3.17.91 \(\int \frac {(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx\) [1691]

3.17.91.1 Optimal result

Integrand size = 19, antiderivative size = 363 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx=-\frac {20 d \sqrt [4]{c+d x}}{77 b^2 (a+b x)^{7/4}}-\frac {20 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}}-\frac {10 \sqrt {2} d^{11/4} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{231 b^{9/4} \sqrt {b c-a d} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]

output

-20/77*d*(d*x+c)^(1/4)/b^2/(b*x+a)^(7/4)-20/231*d^2*(d*x+c)^(1/4)/b^2/(-a* 
d+b*c)/(b*x+a)^(3/4)-4/11*(d*x+c)^(5/4)/b/(b*x+a)^(11/4)-10/231*d^(11/4)*( 
(b*x+a)*(d*x+c))^(3/4)*(cos(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/ 
4)*2^(1/2)/(-a*d+b*c)^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*d^(1/4)*((b*x+ 
a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))*EllipticF(sin(2*arctan(b^(1/4 
)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2))),1/2*2^(1/2))* 
2^(1/2)*(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))*((2*b*d*x 
+a*d+b*c)^2)^(1/2)*((a*d+b*(2*d*x+c))^2/(-a*d+b*c)^2/(1+2*b^(1/2)*d^(1/2)* 
((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))^2)^(1/2)/b^(9/4)/(b*x+a)^(3/4)/(d*x+c) 
^(3/4)/(2*b*d*x+a*d+b*c)/(-a*d+b*c)^(1/2)/((a*d+b*(2*d*x+c))^2)^(1/2)
 

3.17.91.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.20 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx=-\frac {4 (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {5}{4},-\frac {7}{4},\frac {d (a+b x)}{-b c+a d}\right )}{11 b (a+b x)^{11/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \]

input

Integrate[(c + d*x)^(5/4)/(a + b*x)^(15/4),x]
 

output

(-4*(c + d*x)^(5/4)*Hypergeometric2F1[-11/4, -5/4, -7/4, (d*(a + b*x))/(-( 
b*c) + a*d)])/(11*b*(a + b*x)^(11/4)*((b*(c + d*x))/(b*c - a*d))^(5/4))
 

3.17.91.3 Rubi [A] (warning: unable to verify)

Time = 0.33 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.60, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {57, 57, 61, 73, 768, 858, 807, 229}

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.

\(\displaystyle \int \frac {(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {5 d \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{11/4}}dx}{11 b}-\frac {4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}}\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {5 d \left (\frac {d \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}}dx}{7 b}-\frac {4 \sqrt [4]{c+d x}}{7 b (a+b x)^{7/4}}\right )}{11 b}-\frac {4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {5 d \left (\frac {d \left (-\frac {2 d \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}}dx}{3 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 b}-\frac {4 \sqrt [4]{c+d x}}{7 b (a+b x)^{7/4}}\right )}{11 b}-\frac {4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {5 d \left (\frac {d \left (-\frac {8 d \int \frac {1}{\left (c-\frac {a d}{b}+\frac {d (a+b x)}{b}\right )^{3/4}}d\sqrt [4]{a+b x}}{3 b (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 b}-\frac {4 \sqrt [4]{c+d x}}{7 b (a+b x)^{7/4}}\right )}{11 b}-\frac {4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {5 d \left (\frac {d \left (-\frac {8 d (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{(a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4}}d\sqrt [4]{a+b x}}{3 b (b c-a d) \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 b}-\frac {4 \sqrt [4]{c+d x}}{7 b (a+b x)^{7/4}}\right )}{11 b}-\frac {4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {5 d \left (\frac {d \left (\frac {8 d (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{a+b x} \left (\frac {(b c-a d) (a+b x)}{d}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{a+b x}}}{3 b (b c-a d) \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 b}-\frac {4 \sqrt [4]{c+d x}}{7 b (a+b x)^{7/4}}\right )}{11 b}-\frac {4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {5 d \left (\frac {d \left (\frac {4 d (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {a+b x} (b c-a d)}{d}+1\right )^{3/4}}d\sqrt {a+b x}}{3 b (b c-a d) \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 b}-\frac {4 \sqrt [4]{c+d x}}{7 b (a+b x)^{7/4}}\right )}{11 b}-\frac {4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {5 d \left (\frac {d \left (\frac {8 d^{3/2} (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} \sqrt {a+b x}}{\sqrt {d}}\right ),2\right )}{3 b (b c-a d)^{3/2} \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 b}-\frac {4 \sqrt [4]{c+d x}}{7 b (a+b x)^{7/4}}\right )}{11 b}-\frac {4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}}\)

input

Int[(c + d*x)^(5/4)/(a + b*x)^(15/4),x]
 

output

(-4*(c + d*x)^(5/4))/(11*b*(a + b*x)^(11/4)) + (5*d*((-4*(c + d*x)^(1/4))/ 
(7*b*(a + b*x)^(7/4)) + (d*((-4*(c + d*x)^(1/4))/(3*(b*c - a*d)*(a + b*x)^ 
(3/4)) + (8*d^(3/2)*(a + b*x)^(3/4)*(1 + (b*c - a*d)/(d*(a + b*x)))^(3/4)* 
EllipticF[ArcTan[(Sqrt[b*c - a*d]*Sqrt[a + b*x])/Sqrt[d]]/2, 2])/(3*b*(b*c 
 - a*d)^(3/2)*(c - (a*d)/b + (d*(a + b*x))/b)^(3/4))))/(7*b)))/(11*b)
 

3.17.91.3.1 Defintions of rubi rules used

rule 57

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]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

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]
 

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

3.17.91.4 Maple [F]

\[\int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {15}{4}}}d x\]

input

int((d*x+c)^(5/4)/(b*x+a)^(15/4),x)
 

output

int((d*x+c)^(5/4)/(b*x+a)^(15/4),x)
 

3.17.91.5 Fricas [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {15}{4}}} \,d x } \]

input

integrate((d*x+c)^(5/4)/(b*x+a)^(15/4),x, algorithm="fricas")
 

output

integral((b*x + a)^(1/4)*(d*x + c)^(5/4)/(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^ 
2*x^2 + 4*a^3*b*x + a^4), x)
 

3.17.91.6 Sympy [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {15}{4}}}\, dx \]

input

integrate((d*x+c)**(5/4)/(b*x+a)**(15/4),x)
 

output

Integral((c + d*x)**(5/4)/(a + b*x)**(15/4), x)
 

3.17.91.7 Maxima [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {15}{4}}} \,d x } \]

input

integrate((d*x+c)^(5/4)/(b*x+a)^(15/4),x, algorithm="maxima")
 

output

integrate((d*x + c)^(5/4)/(b*x + a)^(15/4), x)
 

3.17.91.8 Giac [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {15}{4}}} \,d x } \]

input

integrate((d*x+c)^(5/4)/(b*x+a)^(15/4),x, algorithm="giac")
 

output

integrate((d*x + c)^(5/4)/(b*x + a)^(15/4), x)
 

3.17.91.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{15/4}} \,d x \]

input

int((c + d*x)^(5/4)/(a + b*x)^(15/4),x)
 

output

int((c + d*x)^(5/4)/(a + b*x)^(15/4), x)