\(\int \frac {(c+d x)^{5/2} (a+b x^2)}{x^4} \, dx\) [487]

Optimal result

Integrand size = 20, antiderivative size = 136 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^4} \, dx=4 b c d \sqrt {c+d x}-\frac {a c^2 \sqrt {c+d x}}{3 x^3}-\frac {13 a c d \sqrt {c+d x}}{12 x^2}-\frac {\left (8 b c^2+11 a d^2\right ) \sqrt {c+d x}}{8 x}+\frac {2}{3} b d (c+d x)^{3/2}-\frac {5 d \left (8 b c^2+a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 \sqrt {c}} \] Output:

4*b*c*d*(d*x+c)^(1/2)-1/3*a*c^2*(d*x+c)^(1/2)/x^3-13/12*a*c*d*(d*x+c)^(1/2 
)/x^2-1/8*(11*a*d^2+8*b*c^2)*(d*x+c)^(1/2)/x+2/3*b*d*(d*x+c)^(3/2)-5/8*d*( 
a*d^2+8*b*c^2)*arctanh((d*x+c)^(1/2)/c^(1/2))/c^(1/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.76 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^4} \, dx=\frac {\sqrt {c+d x} \left (8 b x^2 \left (-3 c^2+14 c d x+2 d^2 x^2\right )-a \left (8 c^2+26 c d x+33 d^2 x^2\right )\right )}{24 x^3}-\frac {5 d \left (8 b c^2+a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 \sqrt {c}} \] Input:

Integrate[((c + d*x)^(5/2)*(a + b*x^2))/x^4,x]
 

Output:

(Sqrt[c + d*x]*(8*b*x^2*(-3*c^2 + 14*c*d*x + 2*d^2*x^2) - a*(8*c^2 + 26*c* 
d*x + 33*d^2*x^2)))/(24*x^3) - (5*d*(8*b*c^2 + a*d^2)*ArcTanh[Sqrt[c + d*x 
]/Sqrt[c]])/(8*Sqrt[c])
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.22, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {517, 1580, 25, 2345, 27, 2345, 1467, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right ) (c+d x)^{5/2}}{x^4} \, dx\)

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 1580

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2345

\(\displaystyle 2 d \left (\frac {1}{6} \left (\frac {\int \frac {3 \left (8 b c (c+d x)^3+8 a c d^2 (c+d x)+3 a c^2 d^2\right )}{d^2 x^2}d\sqrt {c+d x}}{4 c}-\frac {13 a c \sqrt {c+d x}}{4 x^2}\right )-\frac {a c^2 \sqrt {c+d x}}{6 d x^3}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 d \left (\frac {1}{6} \left (\frac {3 \int \frac {8 b c (c+d x)^3+8 a c d^2 (c+d x)+3 a c^2 d^2}{d^2 x^2}d\sqrt {c+d x}}{4 c}-\frac {13 a c \sqrt {c+d x}}{4 x^2}\right )-\frac {a c^2 \sqrt {c+d x}}{6 d x^3}\right )\)

\(\Big \downarrow \) 2345

\(\displaystyle 2 d \left (\frac {1}{6} \left (\frac {3 \left (-\frac {\int -\frac {16 b (c+d x) c^3+16 b (c+d x)^2 c^2+\left (8 b c^2+5 a d^2\right ) c^2}{d x}d\sqrt {c+d x}}{2 c}-\frac {c \sqrt {c+d x} \left (11 a d^2+8 b c^2\right )}{2 d x}\right )}{4 c}-\frac {13 a c \sqrt {c+d x}}{4 x^2}\right )-\frac {a c^2 \sqrt {c+d x}}{6 d x^3}\right )\)

\(\Big \downarrow \) 1467

\(\displaystyle 2 d \left (\frac {1}{6} \left (\frac {3 \left (-\frac {\int \left (-32 b c^3-16 b (c+d x) c^2-\frac {5 \left (8 b c^4+a d^2 c^2\right )}{d x}\right )d\sqrt {c+d x}}{2 c}-\frac {c \sqrt {c+d x} \left (11 a d^2+8 b c^2\right )}{2 d x}\right )}{4 c}-\frac {13 a c \sqrt {c+d x}}{4 x^2}\right )-\frac {a c^2 \sqrt {c+d x}}{6 d x^3}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 d \left (\frac {1}{6} \left (\frac {3 \left (-\frac {5 c^{3/2} \left (a d^2+8 b c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )-32 b c^3 \sqrt {c+d x}-\frac {16}{3} b c^2 (c+d x)^{3/2}}{2 c}-\frac {c \sqrt {c+d x} \left (11 a d^2+8 b c^2\right )}{2 d x}\right )}{4 c}-\frac {13 a c \sqrt {c+d x}}{4 x^2}\right )-\frac {a c^2 \sqrt {c+d x}}{6 d x^3}\right )\)

Input:

Int[((c + d*x)^(5/2)*(a + b*x^2))/x^4,x]
 

Output:

2*d*(-1/6*(a*c^2*Sqrt[c + d*x])/(d*x^3) + ((-13*a*c*Sqrt[c + d*x])/(4*x^2) 
 + (3*(-1/2*(c*(8*b*c^2 + 11*a*d^2)*Sqrt[c + d*x])/(d*x) - (-32*b*c^3*Sqrt 
[c + d*x] - (16*b*c^2*(c + d*x)^(3/2))/3 + 5*c^(3/2)*(8*b*c^2 + a*d^2)*Arc 
Tanh[Sqrt[c + d*x]/Sqrt[c]])/(2*c)))/(4*c))/6)
 

Defintions of rubi rules used

rule 25

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

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 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 1467

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], 
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e 
 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -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 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(-\frac {11 \left (\frac {5 d \,x^{3} \left (a \,d^{2}+8 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{11}+\left (\frac {26 d x \left (-\frac {56 b \,x^{2}}{13}+a \right ) c^{\frac {3}{2}}}{33}+\frac {8 \left (b \,x^{2}+\frac {a}{3}\right ) c^{\frac {5}{2}}}{11}+d^{2} x^{2} \sqrt {c}\, \left (-\frac {16 b \,x^{2}}{33}+a \right )\right ) \sqrt {d x +c}\right )}{8 \sqrt {c}\, x^{3}}\)	\(96\)
risch	\(-\frac {\sqrt {d x +c}\, \left (33 a \,d^{2} x^{2}+24 b \,c^{2} x^{2}+26 a d x c +8 a \,c^{2}\right )}{24 x^{3}}+\frac {d \left (\frac {32 b \left (d x +c \right )^{\frac {3}{2}}}{3}+64 b c \sqrt {d x +c}-\frac {2 \left (5 a \,d^{2}+40 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{16}\)	\(100\)
derivativedivides	\(2 d \left (\frac {b \left (d x +c \right )^{\frac {3}{2}}}{3}+2 b c \sqrt {d x +c}-\frac {\left (\frac {11 a \,d^{2}}{16}+\frac {b \,c^{2}}{2}\right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {5}{6} a \,d^{2} c -b \,c^{3}\right ) \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {1}{2} b \,c^{4}+\frac {5}{16} a \,c^{2} d^{2}\right ) \sqrt {d x +c}}{d^{3} x^{3}}-\frac {5 \left (a \,d^{2}+8 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}\right )\)	\(131\)
default	\(2 d \left (\frac {b \left (d x +c \right )^{\frac {3}{2}}}{3}+2 b c \sqrt {d x +c}-\frac {\left (\frac {11 a \,d^{2}}{16}+\frac {b \,c^{2}}{2}\right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {5}{6} a \,d^{2} c -b \,c^{3}\right ) \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {1}{2} b \,c^{4}+\frac {5}{16} a \,c^{2} d^{2}\right ) \sqrt {d x +c}}{d^{3} x^{3}}-\frac {5 \left (a \,d^{2}+8 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}\right )\)	\(131\)

Input:

int((d*x+c)^(5/2)*(b*x^2+a)/x^4,x,method=_RETURNVERBOSE)
 

Output:

-11/8/c^(1/2)*(5/11*d*x^3*(a*d^2+8*b*c^2)*arctanh((d*x+c)^(1/2)/c^(1/2))+( 
26/33*d*x*(-56/13*b*x^2+a)*c^(3/2)+8/11*(b*x^2+1/3*a)*c^(5/2)+d^2*x^2*c^(1 
/2)*(-16/33*b*x^2+a))*(d*x+c)^(1/2))/x^3
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.66 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^4} \, dx=\left [\frac {15 \, {\left (8 \, b c^{2} d + a d^{3}\right )} \sqrt {c} x^{3} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (16 \, b c d^{2} x^{4} + 112 \, b c^{2} d x^{3} - 26 \, a c^{2} d x - 8 \, a c^{3} - 3 \, {\left (8 \, b c^{3} + 11 \, a c d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{48 \, c x^{3}}, \frac {15 \, {\left (8 \, b c^{2} d + a d^{3}\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (16 \, b c d^{2} x^{4} + 112 \, b c^{2} d x^{3} - 26 \, a c^{2} d x - 8 \, a c^{3} - 3 \, {\left (8 \, b c^{3} + 11 \, a c d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{24 \, c x^{3}}\right ] \] Input:

integrate((d*x+c)^(5/2)*(b*x^2+a)/x^4,x, algorithm="fricas")
 

Output:

[1/48*(15*(8*b*c^2*d + a*d^3)*sqrt(c)*x^3*log((d*x - 2*sqrt(d*x + c)*sqrt( 
c) + 2*c)/x) + 2*(16*b*c*d^2*x^4 + 112*b*c^2*d*x^3 - 26*a*c^2*d*x - 8*a*c^ 
3 - 3*(8*b*c^3 + 11*a*c*d^2)*x^2)*sqrt(d*x + c))/(c*x^3), 1/24*(15*(8*b*c^ 
2*d + a*d^3)*sqrt(-c)*x^3*arctan(sqrt(-c)/sqrt(d*x + c)) + (16*b*c*d^2*x^4 
 + 112*b*c^2*d*x^3 - 26*a*c^2*d*x - 8*a*c^3 - 3*(8*b*c^3 + 11*a*c*d^2)*x^2 
)*sqrt(d*x + c))/(c*x^3)]
 

Sympy [A] (verification not implemented)

Time = 68.13 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.12 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^4} \, dx=- \frac {a c^{3}}{3 \sqrt {d} x^{\frac {7}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {17 a c^{2} \sqrt {d}}{12 x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {35 a c d^{\frac {3}{2}}}{24 x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {a d^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}}{\sqrt {x}} - \frac {3 a d^{\frac {5}{2}}}{8 \sqrt {x} \sqrt {\frac {c}{d x} + 1}} - \frac {5 a d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{8 \sqrt {c}} - b c^{\frac {3}{2}} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )} - \frac {b c^{2} \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{\sqrt {x}} + 2 b c d \left (\begin {cases} \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 \sqrt {c + d x} & \text {for}\: d \neq 0 \\\sqrt {c} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + b d^{2} \left (\begin {cases} \frac {2 \left (c + d x\right )^{\frac {3}{2}}}{3 d} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) \] Input:

integrate((d*x+c)**(5/2)*(b*x**2+a)/x**4,x)
 

Output:

-a*c**3/(3*sqrt(d)*x**(7/2)*sqrt(c/(d*x) + 1)) - 17*a*c**2*sqrt(d)/(12*x** 
(5/2)*sqrt(c/(d*x) + 1)) - 35*a*c*d**(3/2)/(24*x**(3/2)*sqrt(c/(d*x) + 1)) 
 - a*d**(5/2)*sqrt(c/(d*x) + 1)/sqrt(x) - 3*a*d**(5/2)/(8*sqrt(x)*sqrt(c/( 
d*x) + 1)) - 5*a*d**3*asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/(8*sqrt(c)) - b*c** 
(3/2)*d*asinh(sqrt(c)/(sqrt(d)*sqrt(x))) - b*c**2*sqrt(d)*sqrt(c/(d*x) + 1 
)/sqrt(x) + 2*b*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)) + b*d**2*Piecewise((2 
*(c + d*x)**(3/2)/(3*d), Ne(d, 0)), (sqrt(c)*x, True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.48 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^4} \, dx=-\frac {1}{48} \, d^{3} {\left (\frac {2 \, {\left (3 \, {\left (8 \, b c^{2} + 11 \, a d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} - 8 \, {\left (6 \, b c^{3} + 5 \, a c d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 3 \, {\left (8 \, b c^{4} + 5 \, a c^{2} d^{2}\right )} \sqrt {d x + c}\right )}}{{\left (d x + c\right )}^{3} d^{2} - 3 \, {\left (d x + c\right )}^{2} c d^{2} + 3 \, {\left (d x + c\right )} c^{2} d^{2} - c^{3} d^{2}} - \frac {15 \, {\left (8 \, b c^{2} + a d^{2}\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{\sqrt {c} d^{2}} - \frac {32 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b + 6 \, \sqrt {d x + c} b c\right )}}{d^{2}}\right )} \] Input:

integrate((d*x+c)^(5/2)*(b*x^2+a)/x^4,x, algorithm="maxima")
 

Output:

-1/48*d^3*(2*(3*(8*b*c^2 + 11*a*d^2)*(d*x + c)^(5/2) - 8*(6*b*c^3 + 5*a*c* 
d^2)*(d*x + c)^(3/2) + 3*(8*b*c^4 + 5*a*c^2*d^2)*sqrt(d*x + c))/((d*x + c) 
^3*d^2 - 3*(d*x + c)^2*c*d^2 + 3*(d*x + c)*c^2*d^2 - c^3*d^2) - 15*(8*b*c^ 
2 + a*d^2)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/(sqrt( 
c)*d^2) - 32*((d*x + c)^(3/2)*b + 6*sqrt(d*x + c)*b*c)/d^2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.21 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^4} \, dx=\frac {1}{24} \, d^{3} {\left (\frac {15 \, {\left (8 \, b c^{2} + a d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} d^{2}} + \frac {16 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b d^{4} + 6 \, \sqrt {d x + c} b c d^{4}\right )}}{d^{6}} - \frac {24 \, {\left (d x + c\right )}^{\frac {5}{2}} b c^{2} - 48 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{3} + 24 \, \sqrt {d x + c} b c^{4} + 33 \, {\left (d x + c\right )}^{\frac {5}{2}} a d^{2} - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} a c d^{2} + 15 \, \sqrt {d x + c} a c^{2} d^{2}}{d^{5} x^{3}}\right )} \] Input:

integrate((d*x+c)^(5/2)*(b*x^2+a)/x^4,x, algorithm="giac")
 

Output:

1/24*d^3*(15*(8*b*c^2 + a*d^2)*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c)*d^ 
2) + 16*((d*x + c)^(3/2)*b*d^4 + 6*sqrt(d*x + c)*b*c*d^4)/d^6 - (24*(d*x + 
 c)^(5/2)*b*c^2 - 48*(d*x + c)^(3/2)*b*c^3 + 24*sqrt(d*x + c)*b*c^4 + 33*( 
d*x + c)^(5/2)*a*d^2 - 40*(d*x + c)^(3/2)*a*c*d^2 + 15*sqrt(d*x + c)*a*c^2 
*d^2)/(d^5*x^3))
 

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.41 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^4} \, dx=\frac {\left (b\,c^4\,d+\frac {5\,a\,c^2\,d^3}{8}\right )\,\sqrt {c+d\,x}+\left (b\,c^2\,d+\frac {11\,a\,d^3}{8}\right )\,{\left (c+d\,x\right )}^{5/2}-\left (2\,b\,c^3\,d+\frac {5\,a\,c\,d^3}{3}\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,c\,{\left (c+d\,x\right )}^2-3\,c^2\,\left (c+d\,x\right )-{\left (c+d\,x\right )}^3+c^3}+\frac {2\,b\,d\,{\left (c+d\,x\right )}^{3/2}}{3}+4\,b\,c\,d\,\sqrt {c+d\,x}+\frac {d\,\mathrm {atan}\left (\frac {d\,\left (8\,b\,c^2+a\,d^2\right )\,\sqrt {c+d\,x}\,5{}\mathrm {i}}{4\,\sqrt {c}\,\left (10\,b\,c^2\,d+\frac {5\,a\,d^3}{4}\right )}\right )\,\left (8\,b\,c^2+a\,d^2\right )\,5{}\mathrm {i}}{8\,\sqrt {c}} \] Input:

int(((a + b*x^2)*(c + d*x)^(5/2))/x^4,x)
 

Output:

(((5*a*c^2*d^3)/8 + b*c^4*d)*(c + d*x)^(1/2) + ((11*a*d^3)/8 + b*c^2*d)*(c 
 + d*x)^(5/2) - ((5*a*c*d^3)/3 + 2*b*c^3*d)*(c + d*x)^(3/2))/(3*c*(c + d*x 
)^2 - 3*c^2*(c + d*x) - (c + d*x)^3 + c^3) + (2*b*d*(c + d*x)^(3/2))/3 + ( 
d*atan((d*(a*d^2 + 8*b*c^2)*(c + d*x)^(1/2)*5i)/(4*c^(1/2)*((5*a*d^3)/4 + 
10*b*c^2*d)))*(a*d^2 + 8*b*c^2)*5i)/(8*c^(1/2)) + 4*b*c*d*(c + d*x)^(1/2)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.38 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^4} \, dx=\frac {-16 \sqrt {d x +c}\, a \,c^{3}-52 \sqrt {d x +c}\, a \,c^{2} d x -66 \sqrt {d x +c}\, a c \,d^{2} x^{2}-48 \sqrt {d x +c}\, b \,c^{3} x^{2}+224 \sqrt {d x +c}\, b \,c^{2} d \,x^{3}+32 \sqrt {d x +c}\, b c \,d^{2} x^{4}+15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,d^{3} x^{3}+120 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b \,c^{2} d \,x^{3}-15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,d^{3} x^{3}-120 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b \,c^{2} d \,x^{3}}{48 c \,x^{3}} \] Input:

int((d*x+c)^(5/2)*(b*x^2+a)/x^4,x)
 

Output:

( - 16*sqrt(c + d*x)*a*c**3 - 52*sqrt(c + d*x)*a*c**2*d*x - 66*sqrt(c + d* 
x)*a*c*d**2*x**2 - 48*sqrt(c + d*x)*b*c**3*x**2 + 224*sqrt(c + d*x)*b*c**2 
*d*x**3 + 32*sqrt(c + d*x)*b*c*d**2*x**4 + 15*sqrt(c)*log(sqrt(c + d*x) - 
sqrt(c))*a*d**3*x**3 + 120*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*b*c**2*d*x 
**3 - 15*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*d**3*x**3 - 120*sqrt(c)*lo 
g(sqrt(c + d*x) + sqrt(c))*b*c**2*d*x**3)/(48*c*x**3)