\(\int \frac {(e x)^m (c+d x)^3}{(a+b x^2)^{3/2}} \, dx\) [173]

Optimal result

Integrand size = 24, antiderivative size = 251 \[ \int \frac {(e x)^m (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {3 c d^2 (e x)^{1+m}}{b e m \sqrt {a+b x^2}}+\frac {d^3 (e x)^{2+m}}{b e^2 (1+m) \sqrt {a+b x^2}}+\frac {c \left (b c^2 m-3 a d^2 (1+m)\right ) (e x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a b e m (1+m) \sqrt {a+b x^2}}+\frac {d \left (\frac {3 c^2}{a (2+m)}-\frac {d^2}{b+b m}\right ) (e x)^{2+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{e^2 \sqrt {a+b x^2}} \] Output:

3*c*d^2*(e*x)^(1+m)/b/e/m/(b*x^2+a)^(1/2)+d^3*(e*x)^(2+m)/b/e^2/(1+m)/(b*x 
^2+a)^(1/2)+c*(b*c^2*m-3*a*d^2*(1+m))*(e*x)^(1+m)*(1+b*x^2/a)^(1/2)*hyperg 
eom([3/2, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/a/b/e/m/(1+m)/(b*x^2+a)^(1/2)+d 
*(3*c^2/a/(2+m)-d^2/(b*m+b))*(e*x)^(2+m)*(1+b*x^2/a)^(1/2)*hypergeom([3/2, 
 1+1/2*m],[2+1/2*m],-b*x^2/a)/e^2/(b*x^2+a)^(1/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.75 \[ \int \frac {(e x)^m (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {x (e x)^m \sqrt {1+\frac {b x^2}{a}} \left (\frac {c^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{1+m}+d x \left (\frac {3 c^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{2+m}+d x \left (\frac {3 c \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},-\frac {b x^2}{a}\right )}{3+m}+\frac {d x \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {4+m}{2},\frac {6+m}{2},-\frac {b x^2}{a}\right )}{4+m}\right )\right )\right )}{a \sqrt {a+b x^2}} \] Input:

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

Output:

(x*(e*x)^m*Sqrt[1 + (b*x^2)/a]*((c^3*Hypergeometric2F1[3/2, (1 + m)/2, (3 
+ m)/2, -((b*x^2)/a)])/(1 + m) + d*x*((3*c^2*Hypergeometric2F1[3/2, (2 + m 
)/2, (4 + m)/2, -((b*x^2)/a)])/(2 + m) + d*x*((3*c*Hypergeometric2F1[3/2, 
(3 + m)/2, (5 + m)/2, -((b*x^2)/a)])/(3 + m) + (d*x*Hypergeometric2F1[3/2, 
 (4 + m)/2, (6 + m)/2, -((b*x^2)/a)])/(4 + m)))))/(a*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {558, 27, 557, 279, 278}

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.

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

\(\Big \downarrow \) 558

\(\displaystyle \frac {(e x)^{m+1} \left (d x \left (3 c^2-\frac {a d^2}{b}\right )+c \left (c^2-\frac {3 a d^2}{b}\right )\right )}{a e \sqrt {a+b x^2}}-\frac {\int \frac {(e x)^m \left (b c \left (c^2 m-\frac {3 a d^2 (m+1)}{b}\right )+d \left (3 b c^2 (m+1)-a d^2 (m+2)\right ) x\right )}{b \sqrt {b x^2+a}}dx}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(e x)^{m+1} \left (d x \left (3 c^2-\frac {a d^2}{b}\right )+c \left (c^2-\frac {3 a d^2}{b}\right )\right )}{a e \sqrt {a+b x^2}}-\frac {\int \frac {(e x)^m \left (c \left (b c^2 m-3 a d^2 (m+1)\right )+d \left (3 b c^2 (m+1)-a d^2 (m+2)\right ) x\right )}{\sqrt {b x^2+a}}dx}{a b}\)

\(\Big \downarrow \) 557

\(\displaystyle \frac {(e x)^{m+1} \left (d x \left (3 c^2-\frac {a d^2}{b}\right )+c \left (c^2-\frac {3 a d^2}{b}\right )\right )}{a e \sqrt {a+b x^2}}-\frac {c \left (b c^2 m-3 a d^2 (m+1)\right ) \int \frac {(e x)^m}{\sqrt {b x^2+a}}dx+\frac {d \left (3 b c^2 (m+1)-a d^2 (m+2)\right ) \int \frac {(e x)^{m+1}}{\sqrt {b x^2+a}}dx}{e}}{a b}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {(e x)^{m+1} \left (d x \left (3 c^2-\frac {a d^2}{b}\right )+c \left (c^2-\frac {3 a d^2}{b}\right )\right )}{a e \sqrt {a+b x^2}}-\frac {\frac {c \sqrt {\frac {b x^2}{a}+1} \left (b c^2 m-3 a d^2 (m+1)\right ) \int \frac {(e x)^m}{\sqrt {\frac {b x^2}{a}+1}}dx}{\sqrt {a+b x^2}}+\frac {d \sqrt {\frac {b x^2}{a}+1} \left (3 b c^2 (m+1)-a d^2 (m+2)\right ) \int \frac {(e x)^{m+1}}{\sqrt {\frac {b x^2}{a}+1}}dx}{e \sqrt {a+b x^2}}}{a b}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {(e x)^{m+1} \left (d x \left (3 c^2-\frac {a d^2}{b}\right )+c \left (c^2-\frac {3 a d^2}{b}\right )\right )}{a e \sqrt {a+b x^2}}-\frac {\frac {d \sqrt {\frac {b x^2}{a}+1} (e x)^{m+2} \left (3 b c^2 (m+1)-a d^2 (m+2)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-\frac {b x^2}{a}\right )}{e^2 (m+2) \sqrt {a+b x^2}}+\frac {c \sqrt {\frac {b x^2}{a}+1} (e x)^{m+1} \left (b c^2 m-3 a d^2 (m+1)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{e (m+1) \sqrt {a+b x^2}}}{a b}\)

Input:

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

Output:

((e*x)^(1 + m)*(c*(c^2 - (3*a*d^2)/b) + d*(3*c^2 - (a*d^2)/b)*x))/(a*e*Sqr 
t[a + b*x^2]) - ((c*(b*c^2*m - 3*a*d^2*(1 + m))*(e*x)^(1 + m)*Sqrt[1 + (b* 
x^2)/a]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(e*(1 
+ m)*Sqrt[a + b*x^2]) + (d*(3*b*c^2*(1 + m) - a*d^2*(2 + m))*(e*x)^(2 + m) 
*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -((b*x^2 
)/a)])/(e^2*(2 + m)*Sqrt[a + b*x^2]))/(a*b)
 

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 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 557

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

rule 558

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, a + b*x^2, x], f = 
 Coeff[PolynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 0], g = Coeff[Pol 
ynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(-(e*x)^(m + 1))* 
(f + g*x)*((a + b*x^2)^(p + 1)/(2*a*e*(p + 1))), x] + Simp[1/(2*a*(p + 1)) 
  Int[(e*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + f*(m + 2*p + 
 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, m}, x] && IGt 
Q[n, 1] &&  !IntegerQ[m] && LtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

e**m*(int(x**m/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*c**3 + in 
t((x**m*x**3)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*d**3 + 3*i 
nt((x**m*x**2)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*c*d**2 + 
3*int((x**m*x)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*c**2*d)