Integrand size = 19, antiderivative size = 191 \[ \int (c+d x) \left (a x+b x^2\right )^{3/2} \, dx=-\frac {3 a^3 (2 b c-a d) \sqrt {a x+b x^2}}{128 b^3}+\frac {a^2 (2 b c-a d) x \sqrt {a x+b x^2}}{64 b^2}+\frac {3 a (2 b c-a d) x^2 \sqrt {a x+b x^2}}{16 b}+\frac {1}{8} (2 b c-a d) x^3 \sqrt {a x+b x^2}+\frac {d \left (a x+b x^2\right )^{5/2}}{5 b}+\frac {3 a^4 (2 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{128 b^{7/2}} \] Output:
-3/128*a^3*(-a*d+2*b*c)*(b*x^2+a*x)^(1/2)/b^3+1/64*a^2*(-a*d+2*b*c)*x*(b*x ^2+a*x)^(1/2)/b^2+3/16*a*(-a*d+2*b*c)*x^2*(b*x^2+a*x)^(1/2)/b+1/8*(-a*d+2* b*c)*x^3*(b*x^2+a*x)^(1/2)+1/5*d*(b*x^2+a*x)^(5/2)/b+3/128*a^4*(-a*d+2*b*c )*arctanh(b^(1/2)*x/(b*x^2+a*x)^(1/2))/b^(7/2)
Time = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.87 \[ \int (c+d x) \left (a x+b x^2\right )^{3/2} \, dx=\frac {(x (a+b x))^{3/2} \left (\frac {\sqrt {b} \sqrt {x} \left (15 a^4 d-10 a^3 b (3 c+d x)+4 a^2 b^2 x (5 c+2 d x)+32 b^4 x^3 (5 c+4 d x)+16 a b^3 x^2 (15 c+11 d x)\right )}{a+b x}+\frac {30 a^4 (-2 b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{(a+b x)^{3/2}}\right )}{640 b^{7/2} x^{3/2}} \] Input:
Integrate[(c + d*x)*(a*x + b*x^2)^(3/2),x]
Output:
((x*(a + b*x))^(3/2)*((Sqrt[b]*Sqrt[x]*(15*a^4*d - 10*a^3*b*(3*c + d*x) + 4*a^2*b^2*x*(5*c + 2*d*x) + 32*b^4*x^3*(5*c + 4*d*x) + 16*a*b^3*x^2*(15*c + 11*d*x)))/(a + b*x) + (30*a^4*(-2*b*c + a*d)*ArcTanh[(Sqrt[b]*Sqrt[x])/( Sqrt[a] - Sqrt[a + b*x])])/(a + b*x)^(3/2)))/(640*b^(7/2)*x^(3/2))
Time = 0.42 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.71, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1160, 1087, 1087, 1091, 219}
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.
| \(\displaystyle \int \left (a x+b x^2\right )^{3/2} (c+d x) \, dx\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {(2 b c-a d) \int \left (b x^2+a x\right )^{3/2}dx}{2 b}+\frac {d \left (a x+b x^2\right )^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {(2 b c-a d) \left (\frac {(a+2 b x) \left (a x+b x^2\right )^{3/2}}{8 b}-\frac {3 a^2 \int \sqrt {b x^2+a x}dx}{16 b}\right )}{2 b}+\frac {d \left (a x+b x^2\right )^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {(2 b c-a d) \left (\frac {(a+2 b x) \left (a x+b x^2\right )^{3/2}}{8 b}-\frac {3 a^2 \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \int \frac {1}{\sqrt {b x^2+a x}}dx}{8 b}\right )}{16 b}\right )}{2 b}+\frac {d \left (a x+b x^2\right )^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {(2 b c-a d) \left (\frac {(a+2 b x) \left (a x+b x^2\right )^{3/2}}{8 b}-\frac {3 a^2 \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}}{4 b}\right )}{16 b}\right )}{2 b}+\frac {d \left (a x+b x^2\right )^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\left (\frac {(a+2 b x) \left (a x+b x^2\right )^{3/2}}{8 b}-\frac {3 a^2 \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{4 b^{3/2}}\right )}{16 b}\right ) (2 b c-a d)}{2 b}+\frac {d \left (a x+b x^2\right )^{5/2}}{5 b}\) |
Input:
Int[(c + d*x)*(a*x + b*x^2)^(3/2),x]
Output:
(d*(a*x + b*x^2)^(5/2))/(5*b) + ((2*b*c - a*d)*(((a + 2*b*x)*(a*x + b*x^2) ^(3/2))/(8*b) - (3*a^2*(((a + 2*b*x)*Sqrt[a*x + b*x^2])/(4*b) - (a^2*ArcTa nh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(4*b^(3/2))))/(16*b)))/(2*b)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Time = 0.47 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.61
| method | result | size |
| pseudoelliptic | \(-\frac {3 \left (\left (a^{5} d -2 a^{4} b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )-\sqrt {x \left (b x +a \right )}\, \left (-2 \left (\frac {d x}{3}+c \right ) a^{3} b^{\frac {3}{2}}+\frac {4 x \left (\frac {2 d x}{5}+c \right ) a^{2} b^{\frac {5}{2}}}{3}+16 x^{2} a \left (\frac {11 d x}{15}+c \right ) b^{\frac {7}{2}}+\frac {32 x^{3} \left (\frac {4 d x}{5}+c \right ) b^{\frac {9}{2}}}{3}+\sqrt {b}\, a^{4} d \right )\right )}{128 b^{\frac {7}{2}}}\) | \(116\) |
| risch | \(\frac {\left (128 d \,x^{4} b^{4}+176 a \,b^{3} d \,x^{3}+160 b^{4} c \,x^{3}+8 a^{2} b^{2} d \,x^{2}+240 a \,b^{3} c \,x^{2}-10 a^{3} b d x +20 a^{2} b^{2} c x +15 a^{4} d -30 a^{3} b c \right ) x \left (b x +a \right )}{640 b^{3} \sqrt {x \left (b x +a \right )}}-\frac {3 a^{4} \left (a d -2 b c \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{256 b^{\frac {7}{2}}}\) | \(144\) |
| default | \(c \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )+d \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{5 b}-\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{2 b}\right )\) | \(201\) |
Input:
int((d*x+c)*(b*x^2+a*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
-3/128/b^(7/2)*((a^5*d-2*a^4*b*c)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))-(x* (b*x+a))^(1/2)*(-2*(1/3*d*x+c)*a^3*b^(3/2)+4/3*x*(2/5*d*x+c)*a^2*b^(5/2)+1 6*x^2*a*(11/15*d*x+c)*b^(7/2)+32/3*x^3*(4/5*d*x+c)*b^(9/2)+b^(1/2)*a^4*d))
Time = 0.11 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.59 \[ \int (c+d x) \left (a x+b x^2\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (2 \, a^{4} b c - a^{5} d\right )} \sqrt {b} \log \left (2 \, b x + a - 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (128 \, b^{5} d x^{4} - 30 \, a^{3} b^{2} c + 15 \, a^{4} b d + 16 \, {\left (10 \, b^{5} c + 11 \, a b^{4} d\right )} x^{3} + 8 \, {\left (30 \, a b^{4} c + a^{2} b^{3} d\right )} x^{2} + 10 \, {\left (2 \, a^{2} b^{3} c - a^{3} b^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{1280 \, b^{4}}, -\frac {15 \, {\left (2 \, a^{4} b c - a^{5} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) - {\left (128 \, b^{5} d x^{4} - 30 \, a^{3} b^{2} c + 15 \, a^{4} b d + 16 \, {\left (10 \, b^{5} c + 11 \, a b^{4} d\right )} x^{3} + 8 \, {\left (30 \, a b^{4} c + a^{2} b^{3} d\right )} x^{2} + 10 \, {\left (2 \, a^{2} b^{3} c - a^{3} b^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{640 \, b^{4}}\right ] \] Input:
integrate((d*x+c)*(b*x^2+a*x)^(3/2),x, algorithm="fricas")
Output:
[-1/1280*(15*(2*a^4*b*c - a^5*d)*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a* x)*sqrt(b)) - 2*(128*b^5*d*x^4 - 30*a^3*b^2*c + 15*a^4*b*d + 16*(10*b^5*c + 11*a*b^4*d)*x^3 + 8*(30*a*b^4*c + a^2*b^3*d)*x^2 + 10*(2*a^2*b^3*c - a^3 *b^2*d)*x)*sqrt(b*x^2 + a*x))/b^4, -1/640*(15*(2*a^4*b*c - a^5*d)*sqrt(-b) *arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x + a)) - (128*b^5*d*x^4 - 30*a^3*b^ 2*c + 15*a^4*b*d + 16*(10*b^5*c + 11*a*b^4*d)*x^3 + 8*(30*a*b^4*c + a^2*b^ 3*d)*x^2 + 10*(2*a^2*b^3*c - a^3*b^2*d)*x)*sqrt(b*x^2 + a*x))/b^4]
Time = 0.45 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.71 \[ \int (c+d x) \left (a x+b x^2\right )^{3/2} \, dx=\begin {cases} \frac {3 a^{2} \left (a^{2} c - \frac {5 a \left (a^{2} d + 2 a b c - \frac {7 a \left (\frac {11 a b d}{10} + b^{2} c\right )}{8 b}\right )}{6 b}\right ) \left (\begin {cases} \frac {\log {\left (a + 2 \sqrt {b} \sqrt {a x + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: \frac {a^{2}}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b \left (\frac {a}{2 b} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a x + b x^{2}} \left (- \frac {3 a \left (a^{2} c - \frac {5 a \left (a^{2} d + 2 a b c - \frac {7 a \left (\frac {11 a b d}{10} + b^{2} c\right )}{8 b}\right )}{6 b}\right )}{4 b^{2}} + \frac {b d x^{4}}{5} + \frac {x^{3} \cdot \left (\frac {11 a b d}{10} + b^{2} c\right )}{4 b} + \frac {x^{2} \left (a^{2} d + 2 a b c - \frac {7 a \left (\frac {11 a b d}{10} + b^{2} c\right )}{8 b}\right )}{3 b} + \frac {x \left (a^{2} c - \frac {5 a \left (a^{2} d + 2 a b c - \frac {7 a \left (\frac {11 a b d}{10} + b^{2} c\right )}{8 b}\right )}{6 b}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\\frac {2 \left (\frac {c \left (a x\right )^{\frac {5}{2}}}{5} + \frac {d \left (a x\right )^{\frac {7}{2}}}{7 a}\right )}{a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)*(b*x**2+a*x)**(3/2),x)
Output:
Piecewise((3*a**2*(a**2*c - 5*a*(a**2*d + 2*a*b*c - 7*a*(11*a*b*d/10 + b** 2*c)/(8*b))/(6*b))*Piecewise((log(a + 2*sqrt(b)*sqrt(a*x + b*x**2) + 2*b*x )/sqrt(b), Ne(a**2/b, 0)), ((a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True))/(8*b**2) + sqrt(a*x + b*x**2)*(-3*a*(a**2*c - 5*a*(a**2* d + 2*a*b*c - 7*a*(11*a*b*d/10 + b**2*c)/(8*b))/(6*b))/(4*b**2) + b*d*x**4 /5 + x**3*(11*a*b*d/10 + b**2*c)/(4*b) + x**2*(a**2*d + 2*a*b*c - 7*a*(11* a*b*d/10 + b**2*c)/(8*b))/(3*b) + x*(a**2*c - 5*a*(a**2*d + 2*a*b*c - 7*a* (11*a*b*d/10 + b**2*c)/(8*b))/(6*b))/(2*b)), Ne(b, 0)), (2*(c*(a*x)**(5/2) /5 + d*(a*x)**(7/2)/(7*a))/a, Ne(a, 0)), (0, True))
Time = 0.03 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.24 \[ \int (c+d x) \left (a x+b x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} c x - \frac {3 \, \sqrt {b x^{2} + a x} a^{2} c x}{32 \, b} + \frac {3 \, \sqrt {b x^{2} + a x} a^{3} d x}{64 \, b^{2}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} a d x}{8 \, b} + \frac {3 \, a^{4} c \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {5}{2}}} - \frac {3 \, a^{5} d \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{256 \, b^{\frac {7}{2}}} - \frac {3 \, \sqrt {b x^{2} + a x} a^{3} c}{64 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} a c}{8 \, b} + \frac {3 \, \sqrt {b x^{2} + a x} a^{4} d}{128 \, b^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2} d}{16 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} d}{5 \, b} \] Input:
integrate((d*x+c)*(b*x^2+a*x)^(3/2),x, algorithm="maxima")
Output:
1/4*(b*x^2 + a*x)^(3/2)*c*x - 3/32*sqrt(b*x^2 + a*x)*a^2*c*x/b + 3/64*sqrt (b*x^2 + a*x)*a^3*d*x/b^2 - 1/8*(b*x^2 + a*x)^(3/2)*a*d*x/b + 3/128*a^4*c* log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) - 3/256*a^5*d*log(2*b *x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) - 3/64*sqrt(b*x^2 + a*x)*a^3 *c/b^2 + 1/8*(b*x^2 + a*x)^(3/2)*a*c/b + 3/128*sqrt(b*x^2 + a*x)*a^4*d/b^3 - 1/16*(b*x^2 + a*x)^(3/2)*a^2*d/b^2 + 1/5*(b*x^2 + a*x)^(5/2)*d/b
Time = 0.15 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.85 \[ \int (c+d x) \left (a x+b x^2\right )^{3/2} \, dx=\frac {1}{640} \, \sqrt {b x^{2} + a x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, b d x + \frac {10 \, b^{5} c + 11 \, a b^{4} d}{b^{4}}\right )} x + \frac {30 \, a b^{4} c + a^{2} b^{3} d}{b^{4}}\right )} x + \frac {5 \, {\left (2 \, a^{2} b^{3} c - a^{3} b^{2} d\right )}}{b^{4}}\right )} x - \frac {15 \, {\left (2 \, a^{3} b^{2} c - a^{4} b d\right )}}{b^{4}}\right )} - \frac {3 \, {\left (2 \, a^{4} b c - a^{5} d\right )} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{256 \, b^{\frac {7}{2}}} \] Input:
integrate((d*x+c)*(b*x^2+a*x)^(3/2),x, algorithm="giac")
Output:
1/640*sqrt(b*x^2 + a*x)*(2*(4*(2*(8*b*d*x + (10*b^5*c + 11*a*b^4*d)/b^4)*x + (30*a*b^4*c + a^2*b^3*d)/b^4)*x + 5*(2*a^2*b^3*c - a^3*b^2*d)/b^4)*x - 15*(2*a^3*b^2*c - a^4*b*d)/b^4) - 3/256*(2*a^4*b*c - a^5*d)*log(abs(2*(sqr t(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/b^(7/2)
Time = 9.71 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.09 \[ \int (c+d x) \left (a x+b x^2\right )^{3/2} \, dx=\frac {d\,{\left (b\,x^2+a\,x\right )}^{5/2}}{5\,b}-\frac {3\,a^2\,c\,\left (\sqrt {b\,x^2+a\,x}\,\left (\frac {x}{2}+\frac {a}{4\,b}\right )-\frac {a^2\,\ln \left (\frac {\frac {a}{2}+b\,x}{\sqrt {b}}+\sqrt {b\,x^2+a\,x}\right )}{8\,b^{3/2}}\right )}{16\,b}-\frac {a\,d\,\left (\frac {x\,{\left (b\,x^2+a\,x\right )}^{3/2}}{4}+\frac {a\,{\left (b\,x^2+a\,x\right )}^{3/2}}{8\,b}-\frac {3\,a^2\,\left (\frac {\sqrt {b\,x^2+a\,x}\,\left (a+2\,b\,x\right )}{4\,b}-\frac {a^2\,\ln \left (\frac {\frac {a}{2}+b\,x}{\sqrt {b}}+\sqrt {b\,x^2+a\,x}\right )}{8\,b^{3/2}}\right )}{16\,b}\right )}{2\,b}+\frac {c\,{\left (b\,x^2+a\,x\right )}^{3/2}\,\left (\frac {a}{2}+b\,x\right )}{4\,b} \] Input:
int((a*x + b*x^2)^(3/2)*(c + d*x),x)
Output:
(d*(a*x + b*x^2)^(5/2))/(5*b) - (3*a^2*c*((a*x + b*x^2)^(1/2)*(x/2 + a/(4* b)) - (a^2*log((a/2 + b*x)/b^(1/2) + (a*x + b*x^2)^(1/2)))/(8*b^(3/2))))/( 16*b) - (a*d*((x*(a*x + b*x^2)^(3/2))/4 + (a*(a*x + b*x^2)^(3/2))/(8*b) - (3*a^2*(((a*x + b*x^2)^(1/2)*(a + 2*b*x))/(4*b) - (a^2*log((a/2 + b*x)/b^( 1/2) + (a*x + b*x^2)^(1/2)))/(8*b^(3/2))))/(16*b)))/(2*b) + (c*(a*x + b*x^ 2)^(3/2)*(a/2 + b*x))/(4*b)
Time = 0.32 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.14 \[ \int (c+d x) \left (a x+b x^2\right )^{3/2} \, dx=\frac {15 \sqrt {x}\, \sqrt {b x +a}\, a^{4} b d -30 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b^{2} c -10 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b^{2} d x +20 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{3} c x +8 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{3} d \,x^{2}+240 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{4} c \,x^{2}+176 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{4} d \,x^{3}+160 \sqrt {x}\, \sqrt {b x +a}\, b^{5} c \,x^{3}+128 \sqrt {x}\, \sqrt {b x +a}\, b^{5} d \,x^{4}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{5} d +30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{4} b c}{640 b^{4}} \] Input:
int((d*x+c)*(b*x^2+a*x)^(3/2),x)
Output:
(15*sqrt(x)*sqrt(a + b*x)*a**4*b*d - 30*sqrt(x)*sqrt(a + b*x)*a**3*b**2*c - 10*sqrt(x)*sqrt(a + b*x)*a**3*b**2*d*x + 20*sqrt(x)*sqrt(a + b*x)*a**2*b **3*c*x + 8*sqrt(x)*sqrt(a + b*x)*a**2*b**3*d*x**2 + 240*sqrt(x)*sqrt(a + b*x)*a*b**4*c*x**2 + 176*sqrt(x)*sqrt(a + b*x)*a*b**4*d*x**3 + 160*sqrt(x) *sqrt(a + b*x)*b**5*c*x**3 + 128*sqrt(x)*sqrt(a + b*x)*b**5*d*x**4 - 15*sq rt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**5*d + 30*sqrt(b)*l og((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**4*b*c)/(640*b**4)