3.1692 \(\int \frac{(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx\)

Optimal. Leaf size=363 \[ -\frac{10 \sqrt{2} d^{11/4} ((a+b x) (c+d x))^{3/4} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{231 b^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} \sqrt{b c-a d} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{20 d^2 \sqrt [4]{c+d x}}{231 b^2 (a+b x)^{3/4} (b c-a d)}-\frac{20 d \sqrt [4]{c+d x}}{77 b^2 (a+b x)^{7/4}}-\frac{4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}} \]

[Out]

(-20*d*(c + d*x)^(1/4))/(77*b^2*(a + b*x)^(7/4)) - (20*d^2*(c + d*x)^(1/4))/(231
*b^2*(b*c - a*d)*(a + b*x)^(3/4)) - (4*(c + d*x)^(5/4))/(11*b*(a + b*x)^(11/4))
- (10*Sqrt[2]*d^(11/4)*((a + b*x)*(c + d*x))^(3/4)*Sqrt[(b*c + a*d + 2*b*d*x)^2]
*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))*Sqrt[(a*d + b*(
c + 2*d*x))^2/((b*c - a*d)^2*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/
(b*c - a*d))^2)]*EllipticF[2*ArcTan[(Sqrt[2]*b^(1/4)*d^(1/4)*((a + b*x)*(c + d*x
))^(1/4))/Sqrt[b*c - a*d]], 1/2])/(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])

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Rubi [A]  time = 0.829174, antiderivative size = 363, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263 \[ -\frac{10 \sqrt{2} d^{11/4} ((a+b x) (c+d x))^{3/4} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{231 b^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} \sqrt{b c-a d} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{20 d^2 \sqrt [4]{c+d x}}{231 b^2 (a+b x)^{3/4} (b c-a d)}-\frac{20 d \sqrt [4]{c+d x}}{77 b^2 (a+b x)^{7/4}}-\frac{4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}} \]

Warning: Unable to verify antiderivative.

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

[Out]

(-20*d*(c + d*x)^(1/4))/(77*b^2*(a + b*x)^(7/4)) - (20*d^2*(c + d*x)^(1/4))/(231
*b^2*(b*c - a*d)*(a + b*x)^(3/4)) - (4*(c + d*x)^(5/4))/(11*b*(a + b*x)^(11/4))
- (10*Sqrt[2]*d^(11/4)*((a + b*x)*(c + d*x))^(3/4)*Sqrt[(b*c + a*d + 2*b*d*x)^2]
*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))*Sqrt[(a*d + b*(
c + 2*d*x))^2/((b*c - a*d)^2*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/
(b*c - a*d))^2)]*EllipticF[2*ArcTan[(Sqrt[2]*b^(1/4)*d^(1/4)*((a + b*x)*(c + d*x
))^(1/4))/Sqrt[b*c - a*d]], 1/2])/(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])

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Rubi in Sympy [A]  time = 72.1444, size = 411, normalized size = 1.13 \[ - \frac{4 \left (c + d x\right )^{\frac{5}{4}}}{11 b \left (a + b x\right )^{\frac{11}{4}}} + \frac{20 d^{2} \sqrt [4]{c + d x}}{231 b^{2} \left (a + b x\right )^{\frac{3}{4}} \left (a d - b c\right )} - \frac{20 d \sqrt [4]{c + d x}}{77 b^{2} \left (a + b x\right )^{\frac{7}{4}}} + \frac{10 \sqrt{2} d^{\frac{11}{4}} \sqrt{\frac{b d \left (4 a c + 4 b d x^{2} + x \left (4 a d + 4 b c\right )\right ) + \left (a d - b c\right )^{2}}{\left (a d - b c\right )^{2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{a c + b d x^{2} + x \left (a d + b c\right )}}{a d - b c} + 1\right )^{2}}} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{a c + b d x^{2} + x \left (a d + b c\right )}}{a d - b c} + 1\right ) \left (a c + b d x^{2} + x \left (a d + b c\right )\right )^{\frac{3}{4}} \sqrt{\left (a d + b c + 2 b d x\right )^{2}} F\left (2 \operatorname{atan}{\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{a c + b d x^{2} + x \left (a d + b c\right )}}{\sqrt{a d - b c}} \right )}\middle | \frac{1}{2}\right )}{231 b^{\frac{9}{4}} \left (a + b x\right )^{\frac{3}{4}} \left (c + d x\right )^{\frac{3}{4}} \sqrt{a d - b c} \sqrt{b d \left (4 a c + 4 b d x^{2} + x \left (4 a d + 4 b c\right )\right ) + \left (a d - b c\right )^{2}} \left (a d + b c + 2 b d x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-4*(c + d*x)**(5/4)/(11*b*(a + b*x)**(11/4)) + 20*d**2*(c + d*x)**(1/4)/(231*b**
2*(a + b*x)**(3/4)*(a*d - b*c)) - 20*d*(c + d*x)**(1/4)/(77*b**2*(a + b*x)**(7/4
)) + 10*sqrt(2)*d**(11/4)*sqrt((b*d*(4*a*c + 4*b*d*x**2 + x*(4*a*d + 4*b*c)) + (
a*d - b*c)**2)/((a*d - b*c)**2*(2*sqrt(b)*sqrt(d)*sqrt(a*c + b*d*x**2 + x*(a*d +
b*c))/(a*d - b*c) + 1)**2))*(2*sqrt(b)*sqrt(d)*sqrt(a*c + b*d*x**2 + x*(a*d + b
*c))/(a*d - b*c) + 1)*(a*c + b*d*x**2 + x*(a*d + b*c))**(3/4)*sqrt((a*d + b*c +
2*b*d*x)**2)*elliptic_f(2*atan(sqrt(2)*b**(1/4)*d**(1/4)*(a*c + b*d*x**2 + x*(a*
d + b*c))**(1/4)/sqrt(a*d - b*c)), 1/2)/(231*b**(9/4)*(a + b*x)**(3/4)*(c + d*x)
**(3/4)*sqrt(a*d - b*c)*sqrt(b*d*(4*a*c + 4*b*d*x**2 + x*(4*a*d + 4*b*c)) + (a*d
- b*c)**2)*(a*d + b*c + 2*b*d*x))

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Mathematica [C]  time = 0.323713, size = 140, normalized size = 0.39 \[ \frac{4 \sqrt [4]{c+d x} \left (-10 a^2 d^2+10 d^2 (a+b x)^2 \left (\frac{d (a+b x)}{a d-b c}\right )^{3/4} \, _2F_1\left (\frac{1}{4},\frac{3}{4};\frac{5}{4};\frac{b (c+d x)}{b c-a d}\right )-2 a b d (3 c+13 d x)+b^2 \left (21 c^2+36 c d x+5 d^2 x^2\right )\right )}{231 b^2 (a+b x)^{11/4} (a d-b c)} \]

Antiderivative was successfully verified.

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

[Out]

(4*(c + d*x)^(1/4)*(-10*a^2*d^2 - 2*a*b*d*(3*c + 13*d*x) + b^2*(21*c^2 + 36*c*d*
x + 5*d^2*x^2) + 10*d^2*(a + b*x)^2*((d*(a + b*x))/(-(b*c) + a*d))^(3/4)*Hyperge
ometric2F1[1/4, 3/4, 5/4, (b*(c + d*x))/(b*c - a*d)]))/(231*b^2*(-(b*c) + a*d)*(
a + b*x)^(11/4))

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \[ \int{1 \left ( dx+c \right ) ^{{\frac{5}{4}}} \left ( bx+a \right ) ^{-{\frac{15}{4}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (d x + c\right )}^{\frac{5}{4}}}{{\left (b x + a\right )}^{\frac{15}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (d x + c\right )}^{\frac{5}{4}}}{{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}{\left (b x + a\right )}^{\frac{3}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out

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GIAC/XCAS [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out