Integrand size = 19, antiderivative size = 230 \[ \int (a+b x)^{7/2} \sqrt {c+d x} \, dx=-\frac {7 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}{128 b d^4}+\frac {7 (b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{192 b d^3}-\frac {7 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{240 b d^2}+\frac {(b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b d}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}+\frac {7 (b c-a d)^5 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{3/2} d^{9/2}} \]
7/128*(-a*d+b*c)^5*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^ (3/2)/d^(9/2)+7/192*(-a*d+b*c)^3*(b*x+a)^(3/2)*(d*x+c)^(1/2)/b/d^3-7/240*( -a*d+b*c)^2*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b/d^2+1/40*(-a*d+b*c)*(b*x+a)^(7/2 )*(d*x+c)^(1/2)/b/d+1/5*(b*x+a)^(9/2)*(d*x+c)^(1/2)/b-7/128*(-a*d+b*c)^4*( b*x+a)^(1/2)*(d*x+c)^(1/2)/b/d^4
Time = 0.47 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.96 \[ \int (a+b x)^{7/2} \sqrt {c+d x} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (105 a^4 d^4+10 a^3 b d^3 (79 c+121 d x)+2 a^2 b^2 d^2 \left (-448 c^2+289 c d x+1052 d^2 x^2\right )+2 a b^3 d \left (245 c^3-161 c^2 d x+128 c d^2 x^2+744 d^3 x^3\right )+b^4 \left (-105 c^4+70 c^3 d x-56 c^2 d^2 x^2+48 c d^3 x^3+384 d^4 x^4\right )\right )}{1920 b d^4}+\frac {7 (b c-a d)^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{128 b^{3/2} d^{9/2}} \]
(Sqrt[a + b*x]*Sqrt[c + d*x]*(105*a^4*d^4 + 10*a^3*b*d^3*(79*c + 121*d*x) + 2*a^2*b^2*d^2*(-448*c^2 + 289*c*d*x + 1052*d^2*x^2) + 2*a*b^3*d*(245*c^3 - 161*c^2*d*x + 128*c*d^2*x^2 + 744*d^3*x^3) + b^4*(-105*c^4 + 70*c^3*d*x - 56*c^2*d^2*x^2 + 48*c*d^3*x^3 + 384*d^4*x^4)))/(1920*b*d^4) + (7*(b*c - a*d)^5*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(128*b^( 3/2)*d^(9/2))
Time = 0.29 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.03, 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.
| \(\displaystyle \int (a+b x)^{7/2} \sqrt {c+d x} \, dx\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(b c-a d) \int \frac {(a+b x)^{7/2}}{\sqrt {c+d x}}dx}{10 b}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(b c-a d) \left (\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}-\frac {7 (b c-a d) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}}dx}{8 d}\right )}{10 b}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(b c-a d) \left (\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}-\frac {7 (b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{6 d}\right )}{8 d}\right )}{10 b}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(b c-a d) \left (\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}-\frac {7 (b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{6 d}\right )}{8 d}\right )}{10 b}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(b c-a d) \left (\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}-\frac {7 (b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{6 d}\right )}{8 d}\right )}{10 b}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {(b c-a d) \left (\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}-\frac {7 (b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{6 d}\right )}{8 d}\right )}{10 b}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(b c-a d) \left (\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}-\frac {7 (b c-a d) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{6 d}\right )}{8 d}\right )}{10 b}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}\) |
((a + b*x)^(9/2)*Sqrt[c + d*x])/(5*b) + ((b*c - a*d)*(((a + b*x)^(7/2)*Sqr t[c + d*x])/(4*d) - (7*(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*d))) /(10*b)
3.15.60.3.1 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 = 0.26 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.04
| method | result | size |
| default | \(\frac {\left (b x +a \right )^{\frac {7}{2}} \left (d x +c \right )^{\frac {3}{2}}}{5 d}-\frac {7 \left (-a d +b c \right ) \left (\frac {\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {3}{2}}}{4 d}-\frac {5 \left (-a d +b c \right ) \left (\frac {\left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}}}{3 d}-\frac {\left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{2}}}{2 d}-\frac {\left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {d x +c}}{b}-\frac {\left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 b \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}}\right )}{4 d}\right )}{2 d}\right )}{8 d}\right )}{10 d}\) | \(239\) |
1/5/d*(b*x+a)^(7/2)*(d*x+c)^(3/2)-7/10*(-a*d+b*c)/d*(1/4/d*(b*x+a)^(5/2)*( d*x+c)^(3/2)-5/8*(-a*d+b*c)/d*(1/3/d*(b*x+a)^(3/2)*(d*x+c)^(3/2)-1/2*(-a*d +b*c)/d*(1/2/d*(b*x+a)^(1/2)*(d*x+c)^(3/2)-1/4*(-a*d+b*c)/d*((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 = 3.05 \[ \int (a+b x)^{7/2} \sqrt {c+d x} \, dx=\left [-\frac {105 \, {\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (384 \, b^{5} d^{5} x^{4} - 105 \, b^{5} c^{4} d + 490 \, a b^{4} c^{3} d^{2} - 896 \, a^{2} b^{3} c^{2} d^{3} + 790 \, a^{3} b^{2} c d^{4} + 105 \, a^{4} b d^{5} + 48 \, {\left (b^{5} c d^{4} + 31 \, a b^{4} d^{5}\right )} x^{3} - 8 \, {\left (7 \, b^{5} c^{2} d^{3} - 32 \, a b^{4} c d^{4} - 263 \, a^{2} b^{3} d^{5}\right )} x^{2} + 2 \, {\left (35 \, b^{5} c^{3} d^{2} - 161 \, a b^{4} c^{2} d^{3} + 289 \, a^{2} b^{3} c d^{4} + 605 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{7680 \, b^{2} d^{5}}, -\frac {105 \, {\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (384 \, b^{5} d^{5} x^{4} - 105 \, b^{5} c^{4} d + 490 \, a b^{4} c^{3} d^{2} - 896 \, a^{2} b^{3} c^{2} d^{3} + 790 \, a^{3} b^{2} c d^{4} + 105 \, a^{4} b d^{5} + 48 \, {\left (b^{5} c d^{4} + 31 \, a b^{4} d^{5}\right )} x^{3} - 8 \, {\left (7 \, b^{5} c^{2} d^{3} - 32 \, a b^{4} c d^{4} - 263 \, a^{2} b^{3} d^{5}\right )} x^{2} + 2 \, {\left (35 \, b^{5} c^{3} d^{2} - 161 \, a b^{4} c^{2} d^{3} + 289 \, a^{2} b^{3} c d^{4} + 605 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3840 \, b^{2} d^{5}}\right ] \]
[-1/7680*(105*(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*(384*b^5*d^5*x^4 - 105*b^5*c^4*d + 490*a*b^4*c^3*d^2 - 896*a^2*b^3*c^2*d^3 + 790*a^3*b^2*c*d^4 + 105*a^4*b* d^5 + 48*(b^5*c*d^4 + 31*a*b^4*d^5)*x^3 - 8*(7*b^5*c^2*d^3 - 32*a*b^4*c*d^ 4 - 263*a^2*b^3*d^5)*x^2 + 2*(35*b^5*c^3*d^2 - 161*a*b^4*c^2*d^3 + 289*a^2 *b^3*c*d^4 + 605*a^3*b^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^5), - 1/3840*(105*(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*(384*b^5*d^5*x^4 - 105*b^5*c^4*d + 490*a*b^4*c^3*d^2 - 896*a^2*b^3*c^2*d^3 + 790*a^3*b^2*c*d^4 + 105*a^4*b*d^5 + 48*(b^5*c*d^4 + 31*a*b^4*d^5)*x^3 - 8*(7*b^5*c^2*d^3 - 32*a*b^4*c*d^4 - 263*a^2*b^3*d^5)* x^2 + 2*(35*b^5*c^3*d^2 - 161*a*b^4*c^2*d^3 + 289*a^2*b^3*c*d^4 + 605*a^3* b^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^5)]
\[ \int (a+b x)^{7/2} \sqrt {c+d x} \, dx=\int \left (a + b x\right )^{\frac {7}{2}} \sqrt {c + d x}\, dx \]
Exception generated. \[ \int (a+b x)^{7/2} \sqrt {c+d x} \, dx=\text {Exception raised: ValueError} \]
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. 1107 vs. \(2 (186) = 372\).
Time = 0.47 (sec) , antiderivative size = 1107, normalized size of antiderivative = 4.81 \[ \int (a+b x)^{7/2} \sqrt {c+d x} \, dx=\text {Too large to display} \]
1/1920*(480*(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^2*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^4* abs(b)/b^2 + 40*(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))*a*b*abs(b) + (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(4*(b*x + a)*(6*(b*x + a)*(8*(b*x + a)/b^4 + (b^20*c *d^7 - 41*a*b^19*d^8)/(b^23*d^8)) - (7*b^21*c^2*d^6 + 26*a*b^20*c*d^7 - 51 3*a^2*b^19*d^8)/(b^23*d^8)) + 5*(7*b^22*c^3*d^5 + 19*a*b^21*c^2*d^6 + 37*a ^2*b^20*c*d^7 - 447*a^3*b^19*d^8)/(b^23*d^8))*(b*x + a) - 15*(7*b^23*c^4*d ^4 + 12*a*b^22*c^3*d^5 + 18*a^2*b^21*c^2*d^6 + 28*a^3*b^20*c*d^7 - 193*a^4 *b^19*d^8)/(b^23*d^8))*sqrt(b*x + a) - 15*(7*b^5*c^5 + 5*a*b^4*c^4*d + ...
Timed out. \[ \int (a+b x)^{7/2} \sqrt {c+d x} \, dx=\int {\left (a+b\,x\right )}^{7/2}\,\sqrt {c+d\,x} \,d x \]