\(\int \frac {(c+d x)^{5/2} (a+b x^2)}{x^2} \, dx\) [485]

Optimal result

Integrand size = 20, antiderivative size = 90 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^2} \, dx=4 a c d \sqrt {c+d x}-\frac {a c^2 \sqrt {c+d x}}{x}+\frac {2}{3} a d (c+d x)^{3/2}+\frac {2 b (c+d x)^{7/2}}{7 d}-5 a c^{3/2} d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \] Output:

4*a*c*d*(d*x+c)^(1/2)-a*c^2*(d*x+c)^(1/2)/x+2/3*a*d*(d*x+c)^(3/2)+2/7*b*(d 
*x+c)^(7/2)/d-5*a*c^(3/2)*d*arctanh((d*x+c)^(1/2)/c^(1/2))
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.83 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^2} \, dx=\frac {1}{21} \sqrt {c+d x} \left (\frac {6 b (c+d x)^3}{d}+7 a \left (14 c d-\frac {3 c^2}{x}+2 d^2 x\right )\right )-5 a c^{3/2} d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \] Input:

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

Output:

(Sqrt[c + d*x]*((6*b*(c + d*x)^3)/d + 7*a*(14*c*d - (3*c^2)/x + 2*d^2*x))) 
/21 - 5*a*c^(3/2)*d*ArcTanh[Sqrt[c + d*x]/Sqrt[c]]
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {517, 1580, 25, 2341, 2009}

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.

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 1580

\(\displaystyle \frac {2 \left (\frac {1}{2} \int \frac {2 b (c+d x)^4-2 b c (c+d x)^3+2 a d^2 (c+d x)^2+2 a c d^2 (c+d x)+a c^2 d^2}{d x}d\sqrt {c+d x}-\frac {a c^2 d \sqrt {c+d x}}{2 x}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (-\frac {1}{2} \int -\frac {2 b (c+d x)^4-2 b c (c+d x)^3+2 a d^2 (c+d x)^2+2 a c d^2 (c+d x)+a c^2 d^2}{d x}d\sqrt {c+d x}-\frac {a c^2 d \sqrt {c+d x}}{2 x}\right )}{d}\)

\(\Big \downarrow \) 2341

\(\displaystyle \frac {2 \left (-\frac {1}{2} \int \left (-2 b (c+d x)^3-2 a d^2 (c+d x)-4 a c d^2-\frac {5 a c^2 d}{x}\right )d\sqrt {c+d x}-\frac {a c^2 d \sqrt {c+d x}}{2 x}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (\frac {1}{2} \left (-5 a c^{3/2} d^2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+\frac {2}{3} a d^2 (c+d x)^{3/2}+4 a c d^2 \sqrt {c+d x}+\frac {2}{7} b (c+d x)^{7/2}\right )-\frac {a c^2 d \sqrt {c+d x}}{2 x}\right )}{d}\)

Input:

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

Output:

(2*(-1/2*(a*c^2*d*Sqrt[c + d*x])/x + (4*a*c*d^2*Sqrt[c + d*x] + (2*a*d^2*( 
c + d*x)^(3/2))/3 + (2*b*(c + d*x)^(7/2))/7 - 5*a*c^(3/2)*d^2*ArcTanh[Sqrt 
[c + d*x]/Sqrt[c]])/2))/d
 

Defintions of rubi rules used

rule 25

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

rule 517

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1))   Subst[Int[x^(2*n + 1)*(-c + x^ 
2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr 
eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
 

rule 1580

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_) 
^4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d 
 + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[1/(2*e^(2*p + m/2)* 
(q + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2* 
e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b 
*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b, c, d, e 
}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2341

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.90

method	result	size
risch	\(-\frac {a \,c^{2} \sqrt {d x +c}}{x}+\frac {\frac {2 b \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 a \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+4 a c \,d^{2} \sqrt {d x +c}-5 a \,c^{\frac {3}{2}} d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{d}\)	\(81\)
derivativedivides	\(\frac {\frac {2 b \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 a \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+4 a c \,d^{2} \sqrt {d x +c}-2 a \,c^{2} d^{2} \left (\frac {\sqrt {d x +c}}{2 d x}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{d}\)	\(86\)
default	\(\frac {\frac {2 b \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 a \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+4 a c \,d^{2} \sqrt {d x +c}-2 a \,c^{2} d^{2} \left (\frac {\sqrt {d x +c}}{2 d x}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{d}\)	\(86\)
pseudoelliptic	\(\frac {-5 \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right ) a \,c^{2} d^{2} x +\frac {2 \sqrt {d x +c}\, \left (7 d^{2} x \left (\frac {9 b \,x^{2}}{49}+a \right ) c^{\frac {3}{2}}-\frac {3 d \left (-\frac {6 b \,x^{2}}{7}+a \right ) c^{\frac {5}{2}}}{2}+x \left (\frac {3 b \,c^{\frac {7}{2}}}{7}+d^{3} x \sqrt {c}\, \left (\frac {3 b \,x^{2}}{7}+a \right )\right )\right )}{3}}{d x \sqrt {c}}\)	\(100\)

Input:

int((d*x+c)^(5/2)*(b*x^2+a)/x^2,x,method=_RETURNVERBOSE)
 

Output:

-a*c^2*(d*x+c)^(1/2)/x+1/d*(2/7*b*(d*x+c)^(7/2)+2/3*a*d^2*(d*x+c)^(3/2)+4* 
a*c*d^2*(d*x+c)^(1/2)-5*a*c^(3/2)*d^2*arctanh((d*x+c)^(1/2)/c^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.48 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^2} \, dx=\left [\frac {105 \, a c^{\frac {3}{2}} d^{2} x \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (6 \, b d^{3} x^{4} + 18 \, b c d^{2} x^{3} - 21 \, a c^{2} d + 2 \, {\left (9 \, b c^{2} d + 7 \, a d^{3}\right )} x^{2} + 2 \, {\left (3 \, b c^{3} + 49 \, a c d^{2}\right )} x\right )} \sqrt {d x + c}}{42 \, d x}, \frac {105 \, a \sqrt {-c} c d^{2} x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (6 \, b d^{3} x^{4} + 18 \, b c d^{2} x^{3} - 21 \, a c^{2} d + 2 \, {\left (9 \, b c^{2} d + 7 \, a d^{3}\right )} x^{2} + 2 \, {\left (3 \, b c^{3} + 49 \, a c d^{2}\right )} x\right )} \sqrt {d x + c}}{21 \, d x}\right ] \] Input:

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

Output:

[1/42*(105*a*c^(3/2)*d^2*x*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 
2*(6*b*d^3*x^4 + 18*b*c*d^2*x^3 - 21*a*c^2*d + 2*(9*b*c^2*d + 7*a*d^3)*x^2 
 + 2*(3*b*c^3 + 49*a*c*d^2)*x)*sqrt(d*x + c))/(d*x), 1/21*(105*a*sqrt(-c)* 
c*d^2*x*arctan(sqrt(-c)/sqrt(d*x + c)) + (6*b*d^3*x^4 + 18*b*c*d^2*x^3 - 2 
1*a*c^2*d + 2*(9*b*c^2*d + 7*a*d^3)*x^2 + 2*(3*b*c^3 + 49*a*c*d^2)*x)*sqrt 
(d*x + c))/(d*x)]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 17.73 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.99 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^2} \, dx=- a c^{\frac {3}{2}} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )} - \frac {a c^{2} \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{\sqrt {x}} + 2 a c d \left (\begin {cases} \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 \sqrt {c + d x} & \text {for}\: d \neq 0 \\\sqrt {c} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + a d^{2} \left (\begin {cases} \frac {2 \left (c + d x\right )^{\frac {3}{2}}}{3 d} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) + b c^{2} \left (\begin {cases} \frac {2 \left (c + d x\right )^{\frac {3}{2}}}{3 d} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) + 2 b c d \left (\begin {cases} - \frac {2 c \left (c + d x\right )^{\frac {3}{2}}}{3 d^{2}} + \frac {2 \left (c + d x\right )^{\frac {5}{2}}}{5 d^{2}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + b d^{2} \left (\begin {cases} \frac {2 c^{2} \left (c + d x\right )^{\frac {3}{2}}}{3 d^{3}} - \frac {4 c \left (c + d x\right )^{\frac {5}{2}}}{5 d^{3}} + \frac {2 \left (c + d x\right )^{\frac {7}{2}}}{7 d^{3}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

-a*c**(3/2)*d*asinh(sqrt(c)/(sqrt(d)*sqrt(x))) - a*c**2*sqrt(d)*sqrt(c/(d* 
x) + 1)/sqrt(x) + 2*a*c*d*Piecewise((2*c*atan(sqrt(c + d*x)/sqrt(-c))/sqrt 
(-c) + 2*sqrt(c + d*x), Ne(d, 0)), (sqrt(c)*log(x), True)) + a*d**2*Piecew 
ise((2*(c + d*x)**(3/2)/(3*d), Ne(d, 0)), (sqrt(c)*x, True)) + b*c**2*Piec 
ewise((2*(c + d*x)**(3/2)/(3*d), Ne(d, 0)), (sqrt(c)*x, True)) + 2*b*c*d*P 
iecewise((-2*c*(c + d*x)**(3/2)/(3*d**2) + 2*(c + d*x)**(5/2)/(5*d**2), Ne 
(d, 0)), (sqrt(c)*x**2/2, True)) + b*d**2*Piecewise((2*c**2*(c + d*x)**(3/ 
2)/(3*d**3) - 4*c*(c + d*x)**(5/2)/(5*d**3) + 2*(c + d*x)**(7/2)/(7*d**3), 
 Ne(d, 0)), (sqrt(c)*x**3/3, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^2} \, dx=\frac {1}{42} \, {\left (105 \, a c^{\frac {3}{2}} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right ) - \frac {42 \, \sqrt {d x + c} a c^{2}}{d x} + \frac {4 \, {\left (3 \, {\left (d x + c\right )}^{\frac {7}{2}} b + 7 \, {\left (d x + c\right )}^{\frac {3}{2}} a d^{2} + 42 \, \sqrt {d x + c} a c d^{2}\right )}}{d^{2}}\right )} d \] Input:

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

Output:

1/42*(105*a*c^(3/2)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c) 
)) - 42*sqrt(d*x + c)*a*c^2/(d*x) + 4*(3*(d*x + c)^(7/2)*b + 7*(d*x + c)^( 
3/2)*a*d^2 + 42*sqrt(d*x + c)*a*c*d^2)/d^2)*d
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.04 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^2} \, dx=\frac {1}{21} \, {\left (\frac {105 \, a c^{2} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {21 \, \sqrt {d x + c} a c^{2}}{d x} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {7}{2}} b d^{12} + 7 \, {\left (d x + c\right )}^{\frac {3}{2}} a d^{14} + 42 \, \sqrt {d x + c} a c d^{14}\right )}}{d^{14}}\right )} d \] Input:

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

Output:

1/21*(105*a*c^2*arctan(sqrt(d*x + c)/sqrt(-c))/sqrt(-c) - 21*sqrt(d*x + c) 
*a*c^2/(d*x) + 2*(3*(d*x + c)^(7/2)*b*d^12 + 7*(d*x + c)^(3/2)*a*d^14 + 42 
*sqrt(d*x + c)*a*c*d^14)/d^14)*d
 

Mupad [B] (verification not implemented)

Time = 7.76 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.39 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^2} \, dx=\left (\frac {2\,b\,c^2+2\,a\,d^2}{3\,d}-\frac {2\,b\,c^2}{3\,d}\right )\,{\left (c+d\,x\right )}^{3/2}+2\,c\,\left (\frac {2\,b\,c^2+2\,a\,d^2}{d}-\frac {2\,b\,c^2}{d}\right )\,\sqrt {c+d\,x}+\frac {2\,b\,{\left (c+d\,x\right )}^{7/2}}{7\,d}-\frac {a\,c^2\,\sqrt {c+d\,x}}{x}+a\,c^{3/2}\,d\,\mathrm {atan}\left (\frac {\sqrt {c+d\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,5{}\mathrm {i} \] Input:

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

Output:

((2*a*d^2 + 2*b*c^2)/(3*d) - (2*b*c^2)/(3*d))*(c + d*x)^(3/2) + 2*c*((2*a* 
d^2 + 2*b*c^2)/d - (2*b*c^2)/d)*(c + d*x)^(1/2) + (2*b*(c + d*x)^(7/2))/(7 
*d) - (a*c^2*(c + d*x)^(1/2))/x + a*c^(3/2)*d*atan(((c + d*x)^(1/2)*1i)/c^ 
(1/2))*5i
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.70 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^2} \, dx=\frac {-42 \sqrt {d x +c}\, a \,c^{2} d +196 \sqrt {d x +c}\, a c \,d^{2} x +28 \sqrt {d x +c}\, a \,d^{3} x^{2}+12 \sqrt {d x +c}\, b \,c^{3} x +36 \sqrt {d x +c}\, b \,c^{2} d \,x^{2}+36 \sqrt {d x +c}\, b c \,d^{2} x^{3}+12 \sqrt {d x +c}\, b \,d^{3} x^{4}+105 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a c \,d^{2} x -105 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a c \,d^{2} x}{42 d x} \] Input:

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

Output:

( - 42*sqrt(c + d*x)*a*c**2*d + 196*sqrt(c + d*x)*a*c*d**2*x + 28*sqrt(c + 
 d*x)*a*d**3*x**2 + 12*sqrt(c + d*x)*b*c**3*x + 36*sqrt(c + d*x)*b*c**2*d* 
x**2 + 36*sqrt(c + d*x)*b*c*d**2*x**3 + 12*sqrt(c + d*x)*b*d**3*x**4 + 105 
*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a*c*d**2*x - 105*sqrt(c)*log(sqrt(c 
+ d*x) + sqrt(c))*a*c*d**2*x)/(42*d*x)