Integrand size = 19, antiderivative size = 220 \[ \int (a+b x)^{3/2} (c+d x)^{5/4} \, dx=-\frac {8 (b c-a d)^3 \sqrt {a+b x} \sqrt [4]{c+d x}}{231 b^2 d^2}+\frac {4 (b c-a d)^2 (a+b x)^{3/2} \sqrt [4]{c+d x}}{231 b^2 d}+\frac {4 (b c-a d) (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 b^2}+\frac {4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}+\frac {16 (b c-a d)^{17/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{231 b^{9/4} d^3 \sqrt {a+b x}} \] Output:
-8/231*(-a*d+b*c)^3*(b*x+a)^(1/2)*(d*x+c)^(1/4)/b^2/d^2+4/231*(-a*d+b*c)^2 *(b*x+a)^(3/2)*(d*x+c)^(1/4)/b^2/d+4/33*(-a*d+b*c)*(b*x+a)^(5/2)*(d*x+c)^( 1/4)/b^2+4/15*(b*x+a)^(5/2)*(d*x+c)^(5/4)/b+16/231*(-a*d+b*c)^(17/4)*(-d*( b*x+a)/(-a*d+b*c))^(1/2)*EllipticF(b^(1/4)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4), I)/b^(9/4)/d^3/(b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.33 \[ \int (a+b x)^{3/2} (c+d x)^{5/4} \, dx=\frac {2 (a+b x)^{5/2} (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {5}{2},\frac {7}{2},\frac {d (a+b x)}{-b c+a d}\right )}{5 b \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \] Input:
Integrate[(a + b*x)^(3/2)*(c + d*x)^(5/4),x]
Output:
(2*(a + b*x)^(5/2)*(c + d*x)^(5/4)*Hypergeometric2F1[-5/4, 5/2, 7/2, (d*(a + b*x))/(-(b*c) + a*d)])/(5*b*((b*(c + d*x))/(b*c - a*d))^(5/4))
Time = 0.31 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.14, 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, 73, 765, 762}
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.
\(\displaystyle \int (a+b x)^{3/2} (c+d x)^{5/4} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \int (a+b x)^{3/2} \sqrt [4]{c+d x}dx}{3 b}+\frac {4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \left (\frac {(b c-a d) \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}}dx}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\right )}{3 b}+\frac {4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \left (\frac {(b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}}dx}{7 d}\right )}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\right )}{3 b}+\frac {4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \left (\frac {(b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {2 (b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{3 d}\right )}{7 d}\right )}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\right )}{3 b}+\frac {4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(b c-a d) \left (\frac {(b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{3 d^2}\right )}{7 d}\right )}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\right )}{3 b}+\frac {4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle \frac {(b c-a d) \left (\frac {(b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{3 d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{7 d}\right )}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\right )}{3 b}+\frac {4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {(b c-a d) \left (\frac {(b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d)^{5/4} \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 \sqrt [4]{b} d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{7 d}\right )}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\right )}{3 b}+\frac {4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}\) |
Input:
Int[(a + b*x)^(3/2)*(c + d*x)^(5/4),x]
Output:
(4*(a + b*x)^(5/2)*(c + d*x)^(5/4))/(15*b) + ((b*c - a*d)*((4*(a + b*x)^(5 /2)*(c + d*x)^(1/4))/(11*b) + ((b*c - a*d)*((4*(a + b*x)^(3/2)*(c + d*x)^( 1/4))/(7*d) - (6*(b*c - a*d)*((4*Sqrt[a + b*x]*(c + d*x)^(1/4))/(3*d) - (8 *(b*c - a*d)^(5/4)*Sqrt[1 - (b*(c + d*x))/(b*c - a*d)]*EllipticF[ArcSin[(b ^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(3*b^(1/4)*d^2*Sqrt[a - ( b*c)/d + (b*(c + d*x))/d])))/(7*d)))/(11*b)))/(3*b)
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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt [a + b*x^4] Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ [b/a] && !GtQ[a, 0]
\[\int \left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {5}{4}}d x\]
Input:
int((b*x+a)^(3/2)*(d*x+c)^(5/4),x)
Output:
int((b*x+a)^(3/2)*(d*x+c)^(5/4),x)
\[ \int (a+b x)^{3/2} (c+d x)^{5/4} \, dx=\int { {\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {5}{4}} \,d x } \] Input:
integrate((b*x+a)^(3/2)*(d*x+c)^(5/4),x, algorithm="fricas")
Output:
integral((b*d*x^2 + a*c + (b*c + a*d)*x)*sqrt(b*x + a)*(d*x + c)^(1/4), x)
\[ \int (a+b x)^{3/2} (c+d x)^{5/4} \, dx=\int \left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{4}}\, dx \] Input:
integrate((b*x+a)**(3/2)*(d*x+c)**(5/4),x)
Output:
Integral((a + b*x)**(3/2)*(c + d*x)**(5/4), x)
\[ \int (a+b x)^{3/2} (c+d x)^{5/4} \, dx=\int { {\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {5}{4}} \,d x } \] Input:
integrate((b*x+a)^(3/2)*(d*x+c)^(5/4),x, algorithm="maxima")
Output:
integrate((b*x + a)^(3/2)*(d*x + c)^(5/4), x)
\[ \int (a+b x)^{3/2} (c+d x)^{5/4} \, dx=\int { {\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {5}{4}} \,d x } \] Input:
integrate((b*x+a)^(3/2)*(d*x+c)^(5/4),x, algorithm="giac")
Output:
integrate((b*x + a)^(3/2)*(d*x + c)^(5/4), x)
Timed out. \[ \int (a+b x)^{3/2} (c+d x)^{5/4} \, dx=\int {\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/4} \,d x \] Input:
int((a + b*x)^(3/2)*(c + d*x)^(5/4),x)
Output:
int((a + b*x)^(3/2)*(c + d*x)^(5/4), x)
\[ \int (a+b x)^{3/2} (c+d x)^{5/4} \, dx =\text {Too large to display} \] Input:
int((b*x+a)^(3/2)*(d*x+c)^(5/4),x)
Output:
(4*( - 48*(c + d*x)**(1/4)*sqrt(a + b*x)*a**3*c*d**2 + 12*(c + d*x)**(1/4) *sqrt(a + b*x)*a**3*d**3*x + 299*(c + d*x)**(1/4)*sqrt(a + b*x)*a**2*b*c** 2*d + 238*(c + d*x)**(1/4)*sqrt(a + b*x)*a**2*b*c*d**2*x + 119*(c + d*x)** (1/4)*sqrt(a + b*x)*a**2*b*d**3*x**2 - 20*(c + d*x)**(1/4)*sqrt(a + b*x)*a *b**2*c**3 + 433*(c + d*x)**(1/4)*sqrt(a + b*x)*a*b**2*c**2*d*x + 350*(c + d*x)**(1/4)*sqrt(a + b*x)*a*b**2*c*d**2*x**2 + 77*(c + d*x)**(1/4)*sqrt(a + b*x)*a*b**2*d**3*x**3 + 10*(c + d*x)**(1/4)*sqrt(a + b*x)*b**3*c**3*x + 224*(c + d*x)**(1/4)*sqrt(a + b*x)*b**3*c**2*d*x**2 + 154*(c + d*x)**(1/4 )*sqrt(a + b*x)*b**3*c*d**2*x**3 - 15*int(((c + d*x)**(1/4)*sqrt(a + b*x)* x)/(a**2*c*d + a**2*d**2*x + 2*a*b*c**2 + 3*a*b*c*d*x + a*b*d**2*x**2 + 2* b**2*c**2*x + 2*b**2*c*d*x**2),x)*a**5*d**5 + 30*int(((c + d*x)**(1/4)*sqr t(a + b*x)*x)/(a**2*c*d + a**2*d**2*x + 2*a*b*c**2 + 3*a*b*c*d*x + a*b*d** 2*x**2 + 2*b**2*c**2*x + 2*b**2*c*d*x**2),x)*a**4*b*c*d**4 + 30*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**2*c*d + a**2*d**2*x + 2*a*b*c**2 + 3*a*b* c*d*x + a*b*d**2*x**2 + 2*b**2*c**2*x + 2*b**2*c*d*x**2),x)*a**3*b**2*c**2 *d**3 - 120*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**2*c*d + a**2*d**2*x + 2*a*b*c**2 + 3*a*b*c*d*x + a*b*d**2*x**2 + 2*b**2*c**2*x + 2*b**2*c*d*x **2),x)*a**2*b**3*c**3*d**2 + 105*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/( a**2*c*d + a**2*d**2*x + 2*a*b*c**2 + 3*a*b*c*d*x + a*b*d**2*x**2 + 2*b**2 *c**2*x + 2*b**2*c*d*x**2),x)*a*b**4*c**4*d - 30*int(((c + d*x)**(1/4)*...