3.1509 \(\int \frac{1}{(a+b x)^{7/2} (c+d x)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0270072, 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} \[ -\frac{32 d^3 \sqrt{a+b x}}{5 \sqrt{c+d x} (b c-a d)^4}-\frac{16 d^2}{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^3}+\frac{4 d}{5 (a+b x)^{3/2} \sqrt{c+d x} (b c-a d)^2}-\frac{2}{5 (a+b x)^{5/2} \sqrt{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(7/2)*(c + d*x)^(3/2)),x]

[Out]

-2/(5*(b*c - a*d)*(a + b*x)^(5/2)*Sqrt[c + d*x]) + (4*d)/(5*(b*c - a*d)^2*(a + b*x)^(3/2)*Sqrt[c + d*x]) - (16
*d^2)/(5*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x]) - (32*d^3*Sqrt[a + b*x])/(5*(b*c - a*d)^4*Sqrt[c + d*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)^{7/2} (c+d x)^{3/2}} \, dx &=-\frac{2}{5 (b c-a d) (a+b x)^{5/2} \sqrt{c+d x}}-\frac{(6 d) \int \frac{1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx}{5 (b c-a d)}\\ &=-\frac{2}{5 (b c-a d) (a+b x)^{5/2} \sqrt{c+d x}}+\frac{4 d}{5 (b c-a d)^2 (a+b x)^{3/2} \sqrt{c+d x}}+\frac{\left (8 d^2\right ) \int \frac{1}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx}{5 (b c-a d)^2}\\ &=-\frac{2}{5 (b c-a d) (a+b x)^{5/2} \sqrt{c+d x}}+\frac{4 d}{5 (b c-a d)^2 (a+b x)^{3/2} \sqrt{c+d x}}-\frac{16 d^2}{5 (b c-a d)^3 \sqrt{a+b x} \sqrt{c+d x}}-\frac{\left (16 d^3\right ) \int \frac{1}{\sqrt{a+b x} (c+d x)^{3/2}} \, dx}{5 (b c-a d)^3}\\ &=-\frac{2}{5 (b c-a d) (a+b x)^{5/2} \sqrt{c+d x}}+\frac{4 d}{5 (b c-a d)^2 (a+b x)^{3/2} \sqrt{c+d x}}-\frac{16 d^2}{5 (b c-a d)^3 \sqrt{a+b x} \sqrt{c+d x}}-\frac{32 d^3 \sqrt{a+b x}}{5 (b c-a d)^4 \sqrt{c+d x}}\\ \end{align*}

Mathematica [A]  time = 0.04155, size = 114, normalized size = 0.84 \[ -\frac{2 \left (15 a^2 b d^2 (c+2 d x)+5 a^3 d^3+5 a b^2 d \left (-c^2+4 c d x+8 d^2 x^2\right )+b^3 \left (-2 c^2 d x+c^3+8 c d^2 x^2+16 d^3 x^3\right )\right )}{5 (a+b x)^{5/2} \sqrt{c+d x} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(7/2)*(c + d*x)^(3/2)),x]

[Out]

(-2*(5*a^3*d^3 + 15*a^2*b*d^2*(c + 2*d*x) + 5*a*b^2*d*(-c^2 + 4*c*d*x + 8*d^2*x^2) + b^3*(c^3 - 2*c^2*d*x + 8*
c*d^2*x^2 + 16*d^3*x^3)))/(5*(b*c - a*d)^4*(a + b*x)^(5/2)*Sqrt[c + d*x])

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Maple [A]  time = 0.009, size = 170, normalized size = 1.3 \begin{align*} -{\frac{32\,{b}^{3}{d}^{3}{x}^{3}+80\,a{b}^{2}{d}^{3}{x}^{2}+16\,{b}^{3}c{d}^{2}{x}^{2}+60\,{a}^{2}b{d}^{3}x+40\,a{b}^{2}c{d}^{2}x-4\,{b}^{3}{c}^{2}dx+10\,{a}^{3}{d}^{3}+30\,{a}^{2}bc{d}^{2}-10\,a{b}^{2}{c}^{2}d+2\,{b}^{3}{c}^{3}}{5\,{d}^{4}{a}^{4}-20\,b{d}^{3}c{a}^{3}+30\,{b}^{2}{d}^{2}{c}^{2}{a}^{2}-20\,{b}^{3}d{c}^{3}a+5\,{b}^{4}{c}^{4}} \left ( bx+a \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(7/2)/(d*x+c)^(3/2),x)

[Out]

-2/5*(16*b^3*d^3*x^3+40*a*b^2*d^3*x^2+8*b^3*c*d^2*x^2+30*a^2*b*d^3*x+20*a*b^2*c*d^2*x-2*b^3*c^2*d*x+5*a^3*d^3+
15*a^2*b*c*d^2-5*a*b^2*c^2*d+b^3*c^3)/(b*x+a)^(5/2)/(d*x+c)^(1/2)/(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)^(7/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 9.54481, size = 915, normalized size = 6.73 \begin{align*} -\frac{2 \,{\left (16 \, b^{3} d^{3} x^{3} + b^{3} c^{3} - 5 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3} + 8 \,{\left (b^{3} c d^{2} + 5 \, a b^{2} d^{3}\right )} x^{2} - 2 \,{\left (b^{3} c^{2} d - 10 \, a b^{2} c d^{2} - 15 \, a^{2} b d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{5 \,{\left (a^{3} b^{4} c^{5} - 4 \, a^{4} b^{3} c^{4} d + 6 \, a^{5} b^{2} c^{3} d^{2} - 4 \, a^{6} b c^{2} d^{3} + a^{7} c d^{4} +{\left (b^{7} c^{4} d - 4 \, a b^{6} c^{3} d^{2} + 6 \, a^{2} b^{5} c^{2} d^{3} - 4 \, a^{3} b^{4} c d^{4} + a^{4} b^{3} d^{5}\right )} x^{4} +{\left (b^{7} c^{5} - a b^{6} c^{4} d - 6 \, a^{2} b^{5} c^{3} d^{2} + 14 \, a^{3} b^{4} c^{2} d^{3} - 11 \, a^{4} b^{3} c d^{4} + 3 \, a^{5} b^{2} d^{5}\right )} x^{3} + 3 \,{\left (a b^{6} c^{5} - 3 \, a^{2} b^{5} c^{4} d + 2 \, a^{3} b^{4} c^{3} d^{2} + 2 \, a^{4} b^{3} c^{2} d^{3} - 3 \, a^{5} b^{2} c d^{4} + a^{6} b d^{5}\right )} x^{2} +{\left (3 \, a^{2} b^{5} c^{5} - 11 \, a^{3} b^{4} c^{4} d + 14 \, a^{4} b^{3} c^{3} d^{2} - 6 \, a^{5} b^{2} c^{2} d^{3} - a^{6} b c d^{4} + a^{7} d^{5}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(7/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-2/5*(16*b^3*d^3*x^3 + b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3 + 8*(b^3*c*d^2 + 5*a*b^2*d^3)*x^2
- 2*(b^3*c^2*d - 10*a*b^2*c*d^2 - 15*a^2*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^3*b^4*c^5 - 4*a^4*b^3*c^4*d
+ 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d - 4*a*b^6*c^3*d^2 + 6*a^2*b^5*c^2*d^3 - 4*a^3*b
^4*c*d^4 + a^4*b^3*d^5)*x^4 + (b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b^4*c^2*d^3 - 11*a^4*b^3*c*d
^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3*b^4*c^3*d^2 + 2*a^4*b^3*c^2*d^3 - 3*a^5*b^2*c
*d^4 + a^6*b*d^5)*x^2 + (3*a^2*b^5*c^5 - 11*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*a^5*b^2*c^2*d^3 - a^6*b*c*d
^4 + a^7*d^5)*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}{2}} \left (c + d x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(7/2)/(d*x+c)**(3/2),x)

[Out]

Integral(1/((a + b*x)**(7/2)*(c + d*x)**(3/2)), x)

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Giac [B]  time = 1.82799, size = 1121, normalized size = 8.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(7/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

-2*sqrt(b*x + a)*b^2*d^3/((b^4*c^4*abs(b) - 4*a*b^3*c^3*d*abs(b) + 6*a^2*b^2*c^2*d^2*abs(b) - 4*a^3*b*c*d^3*ab
s(b) + a^4*d^4*abs(b))*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) - 4/5*(11*sqrt(b*d)*b^10*c^4*d^2 - 44*sqrt(b*d)*a*
b^9*c^3*d^3 + 66*sqrt(b*d)*a^2*b^8*c^2*d^4 - 44*sqrt(b*d)*a^3*b^7*c*d^5 + 11*sqrt(b*d)*a^4*b^6*d^6 - 50*sqrt(b
*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^8*c^3*d^2 + 150*sqrt(b*d)*(sqrt(b*d)*s
qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^7*c^2*d^3 - 150*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^6*c*d^4 + 50*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (
b*x + a)*b*d - a*b*d))^2*a^3*b^5*d^5 + 80*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*
b*d))^4*b^6*c^2*d^2 - 160*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^5*c*
d^3 + 80*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^4*d^4 - 30*sqrt(b*d
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^4*c*d^2 + 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b
*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^3*d^3 + 5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c
+ (b*x + a)*b*d - a*b*d))^8*b^2*d^2)/((b^3*c^3*abs(b) - 3*a*b^2*c^2*d*abs(b) + 3*a^2*b*c*d^2*abs(b) - a^3*d^3
*abs(b))*(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)^5)