3.1715 \(\int \frac{1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx\)

Optimal. Leaf size=306 \[ -\frac{2 \sqrt{2} d^{3/4} ((a+b x) (c+d x))^{3/4} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right ),\frac{1}{2}\right )}{3 \sqrt [4]{b} (a+b x)^{3/4} (c+d x)^{3/4} \sqrt{b c-a d} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)} \]

[Out]

(-4*(c + d*x)^(1/4))/(3*(b*c - a*d)*(a + b*x)^(3/4)) - (2*Sqrt[2]*d^(3/4)*((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[(Sq
rt[2]*b^(1/4)*d^(1/4)*((a + b*x)*(c + d*x))^(1/4))/Sqrt[b*c - a*d]], 1/2])/(3*b^(1/4)*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 = 0.21517, antiderivative size = 306, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {51, 62, 623, 220} \[ -\frac{2 \sqrt{2} d^{3/4} ((a+b x) (c+d x))^{3/4} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{b} (a+b x)^{3/4} (c+d x)^{3/4} \sqrt{b c-a d} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(7/4)*(c + d*x)^(3/4)),x]

[Out]

(-4*(c + d*x)^(1/4))/(3*(b*c - a*d)*(a + b*x)^(3/4)) - (2*Sqrt[2]*d^(3/4)*((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[(Sq
rt[2]*b^(1/4)*d^(1/4)*((a + b*x)*(c + d*x))^(1/4))/Sqrt[b*c - a*d]], 1/2])/(3*b^(1/4)*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 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{1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx &=-\frac{4 \sqrt [4]{c+d x}}{3 (b c-a d) (a+b x)^{3/4}}-\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 (b c-a d) (a+b x)^{3/4}}-\frac{\left (2 d ((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}{3 (b c-a d) (a+b x)^{3/4} (c+d x)^{3/4}}\\ &=-\frac{4 \sqrt [4]{c+d x}}{3 (b c-a d) (a+b x)^{3/4}}-\frac{\left (8 d ((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 )}{3 (b c-a d) (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x)}\\ &=-\frac{4 \sqrt [4]{c+d x}}{3 (b c-a d) (a+b x)^{3/4}}-\frac{2 \sqrt{2} d^{3/4} ((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 )}{3 \sqrt [4]{b} \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 = 0.0243104, size = 73, normalized size = 0.24 \[ -\frac{4 \left (\frac{b (c+d x)}{b c-a d}\right )^{3/4} \, _2F_1\left (-\frac{3}{4},\frac{3}{4};\frac{1}{4};\frac{d (a+b x)}{a d-b c}\right )}{3 b (a+b x)^{3/4} (c+d x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(7/4)*(c + d*x)^(3/4)),x]

[Out]

(-4*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[-3/4, 3/4, 1/4, (d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(a
+ b*x)^(3/4)*(c + d*x)^(3/4))

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Maple [F]  time = 0.039, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{-{\frac{7}{4}}} \left ( dx+c \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(7/4)/(d*x+c)^(3/4),x)

[Out]

int(1/(b*x+a)^(7/4)/(d*x+c)^(3/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{7}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(7/4)/(d*x+c)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(7/4)*(d*x + c)^(3/4)), 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{1}{4}}}{b^{2} d x^{3} + a^{2} c +{\left (b^{2} c + 2 \, a b d\right )} x^{2} +{\left (2 \, a b c + a^{2} d\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(7/4)/(d*x+c)^(3/4),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x\right )^{\frac{7}{4}} \left (c + d x\right )^{\frac{3}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(7/4)/(d*x+c)**(3/4),x)

[Out]

Integral(1/((a + b*x)**(7/4)*(c + d*x)**(3/4)), x)

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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(1/(b*x+a)^(7/4)/(d*x+c)^(3/4),x, algorithm="giac")

[Out]

Timed out