Integrand size = 28, antiderivative size = 109 \[ \int \left (a+b \sqrt {x}\right )^p \left (c+d \sqrt {x}\right )^q (e x)^m \, dx=\frac {\left (a+b \sqrt {x}\right )^p \left (1+\frac {b \sqrt {x}}{a}\right )^{-p} \left (c+d \sqrt {x}\right )^q \left (1+\frac {d \sqrt {x}}{c}\right )^{-q} (e x)^{1+m} \operatorname {AppellF1}\left (2 (1+m),-p,-q,3+2 m,-\frac {b \sqrt {x}}{a},-\frac {d \sqrt {x}}{c}\right )}{e (1+m)} \] Output:
(a+b*x^(1/2))^p*(c+d*x^(1/2))^q*(e*x)^(1+m)*AppellF1(2+2*m,-p,-q,3+2*m,-b* x^(1/2)/a,-d*x^(1/2)/c)/e/(1+m)/((1+b*x^(1/2)/a)^p)/((1+d*x^(1/2)/c)^q)
Time = 0.21 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.96 \[ \int \left (a+b \sqrt {x}\right )^p \left (c+d \sqrt {x}\right )^q (e x)^m \, dx=\frac {\left (a+b \sqrt {x}\right )^p \left (1+\frac {b \sqrt {x}}{a}\right )^{-p} \left (c+d \sqrt {x}\right )^q \left (1+\frac {d \sqrt {x}}{c}\right )^{-q} x (e x)^m \operatorname {AppellF1}\left (2+2 m,-p,-q,3+2 m,-\frac {b \sqrt {x}}{a},-\frac {d \sqrt {x}}{c}\right )}{1+m} \] Input:
Integrate[(a + b*Sqrt[x])^p*(c + d*Sqrt[x])^q*(e*x)^m,x]
Output:
((a + b*Sqrt[x])^p*(c + d*Sqrt[x])^q*x*(e*x)^m*AppellF1[2 + 2*m, -p, -q, 3 + 2*m, -((b*Sqrt[x])/a), -((d*Sqrt[x])/c)])/((1 + m)*(1 + (b*Sqrt[x])/a)^ p*(1 + (d*Sqrt[x])/c)^q)
Time = 0.42 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1001, 1000, 152, 152, 150}
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.
\(\displaystyle \int (e x)^m \left (a+b \sqrt {x}\right )^p \left (c+d \sqrt {x}\right )^q \, dx\) |
\(\Big \downarrow \) 1001 |
\(\displaystyle x^{-m} (e x)^m \int \left (a+b \sqrt {x}\right )^p \left (c+d \sqrt {x}\right )^q x^mdx\) |
\(\Big \downarrow \) 1000 |
\(\displaystyle 2 x^{-m} (e x)^m \int \left (a+b \sqrt {x}\right )^p \left (c+d \sqrt {x}\right )^q x^{\frac {1}{2} (2 m+1)}d\sqrt {x}\) |
\(\Big \downarrow \) 152 |
\(\displaystyle 2 x^{-m} (e x)^m \left (a+b \sqrt {x}\right )^p \left (\frac {b \sqrt {x}}{a}+1\right )^{-p} \int \left (\frac {\sqrt {x} b}{a}+1\right )^p \left (c+d \sqrt {x}\right )^q x^{\frac {1}{2} (2 m+1)}d\sqrt {x}\) |
\(\Big \downarrow \) 152 |
\(\displaystyle 2 x^{-m} (e x)^m \left (a+b \sqrt {x}\right )^p \left (\frac {b \sqrt {x}}{a}+1\right )^{-p} \left (c+d \sqrt {x}\right )^q \left (\frac {d \sqrt {x}}{c}+1\right )^{-q} \int \left (\frac {\sqrt {x} b}{a}+1\right )^p \left (\frac {\sqrt {x} d}{c}+1\right )^q x^{\frac {1}{2} (2 m+1)}d\sqrt {x}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {x (e x)^m \left (a+b \sqrt {x}\right )^p \left (\frac {b \sqrt {x}}{a}+1\right )^{-p} \left (c+d \sqrt {x}\right )^q \left (\frac {d \sqrt {x}}{c}+1\right )^{-q} \operatorname {AppellF1}\left (2 (m+1),-p,-q,2 m+3,-\frac {b \sqrt {x}}{a},-\frac {d \sqrt {x}}{c}\right )}{m+1}\) |
Input:
Int[(a + b*Sqrt[x])^p*(c + d*Sqrt[x])^q*(e*x)^m,x]
Output:
((a + b*Sqrt[x])^p*(c + d*Sqrt[x])^q*x*(e*x)^m*AppellF1[2*(1 + m), -p, -q, 3 + 2*m, -((b*Sqrt[x])/a), -((d*Sqrt[x])/c)])/((1 + m)*(1 + (b*Sqrt[x])/a )^p*(1 + (d*Sqrt[x])/c)^q)
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{g = Denominator[n]}, Simp[g Subst[Int[x^(g*(m + 1) - 1)*(a + b*x^(g*n))^p*(c + d*x^(g*n))^q, x], x, x^(1/g)], x]] /; FreeQ[{a, b , c, d, m, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[n]
Int[((e_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)) ^(q_), x_Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m]) I nt[x^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, p, q }, x] && NeQ[b*c - a*d, 0] && FractionQ[n]
\[\int \left (a +b \sqrt {x}\right )^{p} \left (c +d \sqrt {x}\right )^{q} \left (e x \right )^{m}d x\]
Input:
int((a+b*x^(1/2))^p*(c+d*x^(1/2))^q*(e*x)^m,x)
Output:
int((a+b*x^(1/2))^p*(c+d*x^(1/2))^q*(e*x)^m,x)
Exception generated. \[ \int \left (a+b \sqrt {x}\right )^p \left (c+d \sqrt {x}\right )^q (e x)^m \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*x^(1/2))^p*(c+d*x^(1/2))^q*(e*x)^m,x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: algl ogextint: unimplemented
Exception generated. \[ \int \left (a+b \sqrt {x}\right )^p \left (c+d \sqrt {x}\right )^q (e x)^m \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:
integrate((a+b*x**(1/2))**p*(c+d*x**(1/2))**q*(e*x)**m,x)
Output:
Exception raised: HeuristicGCDFailed >> no luck
\[ \int \left (a+b \sqrt {x}\right )^p \left (c+d \sqrt {x}\right )^q (e x)^m \, dx=\int { \left (e x\right )^{m} {\left (b \sqrt {x} + a\right )}^{p} {\left (d \sqrt {x} + c\right )}^{q} \,d x } \] Input:
integrate((a+b*x^(1/2))^p*(c+d*x^(1/2))^q*(e*x)^m,x, algorithm="maxima")
Output:
integrate((e*x)^m*(b*sqrt(x) + a)^p*(d*sqrt(x) + c)^q, x)
\[ \int \left (a+b \sqrt {x}\right )^p \left (c+d \sqrt {x}\right )^q (e x)^m \, dx=\int { \left (e x\right )^{m} {\left (b \sqrt {x} + a\right )}^{p} {\left (d \sqrt {x} + c\right )}^{q} \,d x } \] Input:
integrate((a+b*x^(1/2))^p*(c+d*x^(1/2))^q*(e*x)^m,x, algorithm="giac")
Output:
integrate((e*x)^m*(b*sqrt(x) + a)^p*(d*sqrt(x) + c)^q, x)
Timed out. \[ \int \left (a+b \sqrt {x}\right )^p \left (c+d \sqrt {x}\right )^q (e x)^m \, dx=\int {\left (e\,x\right )}^m\,{\left (a+b\,\sqrt {x}\right )}^p\,{\left (c+d\,\sqrt {x}\right )}^q \,d x \] Input:
int((e*x)^m*(a + b*x^(1/2))^p*(c + d*x^(1/2))^q,x)
Output:
int((e*x)^m*(a + b*x^(1/2))^p*(c + d*x^(1/2))^q, x)
\[ \int \left (a+b \sqrt {x}\right )^p \left (c+d \sqrt {x}\right )^q (e x)^m \, dx=\text {too large to display} \] Input:
int((a+b*x^(1/2))^p*(c+d*x^(1/2))^q*(e*x)^m,x)
Output:
(e**m*(4*x**((2*m + 1)/2)*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a**2*d**2* m*p + 2*x**((2*m + 1)/2)*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a**2*d**2*p *q + 4*x**((2*m + 1)/2)*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a*b*c*d*m*p + 4*x**((2*m + 1)/2)*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a*b*c*d*m*q + 2 *x**((2*m + 1)/2)*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a*b*c*d*p**2 + 2*x **((2*m + 1)/2)*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a*b*c*d*q**2 + 4*x** ((2*m + 1)/2)*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*b**2*c**2*m*q + 2*x**( (2*m + 1)/2)*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*b**2*c**2*p*q - 4*x**m* (sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a**2*c*d*m*p - 2*x**m*(sqrt(x)*d + c )**q*(sqrt(x)*b + a)**p*a**2*c*d*p - 4*x**m*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a*b*c**2*m*q - 2*x**m*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a*b*c* *2*q + 8*x**m*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a*b*d**2*m**2*x + 4*x* *m*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a*b*d**2*m*p*x + 8*x**m*(sqrt(x)* d + c)**q*(sqrt(x)*b + a)**p*a*b*d**2*m*q*x + 4*x**m*(sqrt(x)*d + c)**q*(s qrt(x)*b + a)**p*a*b*d**2*m*x + 2*x**m*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)* *p*a*b*d**2*p*q*x + 2*x**m*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a*b*d**2* q**2*x + 2*x**m*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*a*b*d**2*q*x + 8*x** m*(sqrt(x)*d + c)**q*(sqrt(x)*b + a)**p*b**2*c*d*m**2*x + 8*x**m*(sqrt(x)* d + c)**q*(sqrt(x)*b + a)**p*b**2*c*d*m*p*x + 4*x**m*(sqrt(x)*d + c)**q*(s qrt(x)*b + a)**p*b**2*c*d*m*q*x + 4*x**m*(sqrt(x)*d + c)**q*(sqrt(x)*b ...