3.1692 \(\int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx\)

Optimal. Leaf size=401 \[ \frac {4 \sqrt {2} d^{15/4} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{231 b^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} (b c-a d)^{3/2} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (a+b x)^{3/4} (b c-a d)^2}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (a+b x)^{7/4} (b c-a d)}-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}} \]

[Out]

-4/33*d*(d*x+c)^(1/4)/b^2/(b*x+a)^(11/4)-4/231*d^2*(d*x+c)^(1/4)/b^2/(-a*d+b*c)/(b*x+a)^(7/4)+8/231*d^3*(d*x+c
)^(1/4)/b^2/(-a*d+b*c)^2/(b*x+a)^(3/4)-4/15*(d*x+c)^(5/4)/b/(b*x+a)^(15/4)+4/231*d^(15/4)*((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)/(-a*d+b*c)^(3/2)/(b*x+a)^(3/4)/(d*x+c)^(3/4)/(2*b*d*x+a*d+b*c)/((a*d+b*(
2*d*x+c))^2)^(1/2)

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Rubi [A]  time = 0.43, antiderivative size = 401, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {47, 51, 62, 623, 220} \[ \frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (a+b x)^{3/4} (b c-a d)^2}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (a+b x)^{7/4} (b c-a d)}+\frac {4 \sqrt {2} d^{15/4} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{231 b^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} (b c-a d)^{3/2} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/4)/(a + b*x)^(19/4),x]

[Out]

(-4*d*(c + d*x)^(1/4))/(33*b^2*(a + b*x)^(11/4)) - (4*d^2*(c + d*x)^(1/4))/(231*b^2*(b*c - a*d)*(a + b*x)^(7/4
)) + (8*d^3*(c + d*x)^(1/4))/(231*b^2*(b*c - a*d)^2*(a + b*x)^(3/4)) - (4*(c + d*x)^(5/4))/(15*b*(a + b*x)^(15
/4)) + (4*Sqrt[2]*d^(15/4)*((a + b*x)*(c + d*x))^(3/4)*Sqrt[(b*c + a*d + 2*b*d*x)^2]*(1 + (2*Sqrt[b]*Sqrt[d]*S
qrt[(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]*Sqr
t[(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)*(b*c - a*d)^(3/2)*(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])

Rule 47

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] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && 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 51

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] - Dist[(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] && NeQ[b*c - a*d, 0] && 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 62

Int[((a_.) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Dist[((a + b*x)^m*(c + d*x)^m)/((a + b*x)
*(c + d*x))^m, Int[(a*c + (b*c + a*d)*x + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] &&
LtQ[-1, m, 0] && LeQ[3, Denominator[m], 4]

Rule 220

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

Rule 623

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[(d*Sqrt[(b + 2*c*x)
^2])/(b + 2*c*x), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rubi steps

\begin {align*} \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx &=-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {d \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{15/4}} \, dx}{3 b}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {d^2 \int \frac {1}{(a+b x)^{11/4} (c+d x)^{3/4}} \, dx}{33 b^2}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}-\frac {\left (2 d^3\right ) \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx}{77 b^2 (b c-a d)}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (b c-a d)^2 (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {\left (4 d^4\right ) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx}{231 b^2 (b c-a d)^2}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (b c-a d)^2 (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {\left (4 d^4 ((a+b x) (c+d x))^{3/4}\right ) \int \frac {1}{\left (a c+(b c+a d) x+b d x^2\right )^{3/4}} \, dx}{231 b^2 (b c-a d)^2 (a+b x)^{3/4} (c+d x)^{3/4}}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (b c-a d)^2 (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {\left (16 d^4 ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{231 b^2 (b c-a d)^2 (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x)}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (b c-a d)^2 (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {4 \sqrt {2} d^{15/4} ((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}} 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} (b c-a d)^{3/2} (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}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 73, normalized size = 0.18 \[ -\frac {4 (c+d x)^{5/4} \, _2F_1\left (-\frac {15}{4},-\frac {5}{4};-\frac {11}{4};\frac {d (a+b x)}{a d-b c}\right )}{15 b (a+b x)^{15/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/4)/(a + b*x)^(19/4),x]

[Out]

(-4*(c + d*x)^(5/4)*Hypergeometric2F1[-15/4, -5/4, -11/4, (d*(a + b*x))/(-(b*c) + a*d)])/(15*b*(a + b*x)^(15/4
)*((b*(c + d*x))/(b*c - a*d))^(5/4))

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {5}{4}}}{b^{5} x^{5} + 5 \, a b^{4} x^{4} + 10 \, a^{2} b^{3} x^{3} + 10 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x + a^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(19/4),x, algorithm="fricas")

[Out]

integral((b*x + a)^(1/4)*(d*x + c)^(5/4)/(b^5*x^5 + 5*a*b^4*x^4 + 10*a^2*b^3*x^3 + 10*a^3*b^2*x^2 + 5*a^4*b*x
+ a^5), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {19}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(19/4),x, algorithm="giac")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(19/4), x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {19}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/4)/(b*x+a)^(19/4),x)

[Out]

int((d*x+c)^(5/4)/(b*x+a)^(19/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {19}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(19/4),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(19/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{19/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/4)/(a + b*x)^(19/4),x)

[Out]

int((c + d*x)^(5/4)/(a + b*x)^(19/4), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/4)/(b*x+a)**(19/4),x)

[Out]

Timed out

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