\(\int \frac {(c+d x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{13/2}} \, dx\) [883]

Optimal result

Integrand size = 36, antiderivative size = 246 \[ \int \frac {(c+d x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{13/2}} \, dx=-\frac {2 c^2 \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b e (e x)^{11/2} (c+d x)^{3/2}}-\frac {82 c d \left (b c^2-b d^2 x^2\right )^{3/2}}{99 b e^2 (e x)^{9/2} (c+d x)^{3/2}}-\frac {362 d^2 \left (b c^2-b d^2 x^2\right )^{3/2}}{231 b e^3 (e x)^{7/2} (c+d x)^{3/2}}-\frac {382 d^3 \left (b c^2-b d^2 x^2\right )^{3/2}}{231 b c e^4 (e x)^{5/2} (c+d x)^{3/2}}-\frac {764 d^4 \left (b c^2-b d^2 x^2\right )^{3/2}}{693 b c^2 e^5 (e x)^{3/2} (c+d x)^{3/2}} \] Output:

-2/11*c^2*(-b*d^2*x^2+b*c^2)^(3/2)/b/e/(e*x)^(11/2)/(d*x+c)^(3/2)-82/99*c* 
d*(-b*d^2*x^2+b*c^2)^(3/2)/b/e^2/(e*x)^(9/2)/(d*x+c)^(3/2)-362/231*d^2*(-b 
*d^2*x^2+b*c^2)^(3/2)/b/e^3/(e*x)^(7/2)/(d*x+c)^(3/2)-382/231*d^3*(-b*d^2* 
x^2+b*c^2)^(3/2)/b/c/e^4/(e*x)^(5/2)/(d*x+c)^(3/2)-764/693*d^4*(-b*d^2*x^2 
+b*c^2)^(3/2)/b/c^2/e^5/(e*x)^(3/2)/(d*x+c)^(3/2)
 

Mathematica [A] (verified)

Time = 7.36 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.40 \[ \int \frac {(c+d x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{13/2}} \, dx=-\frac {2 \sqrt {e x} \sqrt {b \left (c^2-d^2 x^2\right )} \left (63 c^5+224 c^4 d x+256 c^3 d^2 x^2+30 c^2 d^3 x^3-191 c d^4 x^4-382 d^5 x^5\right )}{693 c^2 e^7 x^6 \sqrt {c+d x}} \] Input:

Integrate[((c + d*x)^(5/2)*Sqrt[b*c^2 - b*d^2*x^2])/(e*x)^(13/2),x]
 

Output:

(-2*Sqrt[e*x]*Sqrt[b*(c^2 - d^2*x^2)]*(63*c^5 + 224*c^4*d*x + 256*c^3*d^2* 
x^2 + 30*c^2*d^3*x^3 - 191*c*d^4*x^4 - 382*d^5*x^5))/(693*c^2*e^7*x^6*Sqrt 
[c + d*x])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {586, 107, 105, 100, 27, 87, 48}

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.

\(\displaystyle \int \frac {(c+d x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{13/2}} \, dx\)

\(\Big \downarrow \) 586

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

\(\Big \downarrow \) 107

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {5 d \int \frac {(c+d x)^3 \sqrt {b c-b d x}}{(e x)^{11/2}}dx}{11 c e}-\frac {2 (c+d x)^4 (b c-b d x)^{3/2}}{11 b c^2 e (e x)^{11/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 105

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {5 d \left (\frac {4 d \int \frac {(c+d x)^2 \sqrt {b c-b d x}}{(e x)^{9/2}}dx}{3 e}-\frac {2 (c+d x)^3 (b c-b d x)^{3/2}}{9 b c e (e x)^{9/2}}\right )}{11 c e}-\frac {2 (c+d x)^4 (b c-b d x)^{3/2}}{11 b c^2 e (e x)^{11/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 100

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {5 d \left (\frac {4 d \left (\frac {2 \int \frac {b c d e^2 (18 c+7 d x) \sqrt {b c-b d x}}{2 (e x)^{7/2}}dx}{7 b c e^3}-\frac {2 c (b c-b d x)^{3/2}}{7 b e (e x)^{7/2}}\right )}{3 e}-\frac {2 (c+d x)^3 (b c-b d x)^{3/2}}{9 b c e (e x)^{9/2}}\right )}{11 c e}-\frac {2 (c+d x)^4 (b c-b d x)^{3/2}}{11 b c^2 e (e x)^{11/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {5 d \left (\frac {4 d \left (\frac {d \int \frac {(18 c+7 d x) \sqrt {b c-b d x}}{(e x)^{7/2}}dx}{7 e}-\frac {2 c (b c-b d x)^{3/2}}{7 b e (e x)^{7/2}}\right )}{3 e}-\frac {2 (c+d x)^3 (b c-b d x)^{3/2}}{9 b c e (e x)^{9/2}}\right )}{11 c e}-\frac {2 (c+d x)^4 (b c-b d x)^{3/2}}{11 b c^2 e (e x)^{11/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {5 d \left (\frac {4 d \left (\frac {d \left (\frac {71 d \int \frac {\sqrt {b c-b d x}}{(e x)^{5/2}}dx}{5 e}-\frac {36 (b c-b d x)^{3/2}}{5 b e (e x)^{5/2}}\right )}{7 e}-\frac {2 c (b c-b d x)^{3/2}}{7 b e (e x)^{7/2}}\right )}{3 e}-\frac {2 (c+d x)^3 (b c-b d x)^{3/2}}{9 b c e (e x)^{9/2}}\right )}{11 c e}-\frac {2 (c+d x)^4 (b c-b d x)^{3/2}}{11 b c^2 e (e x)^{11/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {5 d \left (\frac {4 d \left (\frac {d \left (-\frac {142 d (b c-b d x)^{3/2}}{15 b c e^2 (e x)^{3/2}}-\frac {36 (b c-b d x)^{3/2}}{5 b e (e x)^{5/2}}\right )}{7 e}-\frac {2 c (b c-b d x)^{3/2}}{7 b e (e x)^{7/2}}\right )}{3 e}-\frac {2 (c+d x)^3 (b c-b d x)^{3/2}}{9 b c e (e x)^{9/2}}\right )}{11 c e}-\frac {2 (c+d x)^4 (b c-b d x)^{3/2}}{11 b c^2 e (e x)^{11/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

Input:

Int[((c + d*x)^(5/2)*Sqrt[b*c^2 - b*d^2*x^2])/(e*x)^(13/2),x]
 

Output:

(Sqrt[b*c^2 - b*d^2*x^2]*((-2*(c + d*x)^4*(b*c - b*d*x)^(3/2))/(11*b*c^2*e 
*(e*x)^(11/2)) + (5*d*((-2*(c + d*x)^3*(b*c - b*d*x)^(3/2))/(9*b*c*e*(e*x) 
^(9/2)) + (4*d*((-2*c*(b*c - b*d*x)^(3/2))/(7*b*e*(e*x)^(7/2)) + (d*((-36* 
(b*c - b*d*x)^(3/2))/(5*b*e*(e*x)^(5/2)) - (142*d*(b*c - b*d*x)^(3/2))/(15 
*b*c*e^2*(e*x)^(3/2))))/(7*e)))/(3*e)))/(11*c*e)))/(Sqrt[c + d*x]*Sqrt[b*c 
 - b*d*x])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 48

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp 
[(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ 
a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 105

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

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

rule 586

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_) 
, x_Symbol] :> Simp[(a + b*x^2)^FracPart[p]/((c + d*x)^FracPart[p]*(a/c + ( 
b*x)/d)^FracPart[p])   Int[(e*x)^m*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0]
 

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.34

method	result	size
gosper	\(-\frac {2 \left (-d x +c \right ) x \left (382 d^{4} x^{4}+573 c \,d^{3} x^{3}+543 d^{2} c^{2} x^{2}+287 c^{3} d x +63 c^{4}\right ) \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}{693 c^{2} \sqrt {d x +c}\, \left (e x \right )^{\frac {13}{2}}}\)	\(83\)
orering	\(-\frac {2 \left (-d x +c \right ) x \left (382 d^{4} x^{4}+573 c \,d^{3} x^{3}+543 d^{2} c^{2} x^{2}+287 c^{3} d x +63 c^{4}\right ) \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}{693 c^{2} \sqrt {d x +c}\, \left (e x \right )^{\frac {13}{2}}}\)	\(83\)
default	\(-\frac {2 \sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (-d x +c \right ) \left (382 d^{4} x^{4}+573 c \,d^{3} x^{3}+543 d^{2} c^{2} x^{2}+287 c^{3} d x +63 c^{4}\right )}{693 \sqrt {d x +c}\, e^{6} x^{5} \sqrt {e x}\, c^{2}}\)	\(87\)

Input:

int((d*x+c)^(5/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(13/2),x,method=_RETURNVE 
RBOSE)
 

Output:

-2/693*(-d*x+c)*x*(382*d^4*x^4+573*c*d^3*x^3+543*c^2*d^2*x^2+287*c^3*d*x+6 
3*c^4)*(-b*d^2*x^2+b*c^2)^(1/2)/c^2/(d*x+c)^(1/2)/(e*x)^(13/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.43 \[ \int \frac {(c+d x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{13/2}} \, dx=\frac {2 \, {\left (382 \, d^{5} x^{5} + 191 \, c d^{4} x^{4} - 30 \, c^{2} d^{3} x^{3} - 256 \, c^{3} d^{2} x^{2} - 224 \, c^{4} d x - 63 \, c^{5}\right )} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} \sqrt {e x}}{693 \, {\left (c^{2} d e^{7} x^{7} + c^{3} e^{7} x^{6}\right )}} \] Input:

integrate((d*x+c)^(5/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(13/2),x, algorithm 
="fricas")
 

Output:

2/693*(382*d^5*x^5 + 191*c*d^4*x^4 - 30*c^2*d^3*x^3 - 256*c^3*d^2*x^2 - 22 
4*c^4*d*x - 63*c^5)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*sqrt(e*x)/(c^2* 
d*e^7*x^7 + c^3*e^7*x^6)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{13/2}} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**(5/2)*(-b*d**2*x**2+b*c**2)**(1/2)/(e*x)**(13/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c+d x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{13/2}} \, dx=\int { \frac {\sqrt {-b d^{2} x^{2} + b c^{2}} {\left (d x + c\right )}^{\frac {5}{2}}}{\left (e x\right )^{\frac {13}{2}}} \,d x } \] Input:

integrate((d*x+c)^(5/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(13/2),x, algorithm 
="maxima")
 

Output:

integrate(sqrt(-b*d^2*x^2 + b*c^2)*(d*x + c)^(5/2)/(e*x)^(13/2), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.67 \[ \int \frac {(c+d x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{13/2}} \, dx=\frac {2 \, {\left (1848 \, b^{9} c^{2} d^{11} e^{5} + {\left (4620 \, b^{8} c d^{11} e^{5} + {\left (4554 \, b^{7} d^{11} e^{5} + 191 \, {\left (\frac {11 \, b^{6} d^{11} e^{5}}{c} + \frac {2 \, {\left (b d x - b c\right )} b^{5} d^{11} e^{5}}{c^{2}}\right )} {\left (b d x - b c\right )}\right )} {\left (b d x - b c\right )}\right )} {\left (b d x - b c\right )}\right )} {\left (b d x - b c\right )} \sqrt {-b d x + b c} {\left | b \right |} {\left | d \right |}}{693 \, {\left (b^{2} c d e + {\left (b d x - b c\right )} b d e\right )}^{\frac {11}{2}} d e^{6}} \] Input:

integrate((d*x+c)^(5/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(13/2),x, algorithm 
="giac")
 

Output:

2/693*(1848*b^9*c^2*d^11*e^5 + (4620*b^8*c*d^11*e^5 + (4554*b^7*d^11*e^5 + 
 191*(11*b^6*d^11*e^5/c + 2*(b*d*x - b*c)*b^5*d^11*e^5/c^2)*(b*d*x - b*c)) 
*(b*d*x - b*c))*(b*d*x - b*c))*(b*d*x - b*c)*sqrt(-b*d*x + b*c)*abs(b)*abs 
(d)/((b^2*c*d*e + (b*d*x - b*c)*b*d*e)^(11/2)*d*e^6)
 

Mupad [B] (verification not implemented)

Time = 7.11 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.63 \[ \int \frac {(c+d x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{13/2}} \, dx=-\frac {\sqrt {b\,c^2-b\,d^2\,x^2}\,\left (\frac {2\,c^3\,\sqrt {c+d\,x}}{11\,d\,e^6}+\frac {20\,d^2\,x^3\,\sqrt {c+d\,x}}{231\,e^6}+\frac {64\,c^2\,x\,\sqrt {c+d\,x}}{99\,e^6}+\frac {512\,c\,d\,x^2\,\sqrt {c+d\,x}}{693\,e^6}-\frac {382\,d^3\,x^4\,\sqrt {c+d\,x}}{693\,c\,e^6}-\frac {764\,d^4\,x^5\,\sqrt {c+d\,x}}{693\,c^2\,e^6}\right )}{x^6\,\sqrt {e\,x}+\frac {c\,x^5\,\sqrt {e\,x}}{d}} \] Input:

int(((b*c^2 - b*d^2*x^2)^(1/2)*(c + d*x)^(5/2))/(e*x)^(13/2),x)
 

Output:

-((b*c^2 - b*d^2*x^2)^(1/2)*((2*c^3*(c + d*x)^(1/2))/(11*d*e^6) + (20*d^2* 
x^3*(c + d*x)^(1/2))/(231*e^6) + (64*c^2*x*(c + d*x)^(1/2))/(99*e^6) + (51 
2*c*d*x^2*(c + d*x)^(1/2))/(693*e^6) - (382*d^3*x^4*(c + d*x)^(1/2))/(693* 
c*e^6) - (764*d^4*x^5*(c + d*x)^(1/2))/(693*c^2*e^6)))/(x^6*(e*x)^(1/2) + 
(c*x^5*(e*x)^(1/2))/d)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.54 \[ \int \frac {(c+d x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{13/2}} \, dx=\frac {2 \sqrt {e}\, \sqrt {b}\, \left (-63 \sqrt {x}\, \sqrt {-d x +c}\, c^{5}-224 \sqrt {x}\, \sqrt {-d x +c}\, c^{4} d x -256 \sqrt {x}\, \sqrt {-d x +c}\, c^{3} d^{2} x^{2}-30 \sqrt {x}\, \sqrt {-d x +c}\, c^{2} d^{3} x^{3}+191 \sqrt {x}\, \sqrt {-d x +c}\, c \,d^{4} x^{4}+382 \sqrt {x}\, \sqrt {-d x +c}\, d^{5} x^{5}-382 \sqrt {d}\, d^{5} i \,x^{6}\right )}{693 c^{2} e^{7} x^{6}} \] Input:

int((d*x+c)^(5/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(13/2),x)
 

Output:

(2*sqrt(e)*sqrt(b)*( - 63*sqrt(x)*sqrt(c - d*x)*c**5 - 224*sqrt(x)*sqrt(c 
- d*x)*c**4*d*x - 256*sqrt(x)*sqrt(c - d*x)*c**3*d**2*x**2 - 30*sqrt(x)*sq 
rt(c - d*x)*c**2*d**3*x**3 + 191*sqrt(x)*sqrt(c - d*x)*c*d**4*x**4 + 382*s 
qrt(x)*sqrt(c - d*x)*d**5*x**5 - 382*sqrt(d)*d**5*i*x**6))/(693*c**2*e**7* 
x**6)