a)^(3/2)/(d*x+c)^(3/4),x)
 

Fricas [F]

\[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((b*x+a)^(3/2)/(d*x+c)^(3/4),x, algorithm="fricas")
 

Output:

integral((b*x + a)^(3/2)/(d*x + c)^(3/4), x)
 

Sympy [F]

\[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\int \frac {\left (a + b x\right )^{\frac {3}{2}}}{\left (c + d x\right )^{\frac {3}{4}}}\, dx \] Input:

integrate((b*x+a)**(3/2)/(d*x+c)**(3/4),x)
 

Output:

Integral((a + b*x)**(3/2)/(c + d*x)**(3/4), x)
 

Maxima [F]

\[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((b*x+a)^(3/2)/(d*x+c)^(3/4),x, algorithm="maxima")
 

Output:

integrate((b*x + a)^(3/2)/(d*x + c)^(3/4), x)
 

Giac [F]

\[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((b*x+a)^(3/2)/(d*x+c)^(3/4),x, algorithm="giac")
 

Output:

integrate((b*x + a)^(3/2)/(d*x + c)^(3/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\int \frac {{\left (a+b\,x\right )}^{3/2}}{{\left (c+d\,x\right )}^{3/4}} \,d x \] Input:

int((a + b*x)^(3/2)/(c + d*x)^(3/4),x)
 

Output:

int((a + b*x)^(3/2)/(c + d*x)^(3/4), x)
 

Reduce [F]

\[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\left (\int \frac {\sqrt {b x +a}}{\left (d x +c \right )^{\frac {3}{4}}}d x \right ) a +\left (\int \frac {\sqrt {b x +a}\, x}{\left (d x +c \right )^{\frac {3}{4}}}d x \right ) b \] Input:

int((b*x+a)^(3/2)/(d*x+c)^(3/4),x)
 

Output:

int(sqrt(a + b*x)/(c + d*x)**(3/4),x)*a + int((sqrt(a + b*x)*x)/(c + d*x)* 
*(3/4),x)*b