3.1691 \(\int \frac{(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx\)

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

[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[(Sqr
t[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 = 0.352148, 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, Rules used = {47, 51, 62, 623, 220} \[ -\frac{20 d^2 \sqrt [4]{c+d x}}{231 b^2 (a+b x)^{3/4} (b c-a d)}-\frac{10 \sqrt{2} d^{11/4} ((a+b x) (c+d x))^{3/4} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{231 b^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} \sqrt{b c-a d} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{20 d \sqrt [4]{c+d x}}{77 b^2 (a+b x)^{7/4}}-\frac{4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}} \]

Antiderivative was successfully verified.

[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[(Sqr
t[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])

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 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]

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]

Rubi steps

\begin{align*} \int \frac{(c+d x)^{5/4}}{(a+b x)^{15/4}} \, dx &=-\frac{4 (c+d x)^{5/4}}{11 b (a+b x)^{11/4}}+\frac{(5 d) \int \frac{\sqrt [4]{c+d x}}{(a+b x)^{11/4}} \, dx}{11 b}\\ &=-\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}}+\frac{\left (5 d^2\right ) \int \frac{1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx}{77 b^2}\\ &=-\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{\left (10 d^3\right ) \int \frac{1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx}{231 b^2 (b c-a d)}\\ &=-\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{\left (10 d^3 ((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) (a+b x)^{3/4} (c+d x)^{3/4}}\\ &=-\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{\left (40 d^3 ((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) (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x)}\\ &=-\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}} 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} \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}}\\ \end{align*}

Mathematica [C]  time = 0.0526712, size = 73, normalized size = 0.2 \[ -\frac{4 (c+d x)^{5/4} \, _2F_1\left (-\frac{11}{4},-\frac{5}{4};-\frac{7}{4};\frac{d (a+b x)}{a d-b c}\right )}{11 b (a+b x)^{11/4} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-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))

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Maple [F]  time = 0.042, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx+c \right ) ^{{\frac{5}{4}}} \left ( bx+a \right ) ^{-{\frac{15}{4}}}}\, dx \end{align*}

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. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{5}{4}}}{{\left (b x + a\right )}^{\frac{15}{4}}}\,{d x} \end{align*}

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. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{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\right ) \end{align*}

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((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)

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

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

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