Integrand size = 19, antiderivative size = 204 \[ \int \frac {(a+b x)^{9/2}}{(c+d x)^{5/2}} \, dx=-\frac {2 (a+b x)^{9/2}}{3 d (c+d x)^{3/2}}-\frac {6 b (a+b x)^{7/2}}{d^2 \sqrt {c+d x}}+\frac {105 b^2 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^5}-\frac {35 b^2 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{4 d^4}+\frac {7 b^2 (a+b x)^{5/2} \sqrt {c+d x}}{d^3}-\frac {105 b^{3/2} (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 d^{11/2}} \] Output:
-2/3*(b*x+a)^(9/2)/d/(d*x+c)^(3/2)-6*b*(b*x+a)^(7/2)/d^2/(d*x+c)^(1/2)+105 /8*b^2*(-a*d+b*c)^2*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^5-35/4*b^2*(-a*d+b*c)*(b *x+a)^(3/2)*(d*x+c)^(1/2)/d^4+7*b^2*(b*x+a)^(5/2)*(d*x+c)^(1/2)/d^3-105/8* b^(3/2)*(-a*d+b*c)^3*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/ d^(11/2)
Time = 0.38 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b x)^{9/2}}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (-16 a^4 d^4-16 a^3 b d^3 (9 c+13 d x)+3 a^2 b^2 d^2 \left (231 c^2+318 c d x+55 d^2 x^2\right )-2 a b^3 d \left (420 c^3+567 c^2 d x+90 c d^2 x^2-25 d^3 x^3\right )+b^4 \left (315 c^4+420 c^3 d x+63 c^2 d^2 x^2-18 c d^3 x^3+8 d^4 x^4\right )\right )}{24 d^5 (c+d x)^{3/2}}-\frac {105 b^{3/2} (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 d^{11/2}} \] Input:
Integrate[(a + b*x)^(9/2)/(c + d*x)^(5/2),x]
Output:
(Sqrt[a + b*x]*(-16*a^4*d^4 - 16*a^3*b*d^3*(9*c + 13*d*x) + 3*a^2*b^2*d^2* (231*c^2 + 318*c*d*x + 55*d^2*x^2) - 2*a*b^3*d*(420*c^3 + 567*c^2*d*x + 90 *c*d^2*x^2 - 25*d^3*x^3) + b^4*(315*c^4 + 420*c^3*d*x + 63*c^2*d^2*x^2 - 1 8*c*d^3*x^3 + 8*d^4*x^4)))/(24*d^5*(c + d*x)^(3/2)) - (105*b^(3/2)*(b*c - a*d)^3*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(8*d^(11/ 2))
Time = 0.24 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {57, 57, 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.
\(\displaystyle \int \frac {(a+b x)^{9/2}}{(c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {3 b \int \frac {(a+b x)^{7/2}}{(c+d x)^{3/2}}dx}{d}-\frac {2 (a+b x)^{9/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {3 b \left (\frac {7 b \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}}dx}{d}-\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}\right )}{d}-\frac {2 (a+b x)^{9/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3 b \left (\frac {7 b \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{6 d}\right )}{d}-\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}\right )}{d}-\frac {2 (a+b x)^{9/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3 b \left (\frac {7 b \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{6 d}\right )}{d}-\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}\right )}{d}-\frac {2 (a+b x)^{9/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3 b \left (\frac {7 b \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{6 d}\right )}{d}-\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}\right )}{d}-\frac {2 (a+b x)^{9/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {3 b \left (\frac {7 b \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{6 d}\right )}{d}-\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}\right )}{d}-\frac {2 (a+b x)^{9/2}}{3 d (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {3 b \left (\frac {7 b \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{6 d}\right )}{d}-\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}\right )}{d}-\frac {2 (a+b x)^{9/2}}{3 d (c+d x)^{3/2}}\) |
Input:
Int[(a + b*x)^(9/2)/(c + d*x)^(5/2),x]
Output:
(-2*(a + b*x)^(9/2))/(3*d*(c + d*x)^(3/2)) + (3*b*((-2*(a + b*x)^(7/2))/(d *Sqrt[c + d*x]) + (7*b*(((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)))/d))/d
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
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]
\[\int \frac {\left (b x +a \right )^{\frac {9}{2}}}{\left (x d +c \right )^{\frac {5}{2}}}d x\]
Input:
int((b*x+a)^(9/2)/(d*x+c)^(5/2),x)
Output:
int((b*x+a)^(9/2)/(d*x+c)^(5/2),x)
Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (164) = 328\).
Time = 0.62 (sec) , antiderivative size = 879, normalized size of antiderivative = 4.31 \[ \int \frac {(a+b x)^{9/2}}{(c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate((b*x+a)^(9/2)/(d*x+c)^(5/2),x, algorithm="fricas")
Output:
[-1/96*(315*(b^4*c^5 - 3*a*b^3*c^4*d + 3*a^2*b^2*c^3*d^2 - a^3*b*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^2 + 2*(b^ 4*c^4*d - 3*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 - a^3*b*c*d^4)*x)*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^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(8*b^4*d^4*x^4 + 315*b^4*c^4 - 840*a*b^3*c^3*d + 693*a^2*b^2*c^2*d^2 - 144*a^3*b*c*d^3 - 16*a^4*d^4 - 2*(9*b^4*c*d^3 - 25*a*b^3*d^4)*x^3 + 3*(21* b^4*c^2*d^2 - 60*a*b^3*c*d^3 + 55*a^2*b^2*d^4)*x^2 + 2*(210*b^4*c^3*d - 56 7*a*b^3*c^2*d^2 + 477*a^2*b^2*c*d^3 - 104*a^3*b*d^4)*x)*sqrt(b*x + a)*sqrt (d*x + c))/(d^7*x^2 + 2*c*d^6*x + c^2*d^5), 1/48*(315*(b^4*c^5 - 3*a*b^3*c ^4*d + 3*a^2*b^2*c^3*d^2 - a^3*b*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^2 + 2*(b^4*c^4*d - 3*a*b^3*c^3*d^2 + 3*a^ 2*b^2*c^2*d^3 - a^3*b*c*d^4)*x)*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d )*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*x^2 + a*b*c + (b^2*c + a*b *d)*x)) + 2*(8*b^4*d^4*x^4 + 315*b^4*c^4 - 840*a*b^3*c^3*d + 693*a^2*b^2*c ^2*d^2 - 144*a^3*b*c*d^3 - 16*a^4*d^4 - 2*(9*b^4*c*d^3 - 25*a*b^3*d^4)*x^3 + 3*(21*b^4*c^2*d^2 - 60*a*b^3*c*d^3 + 55*a^2*b^2*d^4)*x^2 + 2*(210*b^4*c ^3*d - 567*a*b^3*c^2*d^2 + 477*a^2*b^2*c*d^3 - 104*a^3*b*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^7*x^2 + 2*c*d^6*x + c^2*d^5)]
\[ \int \frac {(a+b x)^{9/2}}{(c+d x)^{5/2}} \, dx=\int \frac {\left (a + b x\right )^{\frac {9}{2}}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((b*x+a)**(9/2)/(d*x+c)**(5/2),x)
Output:
Integral((a + b*x)**(9/2)/(c + d*x)**(5/2), x)
Exception generated. \[ \int \frac {(a+b x)^{9/2}}{(c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x+a)^(9/2)/(d*x+c)^(5/2),x, algorithm="maxima")
Output:
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. 500 vs. \(2 (164) = 328\).
Time = 0.23 (sec) , antiderivative size = 500, normalized size of antiderivative = 2.45 \[ \int \frac {(a+b x)^{9/2}}{(c+d x)^{5/2}} \, dx=\frac {{\left ({\left ({\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b^{6} c d^{8} - a b^{5} d^{9}\right )} {\left (b x + a\right )}}{b^{2} c d^{9} {\left | b \right |} - a b d^{10} {\left | b \right |}} - \frac {9 \, {\left (b^{7} c^{2} d^{7} - 2 \, a b^{6} c d^{8} + a^{2} b^{5} d^{9}\right )}}{b^{2} c d^{9} {\left | b \right |} - a b d^{10} {\left | b \right |}}\right )} + \frac {63 \, {\left (b^{8} c^{3} d^{6} - 3 \, a b^{7} c^{2} d^{7} + 3 \, a^{2} b^{6} c d^{8} - a^{3} b^{5} d^{9}\right )}}{b^{2} c d^{9} {\left | b \right |} - a b d^{10} {\left | b \right |}}\right )} {\left (b x + a\right )} + \frac {420 \, {\left (b^{9} c^{4} d^{5} - 4 \, a b^{8} c^{3} d^{6} + 6 \, a^{2} b^{7} c^{2} d^{7} - 4 \, a^{3} b^{6} c d^{8} + a^{4} b^{5} d^{9}\right )}}{b^{2} c d^{9} {\left | b \right |} - a b d^{10} {\left | b \right |}}\right )} {\left (b x + a\right )} + \frac {315 \, {\left (b^{10} c^{5} d^{4} - 5 \, a b^{9} c^{4} d^{5} + 10 \, a^{2} b^{8} c^{3} d^{6} - 10 \, a^{3} b^{7} c^{2} d^{7} + 5 \, a^{4} b^{6} c d^{8} - a^{5} b^{5} d^{9}\right )}}{b^{2} c d^{9} {\left | b \right |} - a b d^{10} {\left | b \right |}}\right )} \sqrt {b x + a}}{24 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} + \frac {105 \, {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{8 \, \sqrt {b d} d^{5} {\left | b \right |}} \] Input:
integrate((b*x+a)^(9/2)/(d*x+c)^(5/2),x, algorithm="giac")
Output:
1/24*(((2*(b*x + a)*(4*(b^6*c*d^8 - a*b^5*d^9)*(b*x + a)/(b^2*c*d^9*abs(b) - a*b*d^10*abs(b)) - 9*(b^7*c^2*d^7 - 2*a*b^6*c*d^8 + a^2*b^5*d^9)/(b^2*c *d^9*abs(b) - a*b*d^10*abs(b))) + 63*(b^8*c^3*d^6 - 3*a*b^7*c^2*d^7 + 3*a^ 2*b^6*c*d^8 - a^3*b^5*d^9)/(b^2*c*d^9*abs(b) - a*b*d^10*abs(b)))*(b*x + a) + 420*(b^9*c^4*d^5 - 4*a*b^8*c^3*d^6 + 6*a^2*b^7*c^2*d^7 - 4*a^3*b^6*c*d^ 8 + a^4*b^5*d^9)/(b^2*c*d^9*abs(b) - a*b*d^10*abs(b)))*(b*x + a) + 315*(b^ 10*c^5*d^4 - 5*a*b^9*c^4*d^5 + 10*a^2*b^8*c^3*d^6 - 10*a^3*b^7*c^2*d^7 + 5 *a^4*b^6*c*d^8 - a^5*b^5*d^9)/(b^2*c*d^9*abs(b) - a*b*d^10*abs(b)))*sqrt(b *x + a)/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) + 105/8*(b^6*c^3 - 3*a*b^5*c ^2*d + 3*a^2*b^4*c*d^2 - a^3*b^3*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + s qrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d^5*abs(b))
Timed out. \[ \int \frac {(a+b x)^{9/2}}{(c+d x)^{5/2}} \, dx=\int \frac {{\left (a+b\,x\right )}^{9/2}}{{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:
int((a + b*x)^(9/2)/(c + d*x)^(5/2),x)
Output:
int((a + b*x)^(9/2)/(c + d*x)^(5/2), x)
Time = 17.11 (sec) , antiderivative size = 1159, normalized size of antiderivative = 5.68 \[ \int \frac {(a+b x)^{9/2}}{(c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:
int((b*x+a)^(9/2)/(d*x+c)^(5/2),x)
Output:
( - 128*sqrt(c + d*x)*sqrt(a + b*x)*a**4*d**5 - 1152*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*c*d**4 - 1664*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*d**5*x + 55 44*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**2*c**2*d**3 + 7632*sqrt(c + d*x)*sq rt(a + b*x)*a**2*b**2*c*d**4*x + 1320*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b** 2*d**5*x**2 - 6720*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*c**3*d**2 - 9072*sqr t(c + d*x)*sqrt(a + b*x)*a*b**3*c**2*d**3*x - 1440*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*c*d**4*x**2 + 400*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*d**5*x**3 + 2520*sqrt(c + d*x)*sqrt(a + b*x)*b**4*c**4*d + 3360*sqrt(c + d*x)*sqrt( a + b*x)*b**4*c**3*d**2*x + 504*sqrt(c + d*x)*sqrt(a + b*x)*b**4*c**2*d**3 *x**2 - 144*sqrt(c + d*x)*sqrt(a + b*x)*b**4*c*d**4*x**3 + 64*sqrt(c + d*x )*sqrt(a + b*x)*b**4*d**5*x**4 + 2520*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*b*c**2*d**3 + 5040*s qrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a* d - b*c))*a**3*b*c*d**4*x + 2520*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x ) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*b*d**5*x**2 - 7560*sqrt(d )*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b *c))*a**2*b**2*c**3*d**2 - 15120*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x ) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b**2*c**2*d**3*x - 7560*s qrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a* d - b*c))*a**2*b**2*c*d**4*x**2 + 7560*sqrt(d)*sqrt(b)*log((sqrt(d)*sqr...