Integrand size = 19, antiderivative size = 227 \[ \int (a+b x)^{5/2} (c+d x)^{3/2} \, dx=\frac {3 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}{128 b^2 d^3}-\frac {(b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{64 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{80 b^2 d}+\frac {3 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b^2}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}-\frac {3 (b c-a d)^5 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{5/2} d^{7/2}} \]
1/5*(b*x+a)^(7/2)*(d*x+c)^(3/2)/b-3/128*(-a*d+b*c)^5*arctanh(d^(1/2)*(b*x+ a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(5/2)/d^(7/2)-1/64*(-a*d+b*c)^3*(b*x+a)^ (3/2)*(d*x+c)^(1/2)/b^2/d^2+1/80*(-a*d+b*c)^2*(b*x+a)^(5/2)*(d*x+c)^(1/2)/ b^2/d+3/40*(-a*d+b*c)*(b*x+a)^(7/2)*(d*x+c)^(1/2)/b^2+3/128*(-a*d+b*c)^4*( b*x+a)^(1/2)*(d*x+c)^(1/2)/b^2/d^3
Time = 0.06 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.97 \[ \int (a+b x)^{5/2} (c+d x)^{3/2} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-15 a^4 d^4+10 a^3 b d^3 (7 c+d x)+2 a^2 b^2 d^2 \left (64 c^2+233 c d x+124 d^2 x^2\right )+2 a b^3 d \left (-35 c^3+23 c^2 d x+256 c d^2 x^2+168 d^3 x^3\right )+b^4 \left (15 c^4-10 c^3 d x+8 c^2 d^2 x^2+176 c d^3 x^3+128 d^4 x^4\right )\right )}{640 b^2 d^3}-\frac {3 (b c-a d)^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{128 b^{5/2} d^{7/2}} \]
(Sqrt[a + b*x]*Sqrt[c + d*x]*(-15*a^4*d^4 + 10*a^3*b*d^3*(7*c + d*x) + 2*a ^2*b^2*d^2*(64*c^2 + 233*c*d*x + 124*d^2*x^2) + 2*a*b^3*d*(-35*c^3 + 23*c^ 2*d*x + 256*c*d^2*x^2 + 168*d^3*x^3) + b^4*(15*c^4 - 10*c^3*d*x + 8*c^2*d^ 2*x^2 + 176*c*d^3*x^3 + 128*d^4*x^4)))/(640*b^2*d^3) - (3*(b*c - a*d)^5*Ar cTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(128*b^(5/2)*d^(7/ 2))
Time = 0.26 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {60, 60, 60, 60, 60, 66, 221}
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.
| \(\displaystyle \int (a+b x)^{5/2} (c+d x)^{3/2} \, dx\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3 (b c-a d) \int (a+b x)^{5/2} \sqrt {c+d x}dx}{10 b}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}}dx}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{6 d}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{6 d}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{6 d}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{6 d}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{6 d}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}\right )}{10 b}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}\) |
((a + b*x)^(7/2)*(c + d*x)^(3/2))/(5*b) + (3*(b*c - a*d)*(((a + b*x)^(7/2) *Sqrt[c + d*x])/(4*b) + ((b*c - a*d)*(((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*d ) - (5*(b*c - a*d)*(((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d) - (3*(b*c - a*d) *((Sqrt[a + b*x]*Sqrt[c + d*x])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*d^(3/2))))/(4*d)))/(6*d)))/(8*b) ))/(10*b)
3.7.58.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Time = 0.55 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.05
| method | result | size |
| default | \(\frac {\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {5}{2}}}{5 d}-\frac {\left (-a d +b c \right ) \left (\frac {\left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {5}{2}}}{4 d}-\frac {3 \left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {5}{2}}}{3 d}-\frac {\left (-a d +b c \right ) \left (\frac {\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}}{2 b}-\frac {3 \left (a d -b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {d x +c}}{b}-\frac {\left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 b \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}}\right )}{4 b}\right )}{6 d}\right )}{8 d}\right )}{2 d}\) | \(239\) |
1/5/d*(b*x+a)^(5/2)*(d*x+c)^(5/2)-1/2*(-a*d+b*c)/d*(1/4/d*(b*x+a)^(3/2)*(d *x+c)^(5/2)-3/8*(-a*d+b*c)/d*(1/3/d*(b*x+a)^(1/2)*(d*x+c)^(5/2)-1/6*(-a*d+ b*c)/d*(1/2*(d*x+c)^(3/2)*(b*x+a)^(1/2)/b-3/4*(a*d-b*c)/b*((b*x+a)^(1/2)*( d*x+c)^(1/2)/b-1/2*(a*d-b*c)/b*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+ a)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(b*d)^(1/2)+(b*d*x^2+(a*d+b*c)*x+a*c)^ (1/2))/(b*d)^(1/2)))))
Time = 0.28 (sec) , antiderivative size = 702, normalized size of antiderivative = 3.09 \[ \int (a+b x)^{5/2} (c+d x)^{3/2} \, dx=\left [-\frac {15 \, {\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (128 \, b^{5} d^{5} x^{4} + 15 \, b^{5} c^{4} d - 70 \, a b^{4} c^{3} d^{2} + 128 \, a^{2} b^{3} c^{2} d^{3} + 70 \, a^{3} b^{2} c d^{4} - 15 \, a^{4} b d^{5} + 16 \, {\left (11 \, b^{5} c d^{4} + 21 \, a b^{4} d^{5}\right )} x^{3} + 8 \, {\left (b^{5} c^{2} d^{3} + 64 \, a b^{4} c d^{4} + 31 \, a^{2} b^{3} d^{5}\right )} x^{2} - 2 \, {\left (5 \, b^{5} c^{3} d^{2} - 23 \, a b^{4} c^{2} d^{3} - 233 \, a^{2} b^{3} c d^{4} - 5 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{2560 \, b^{3} d^{4}}, \frac {15 \, {\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (128 \, b^{5} d^{5} x^{4} + 15 \, b^{5} c^{4} d - 70 \, a b^{4} c^{3} d^{2} + 128 \, a^{2} b^{3} c^{2} d^{3} + 70 \, a^{3} b^{2} c d^{4} - 15 \, a^{4} b d^{5} + 16 \, {\left (11 \, b^{5} c d^{4} + 21 \, a b^{4} d^{5}\right )} x^{3} + 8 \, {\left (b^{5} c^{2} d^{3} + 64 \, a b^{4} c d^{4} + 31 \, a^{2} b^{3} d^{5}\right )} x^{2} - 2 \, {\left (5 \, b^{5} c^{3} d^{2} - 23 \, a b^{4} c^{2} d^{3} - 233 \, a^{2} b^{3} c d^{4} - 5 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{1280 \, b^{3} d^{4}}\right ] \]
[-1/2560*(15*(b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 10*a^3*b^2*c^ 2*d^3 + 5*a^4*b*c*d^4 - a^5*d^5)*sqrt(b*d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6 *a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt( d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(128*b^5*d^5*x^4 + 15*b^5*c^4*d - 70*a*b^4*c^3*d^2 + 128*a^2*b^3*c^2*d^3 + 70*a^3*b^2*c*d^4 - 15*a^4*b*d^5 + 16*(11*b^5*c*d^4 + 21*a*b^4*d^5)*x^3 + 8*(b^5*c^2*d^3 + 64*a*b^4*c*d^4 + 31*a^2*b^3*d^5)*x^2 - 2*(5*b^5*c^3*d^2 - 23*a*b^4*c^2*d^3 - 233*a^2*b^3*c* d^4 - 5*a^3*b^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^4), 1/1280*(15 *(b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 10*a^3*b^2*c^2*d^3 + 5*a^ 4*b*c*d^4 - a^5*d^5)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d )*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2) *x)) + 2*(128*b^5*d^5*x^4 + 15*b^5*c^4*d - 70*a*b^4*c^3*d^2 + 128*a^2*b^3* c^2*d^3 + 70*a^3*b^2*c*d^4 - 15*a^4*b*d^5 + 16*(11*b^5*c*d^4 + 21*a*b^4*d^ 5)*x^3 + 8*(b^5*c^2*d^3 + 64*a*b^4*c*d^4 + 31*a^2*b^3*d^5)*x^2 - 2*(5*b^5* c^3*d^2 - 23*a*b^4*c^2*d^3 - 233*a^2*b^3*c*d^4 - 5*a^3*b^2*d^5)*x)*sqrt(b* x + a)*sqrt(d*x + c))/(b^3*d^4)]
\[ \int (a+b x)^{5/2} (c+d x)^{3/2} \, dx=\int \left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {3}{2}}\, dx \]
Exception generated. \[ \int (a+b x)^{5/2} (c+d x)^{3/2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 1740 vs. \(2 (183) = 366\).
Time = 0.53 (sec) , antiderivative size = 1740, normalized size of antiderivative = 7.67 \[ \int (a+b x)^{5/2} (c+d x)^{3/2} \, dx=\text {Too large to display} \]
1/1920*(240*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a )*(4*(b*x + a)/b^2 + (b^6*c*d^3 - 13*a*b^5*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^ 2 + 2*a*b^6*c*d^3 - 11*a^2*b^5*d^4)/(b^7*d^4)) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b*d^2))*a*c*abs(b) - 1920*((b^2*c - a*b*d)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a* b*d)))/sqrt(b*d) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a))*a^3* c*abs(b)/b^2 + 10*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b* x + a)*(6*(b*x + a)/b^3 + (b^12*c*d^5 - 25*a*b^11*d^6)/(b^14*d^6)) - (5*b^ 13*c^2*d^4 + 14*a*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^6)) + 3*(5*b^14*c ^3*d^3 + 9*a*b^13*c^2*d^4 + 15*a^2*b^12*c*d^5 - 93*a^3*b^11*d^6)/(b^14*d^6 ))*sqrt(b*x + a) + 3*(5*b^4*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 20*a ^3*b*c*d^3 - 35*a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + ( b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^3))*b*c*abs(b) + 30*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^3 + (b^12 *c*d^5 - 25*a*b^11*d^6)/(b^14*d^6)) - (5*b^13*c^2*d^4 + 14*a*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^6)) + 3*(5*b^14*c^3*d^3 + 9*a*b^13*c^2*d^4 + 15* a^2*b^12*c*d^5 - 93*a^3*b^11*d^6)/(b^14*d^6))*sqrt(b*x + a) + 3*(5*b^4*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 35*a^4*d^4)*log(ab s(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqr...
Timed out. \[ \int (a+b x)^{5/2} (c+d x)^{3/2} \, dx=\int {\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{3/2} \,d x \]