Integrand size = 19, antiderivative size = 175 \[ \int (d+e x) \left (b x+c x^2\right )^{5/2} \, dx=\frac {5 b^4 (2 c d-b e) (b+2 c x) \sqrt {b x+c x^2}}{1024 c^4}-\frac {5 b^2 (2 c d-b e) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{384 c^3}+\frac {(2 c d-b e) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{24 c^2}+\frac {e \left (b x+c x^2\right )^{7/2}}{7 c}-\frac {5 b^6 (2 c d-b e) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{1024 c^{9/2}} \]
-5/384*b^2*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c^3+1/24*(-b*e+2*c*d)* (2*c*x+b)*(c*x^2+b*x)^(5/2)/c^2+1/7*e*(c*x^2+b*x)^(7/2)/c-5/1024*b^6*(-b*e +2*c*d)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(9/2)+5/1024*b^4*(-b*e+2*c* d)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^4
Time = 1.72 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.36 \[ \int (d+e x) \left (b x+c x^2\right )^{5/2} \, dx=\frac {\sqrt {x} \sqrt {b+c x} \left (\sqrt {c} \sqrt {x} \sqrt {b+c x} \left (-105 b^6 e+70 b^5 c (3 d+e x)-28 b^4 c^2 x (5 d+2 e x)+16 b^3 c^3 x^2 (7 d+3 e x)+512 c^6 x^5 (7 d+6 e x)+256 b c^5 x^4 (35 d+29 e x)+32 b^2 c^4 x^3 (189 d+148 e x)\right )+420 b^6 c d \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )+210 b^7 e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{21504 c^{9/2} \sqrt {x (b+c x)}} \]
(Sqrt[x]*Sqrt[b + c*x]*(Sqrt[c]*Sqrt[x]*Sqrt[b + c*x]*(-105*b^6*e + 70*b^5 *c*(3*d + e*x) - 28*b^4*c^2*x*(5*d + 2*e*x) + 16*b^3*c^3*x^2*(7*d + 3*e*x) + 512*c^6*x^5*(7*d + 6*e*x) + 256*b*c^5*x^4*(35*d + 29*e*x) + 32*b^2*c^4* x^3*(189*d + 148*e*x)) + 420*b^6*c*d*ArcTanh[(Sqrt[c]*Sqrt[x])/(Sqrt[b] - Sqrt[b + c*x])] + 210*b^7*e*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])]))/(21504*c^(9/2)*Sqrt[x*(b + c*x)])
Time = 0.30 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1160, 1087, 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.
\(\displaystyle \int \left (b x+c x^2\right )^{5/2} (d+e x) \, dx\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {(2 c d-b e) \int \left (c x^2+b x\right )^{5/2}dx}{2 c}+\frac {e \left (b x+c x^2\right )^{7/2}}{7 c}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {5 b^2 \int \left (c x^2+b x\right )^{3/2}dx}{24 c}\right )}{2 c}+\frac {e \left (b x+c x^2\right )^{7/2}}{7 c}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {5 b^2 \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^2+b x}dx}{16 c}\right )}{24 c}\right )}{2 c}+\frac {e \left (b x+c x^2\right )^{7/2}}{7 c}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {5 b^2 \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{16 c}\right )}{24 c}\right )}{2 c}+\frac {e \left (b x+c x^2\right )^{7/2}}{7 c}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {5 b^2 \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{16 c}\right )}{24 c}\right )}{2 c}+\frac {e \left (b x+c x^2\right )^{7/2}}{7 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {5 b^2 \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{16 c}\right )}{24 c}\right ) (2 c d-b e)}{2 c}+\frac {e \left (b x+c x^2\right )^{7/2}}{7 c}\) |
(e*(b*x + c*x^2)^(7/2))/(7*c) + ((2*c*d - b*e)*(((b + 2*c*x)*(b*x + c*x^2) ^(5/2))/(12*c) - (5*b^2*(((b + 2*c*x)*(b*x + c*x^2)^(3/2))/(8*c) - (3*b^2* (((b + 2*c*x)*Sqrt[b*x + c*x^2])/(4*c) - (b^2*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2))))/(16*c)))/(24*c)))/(2*c)
3.4.4.3.1 Defintions of rubi rules used
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 = 1.89 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.10
method | result | size |
risch | \(-\frac {\left (-3072 c^{6} e \,x^{6}-7424 b \,c^{5} e \,x^{5}-3584 c^{6} d \,x^{5}-4736 b^{2} c^{4} e \,x^{4}-8960 b \,c^{5} d \,x^{4}-48 b^{3} c^{3} e \,x^{3}-6048 b^{2} c^{4} d \,x^{3}+56 b^{4} c^{2} e \,x^{2}-112 b^{3} c^{3} d \,x^{2}-70 b^{5} c e x +140 x \,b^{4} c^{2} d +105 b^{6} e -210 b^{5} c d \right ) x \left (c x +b \right )}{21504 c^{4} \sqrt {x \left (c x +b \right )}}+\frac {5 b^{6} \left (b e -2 c d \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2048 c^{\frac {9}{2}}}\) | \(192\) |
default | \(d \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{12 c}-\frac {5 b^{2} \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\right )+e \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{7 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{12 c}-\frac {5 b^{2} \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\right )}{2 c}\right )\) | \(263\) |
-1/21504/c^4*(-3072*c^6*e*x^6-7424*b*c^5*e*x^5-3584*c^6*d*x^5-4736*b^2*c^4 *e*x^4-8960*b*c^5*d*x^4-48*b^3*c^3*e*x^3-6048*b^2*c^4*d*x^3+56*b^4*c^2*e*x ^2-112*b^3*c^3*d*x^2-70*b^5*c*e*x+140*b^4*c^2*d*x+105*b^6*e-210*b^5*c*d)*x *(c*x+b)/(x*(c*x+b))^(1/2)+5/2048*b^6*(b*e-2*c*d)/c^(9/2)*ln((1/2*b+c*x)/c ^(1/2)+(c*x^2+b*x)^(1/2))
Time = 0.76 (sec) , antiderivative size = 397, normalized size of antiderivative = 2.27 \[ \int (d+e x) \left (b x+c x^2\right )^{5/2} \, dx=\left [-\frac {105 \, {\left (2 \, b^{6} c d - b^{7} e\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (3072 \, c^{7} e x^{6} + 210 \, b^{5} c^{2} d - 105 \, b^{6} c e + 256 \, {\left (14 \, c^{7} d + 29 \, b c^{6} e\right )} x^{5} + 128 \, {\left (70 \, b c^{6} d + 37 \, b^{2} c^{5} e\right )} x^{4} + 48 \, {\left (126 \, b^{2} c^{5} d + b^{3} c^{4} e\right )} x^{3} + 56 \, {\left (2 \, b^{3} c^{4} d - b^{4} c^{3} e\right )} x^{2} - 70 \, {\left (2 \, b^{4} c^{3} d - b^{5} c^{2} e\right )} x\right )} \sqrt {c x^{2} + b x}}{43008 \, c^{5}}, \frac {105 \, {\left (2 \, b^{6} c d - b^{7} e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (3072 \, c^{7} e x^{6} + 210 \, b^{5} c^{2} d - 105 \, b^{6} c e + 256 \, {\left (14 \, c^{7} d + 29 \, b c^{6} e\right )} x^{5} + 128 \, {\left (70 \, b c^{6} d + 37 \, b^{2} c^{5} e\right )} x^{4} + 48 \, {\left (126 \, b^{2} c^{5} d + b^{3} c^{4} e\right )} x^{3} + 56 \, {\left (2 \, b^{3} c^{4} d - b^{4} c^{3} e\right )} x^{2} - 70 \, {\left (2 \, b^{4} c^{3} d - b^{5} c^{2} e\right )} x\right )} \sqrt {c x^{2} + b x}}{21504 \, c^{5}}\right ] \]
[-1/43008*(105*(2*b^6*c*d - b^7*e)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(3072*c^7*e*x^6 + 210*b^5*c^2*d - 105*b^6*c*e + 256*(14* c^7*d + 29*b*c^6*e)*x^5 + 128*(70*b*c^6*d + 37*b^2*c^5*e)*x^4 + 48*(126*b^ 2*c^5*d + b^3*c^4*e)*x^3 + 56*(2*b^3*c^4*d - b^4*c^3*e)*x^2 - 70*(2*b^4*c^ 3*d - b^5*c^2*e)*x)*sqrt(c*x^2 + b*x))/c^5, 1/21504*(105*(2*b^6*c*d - b^7* e)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (3072*c^7*e*x^6 + 2 10*b^5*c^2*d - 105*b^6*c*e + 256*(14*c^7*d + 29*b*c^6*e)*x^5 + 128*(70*b*c ^6*d + 37*b^2*c^5*e)*x^4 + 48*(126*b^2*c^5*d + b^3*c^4*e)*x^3 + 56*(2*b^3* c^4*d - b^4*c^3*e)*x^2 - 70*(2*b^4*c^3*d - b^5*c^2*e)*x)*sqrt(c*x^2 + b*x) )/c^5]
Leaf count of result is larger than twice the leaf count of optimal. 570 vs. \(2 (163) = 326\).
Time = 0.56 (sec) , antiderivative size = 570, normalized size of antiderivative = 3.26 \[ \int (d+e x) \left (b x+c x^2\right )^{5/2} \, dx=\begin {cases} - \frac {5 b^{3} \left (b^{3} d - \frac {7 b \left (b^{3} e + 3 b^{2} c d - \frac {9 b \left (3 b^{2} c e + 3 b c^{2} d - \frac {11 b \left (\frac {29 b c^{2} e}{14} + c^{3} d\right )}{12 c}\right )}{10 c}\right )}{8 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 c^{3}} + \sqrt {b x + c x^{2}} \cdot \left (\frac {5 b^{2} \left (b^{3} d - \frac {7 b \left (b^{3} e + 3 b^{2} c d - \frac {9 b \left (3 b^{2} c e + 3 b c^{2} d - \frac {11 b \left (\frac {29 b c^{2} e}{14} + c^{3} d\right )}{12 c}\right )}{10 c}\right )}{8 c}\right )}{8 c^{3}} - \frac {5 b x \left (b^{3} d - \frac {7 b \left (b^{3} e + 3 b^{2} c d - \frac {9 b \left (3 b^{2} c e + 3 b c^{2} d - \frac {11 b \left (\frac {29 b c^{2} e}{14} + c^{3} d\right )}{12 c}\right )}{10 c}\right )}{8 c}\right )}{12 c^{2}} + \frac {c^{2} e x^{6}}{7} + \frac {x^{5} \cdot \left (\frac {29 b c^{2} e}{14} + c^{3} d\right )}{6 c} + \frac {x^{4} \cdot \left (3 b^{2} c e + 3 b c^{2} d - \frac {11 b \left (\frac {29 b c^{2} e}{14} + c^{3} d\right )}{12 c}\right )}{5 c} + \frac {x^{3} \left (b^{3} e + 3 b^{2} c d - \frac {9 b \left (3 b^{2} c e + 3 b c^{2} d - \frac {11 b \left (\frac {29 b c^{2} e}{14} + c^{3} d\right )}{12 c}\right )}{10 c}\right )}{4 c} + \frac {x^{2} \left (b^{3} d - \frac {7 b \left (b^{3} e + 3 b^{2} c d - \frac {9 b \left (3 b^{2} c e + 3 b c^{2} d - \frac {11 b \left (\frac {29 b c^{2} e}{14} + c^{3} d\right )}{12 c}\right )}{10 c}\right )}{8 c}\right )}{3 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {d \left (b x\right )^{\frac {7}{2}}}{7} + \frac {e \left (b x\right )^{\frac {9}{2}}}{9 b}\right )}{b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((-5*b**3*(b**3*d - 7*b*(b**3*e + 3*b**2*c*d - 9*b*(3*b**2*c*e + 3*b*c**2*d - 11*b*(29*b*c**2*e/14 + c**3*d)/(12*c))/(10*c))/(8*c))*Piecewi se((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(16*c**3 ) + sqrt(b*x + c*x**2)*(5*b**2*(b**3*d - 7*b*(b**3*e + 3*b**2*c*d - 9*b*(3 *b**2*c*e + 3*b*c**2*d - 11*b*(29*b*c**2*e/14 + c**3*d)/(12*c))/(10*c))/(8 *c))/(8*c**3) - 5*b*x*(b**3*d - 7*b*(b**3*e + 3*b**2*c*d - 9*b*(3*b**2*c*e + 3*b*c**2*d - 11*b*(29*b*c**2*e/14 + c**3*d)/(12*c))/(10*c))/(8*c))/(12* c**2) + c**2*e*x**6/7 + x**5*(29*b*c**2*e/14 + c**3*d)/(6*c) + x**4*(3*b** 2*c*e + 3*b*c**2*d - 11*b*(29*b*c**2*e/14 + c**3*d)/(12*c))/(5*c) + x**3*( b**3*e + 3*b**2*c*d - 9*b*(3*b**2*c*e + 3*b*c**2*d - 11*b*(29*b*c**2*e/14 + c**3*d)/(12*c))/(10*c))/(4*c) + x**2*(b**3*d - 7*b*(b**3*e + 3*b**2*c*d - 9*b*(3*b**2*c*e + 3*b*c**2*d - 11*b*(29*b*c**2*e/14 + c**3*d)/(12*c))/(1 0*c))/(8*c))/(3*c)), Ne(c, 0)), (2*(d*(b*x)**(7/2)/7 + e*(b*x)**(9/2)/(9*b ))/b, Ne(b, 0)), (0, True))
Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (151) = 302\).
Time = 0.21 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.82 \[ \int (d+e x) \left (b x+c x^2\right )^{5/2} \, dx=\frac {1}{6} \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} d x + \frac {5 \, \sqrt {c x^{2} + b x} b^{4} d x}{256 \, c^{2}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} d x}{96 \, c} - \frac {5 \, \sqrt {c x^{2} + b x} b^{5} e x}{512 \, c^{3}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3} e x}{192 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} b e x}{12 \, c} - \frac {5 \, b^{6} d \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {7}{2}}} + \frac {5 \, b^{7} e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2048 \, c^{\frac {9}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} b^{5} d}{512 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3} d}{192 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} b d}{12 \, c} - \frac {5 \, \sqrt {c x^{2} + b x} b^{6} e}{1024 \, c^{4}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{4} e}{384 \, c^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} b^{2} e}{24 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {7}{2}} e}{7 \, c} \]
1/6*(c*x^2 + b*x)^(5/2)*d*x + 5/256*sqrt(c*x^2 + b*x)*b^4*d*x/c^2 - 5/96*( c*x^2 + b*x)^(3/2)*b^2*d*x/c - 5/512*sqrt(c*x^2 + b*x)*b^5*e*x/c^3 + 5/192 *(c*x^2 + b*x)^(3/2)*b^3*e*x/c^2 - 1/12*(c*x^2 + b*x)^(5/2)*b*e*x/c - 5/10 24*b^6*d*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 5/2048*b^7 *e*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) + 5/512*sqrt(c*x^2 + b*x)*b^5*d/c^3 - 5/192*(c*x^2 + b*x)^(3/2)*b^3*d/c^2 + 1/12*(c*x^2 + b* x)^(5/2)*b*d/c - 5/1024*sqrt(c*x^2 + b*x)*b^6*e/c^4 + 5/384*(c*x^2 + b*x)^ (3/2)*b^4*e/c^3 - 1/24*(c*x^2 + b*x)^(5/2)*b^2*e/c^2 + 1/7*(c*x^2 + b*x)^( 7/2)*e/c
Time = 0.31 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.27 \[ \int (d+e x) \left (b x+c x^2\right )^{5/2} \, dx=\frac {1}{21504} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, {\left (12 \, c^{2} e x + \frac {14 \, c^{8} d + 29 \, b c^{7} e}{c^{6}}\right )} x + \frac {70 \, b c^{7} d + 37 \, b^{2} c^{6} e}{c^{6}}\right )} x + \frac {3 \, {\left (126 \, b^{2} c^{6} d + b^{3} c^{5} e\right )}}{c^{6}}\right )} x + \frac {7 \, {\left (2 \, b^{3} c^{5} d - b^{4} c^{4} e\right )}}{c^{6}}\right )} x - \frac {35 \, {\left (2 \, b^{4} c^{4} d - b^{5} c^{3} e\right )}}{c^{6}}\right )} x + \frac {105 \, {\left (2 \, b^{5} c^{3} d - b^{6} c^{2} e\right )}}{c^{6}}\right )} + \frac {5 \, {\left (2 \, b^{6} c d - b^{7} e\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{2048 \, c^{\frac {9}{2}}} \]
1/21504*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(2*(12*c^2*e*x + (14*c^8*d + 29*b*c^ 7*e)/c^6)*x + (70*b*c^7*d + 37*b^2*c^6*e)/c^6)*x + 3*(126*b^2*c^6*d + b^3* c^5*e)/c^6)*x + 7*(2*b^3*c^5*d - b^4*c^4*e)/c^6)*x - 35*(2*b^4*c^4*d - b^5 *c^3*e)/c^6)*x + 105*(2*b^5*c^3*d - b^6*c^2*e)/c^6) + 5/2048*(2*b^6*c*d - b^7*e)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(9/2)
Timed out. \[ \int (d+e x) \left (b x+c x^2\right )^{5/2} \, dx=\int {\left (c\,x^2+b\,x\right )}^{5/2}\,\left (d+e\,x\right ) \,d x \]