3.1500 \(\int \frac{1}{(a+b x)^{9/2} \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=136 \[ \frac{32 d^3 \sqrt{c+d x}}{35 \sqrt{a+b x} (b c-a d)^4}-\frac{16 d^2 \sqrt{c+d x}}{35 (a+b x)^{3/2} (b c-a d)^3}+\frac{12 d \sqrt{c+d x}}{35 (a+b x)^{5/2} (b c-a d)^2}-\frac{2 \sqrt{c+d x}}{7 (a+b x)^{7/2} (b c-a d)} \]

[Out]

(-2*Sqrt[c + d*x])/(7*(b*c - a*d)*(a + b*x)^(7/2)) + (12*d*Sqrt[c + d*x])/(35*(b*c - a*d)^2*(a + b*x)^(5/2)) -
(16*d^2*Sqrt[c + d*x])/(35*(b*c - a*d)^3*(a + b*x)^(3/2)) + (32*d^3*Sqrt[c + d*x])/(35*(b*c - a*d)^4*Sqrt[a +
b*x])

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Rubi [A]  time = 0.0280244, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {45, 37} \[ \frac{32 d^3 \sqrt{c+d x}}{35 \sqrt{a+b x} (b c-a d)^4}-\frac{16 d^2 \sqrt{c+d x}}{35 (a+b x)^{3/2} (b c-a d)^3}+\frac{12 d \sqrt{c+d x}}{35 (a+b x)^{5/2} (b c-a d)^2}-\frac{2 \sqrt{c+d x}}{7 (a+b x)^{7/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(9/2)*Sqrt[c + d*x]),x]

[Out]

(-2*Sqrt[c + d*x])/(7*(b*c - a*d)*(a + b*x)^(7/2)) + (12*d*Sqrt[c + d*x])/(35*(b*c - a*d)^2*(a + b*x)^(5/2)) -
(16*d^2*Sqrt[c + d*x])/(35*(b*c - a*d)^3*(a + b*x)^(3/2)) + (32*d^3*Sqrt[c + d*x])/(35*(b*c - a*d)^4*Sqrt[a +
b*x])

Rule 45

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*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{(a+b x)^{9/2} \sqrt{c+d x}} \, dx &=-\frac{2 \sqrt{c+d x}}{7 (b c-a d) (a+b x)^{7/2}}-\frac{(6 d) \int \frac{1}{(a+b x)^{7/2} \sqrt{c+d x}} \, dx}{7 (b c-a d)}\\ &=-\frac{2 \sqrt{c+d x}}{7 (b c-a d) (a+b x)^{7/2}}+\frac{12 d \sqrt{c+d x}}{35 (b c-a d)^2 (a+b x)^{5/2}}+\frac{\left (24 d^2\right ) \int \frac{1}{(a+b x)^{5/2} \sqrt{c+d x}} \, dx}{35 (b c-a d)^2}\\ &=-\frac{2 \sqrt{c+d x}}{7 (b c-a d) (a+b x)^{7/2}}+\frac{12 d \sqrt{c+d x}}{35 (b c-a d)^2 (a+b x)^{5/2}}-\frac{16 d^2 \sqrt{c+d x}}{35 (b c-a d)^3 (a+b x)^{3/2}}-\frac{\left (16 d^3\right ) \int \frac{1}{(a+b x)^{3/2} \sqrt{c+d x}} \, dx}{35 (b c-a d)^3}\\ &=-\frac{2 \sqrt{c+d x}}{7 (b c-a d) (a+b x)^{7/2}}+\frac{12 d \sqrt{c+d x}}{35 (b c-a d)^2 (a+b x)^{5/2}}-\frac{16 d^2 \sqrt{c+d x}}{35 (b c-a d)^3 (a+b x)^{3/2}}+\frac{32 d^3 \sqrt{c+d x}}{35 (b c-a d)^4 \sqrt{a+b x}}\\ \end{align*}

Mathematica [A]  time = 0.0456081, size = 116, normalized size = 0.85 \[ \frac{2 \sqrt{c+d x} \left (-35 a^2 b d^2 (c-2 d x)+35 a^3 d^3+7 a b^2 d \left (3 c^2-4 c d x+8 d^2 x^2\right )+b^3 \left (6 c^2 d x-5 c^3-8 c d^2 x^2+16 d^3 x^3\right )\right )}{35 (a+b x)^{7/2} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(9/2)*Sqrt[c + d*x]),x]

[Out]

(2*Sqrt[c + d*x]*(35*a^3*d^3 - 35*a^2*b*d^2*(c - 2*d*x) + 7*a*b^2*d*(3*c^2 - 4*c*d*x + 8*d^2*x^2) + b^3*(-5*c^
3 + 6*c^2*d*x - 8*c*d^2*x^2 + 16*d^3*x^3)))/(35*(b*c - a*d)^4*(a + b*x)^(7/2))

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Maple [A]  time = 0.008, size = 171, normalized size = 1.3 \begin{align*}{\frac{32\,{b}^{3}{d}^{3}{x}^{3}+112\,a{b}^{2}{d}^{3}{x}^{2}-16\,{b}^{3}c{d}^{2}{x}^{2}+140\,{a}^{2}b{d}^{3}x-56\,a{b}^{2}c{d}^{2}x+12\,{b}^{3}{c}^{2}dx+70\,{a}^{3}{d}^{3}-70\,{a}^{2}bc{d}^{2}+42\,a{b}^{2}{c}^{2}d-10\,{b}^{3}{c}^{3}}{35\,{d}^{4}{a}^{4}-140\,b{d}^{3}c{a}^{3}+210\,{b}^{2}{d}^{2}{c}^{2}{a}^{2}-140\,{b}^{3}d{c}^{3}a+35\,{b}^{4}{c}^{4}}\sqrt{dx+c} \left ( bx+a \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(9/2)/(d*x+c)^(1/2),x)

[Out]

2/35*(d*x+c)^(1/2)*(16*b^3*d^3*x^3+56*a*b^2*d^3*x^2-8*b^3*c*d^2*x^2+70*a^2*b*d^3*x-28*a*b^2*c*d^2*x+6*b^3*c^2*
d*x+35*a^3*d^3-35*a^2*b*c*d^2+21*a*b^2*c^2*d-5*b^3*c^3)/(b*x+a)^(7/2)/(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^2*c^2*d^2
-4*a*b^3*c^3*d+b^4*c^4)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(9/2)/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 12.3486, size = 848, normalized size = 6.24 \begin{align*} \frac{2 \,{\left (16 \, b^{3} d^{3} x^{3} - 5 \, b^{3} c^{3} + 21 \, a b^{2} c^{2} d - 35 \, a^{2} b c d^{2} + 35 \, a^{3} d^{3} - 8 \,{\left (b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{2} + 2 \,{\left (3 \, b^{3} c^{2} d - 14 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{35 \,{\left (a^{4} b^{4} c^{4} - 4 \, a^{5} b^{3} c^{3} d + 6 \, a^{6} b^{2} c^{2} d^{2} - 4 \, a^{7} b c d^{3} + a^{8} d^{4} +{\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{4} + 4 \,{\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{3} + 6 \,{\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x^{2} + 4 \,{\left (a^{3} b^{5} c^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(9/2)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

2/35*(16*b^3*d^3*x^3 - 5*b^3*c^3 + 21*a*b^2*c^2*d - 35*a^2*b*c*d^2 + 35*a^3*d^3 - 8*(b^3*c*d^2 - 7*a*b^2*d^3)*
x^2 + 2*(3*b^3*c^2*d - 14*a*b^2*c*d^2 + 35*a^2*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^4*b^4*c^4 - 4*a^5*b^3*
c^3*d + 6*a^6*b^2*c^2*d^2 - 4*a^7*b*c*d^3 + a^8*d^4 + (b^8*c^4 - 4*a*b^7*c^3*d + 6*a^2*b^6*c^2*d^2 - 4*a^3*b^5
*c*d^3 + a^4*b^4*d^4)*x^4 + 4*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 - 4*a^4*b^4*c*d^3 + a^5*b^3*d^4
)*x^3 + 6*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4*a^5*b^3*c*d^3 + a^6*b^2*d^4)*x^2 + 4*(a^3*b^5
*c^4 - 4*a^4*b^4*c^3*d + 6*a^5*b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^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(1/(b*x+a)**(9/2)/(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 1.18414, size = 521, normalized size = 3.83 \begin{align*} \frac{64 \,{\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3} - 7 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} b^{4} c^{2} + 14 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{3} c d - 7 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{2} d^{2} + 21 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{4} b^{2} c - 21 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{4} a b d - 35 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{6}\right )} \sqrt{b d} b^{4} d^{3}}{35 \,{\left (b^{2} c - a b d -{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{7}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(9/2)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

64/35*(b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^2 - a^3*b^3*d^3 - 7*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*
x + a)*b*d - a*b*d))^2*b^4*c^2 + 14*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^3*c*
d - 7*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^2*d^2 + 21*(sqrt(b*d)*sqrt(b*x +
a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^2*c - 21*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^4*a*b*d - 35*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6)*sqrt(b*d)*b^4*d^3/((
b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)^7*abs(b))