3.15.4 \(\int \frac {1}{(a+b x)^{9/2} (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=171 \[ \frac {256 d^4 \sqrt {a+b x}}{35 \sqrt {c+d x} (b c-a d)^5}+\frac {128 d^3}{35 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^4}-\frac {32 d^2}{35 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^3}+\frac {16 d}{35 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)^2}-\frac {2}{7 (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)} \]

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Rubi [A]  time = 0.04, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {45, 37} \begin {gather*} \frac {256 d^4 \sqrt {a+b x}}{35 \sqrt {c+d x} (b c-a d)^5}+\frac {128 d^3}{35 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^4}-\frac {32 d^2}{35 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^3}+\frac {16 d}{35 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)^2}-\frac {2}{7 (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(9/2)*(c + d*x)^(3/2)),x]

[Out]

-2/(7*(b*c - a*d)*(a + b*x)^(7/2)*Sqrt[c + d*x]) + (16*d)/(35*(b*c - a*d)^2*(a + b*x)^(5/2)*Sqrt[c + d*x]) - (
32*d^2)/(35*(b*c - a*d)^3*(a + b*x)^(3/2)*Sqrt[c + d*x]) + (128*d^3)/(35*(b*c - a*d)^4*Sqrt[a + b*x]*Sqrt[c +
d*x]) + (256*d^4*Sqrt[a + b*x])/(35*(b*c - a*d)^5*Sqrt[c + d*x])

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]

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])

Rubi steps

\begin {align*} \int \frac {1}{(a+b x)^{9/2} (c+d x)^{3/2}} \, dx &=-\frac {2}{7 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}-\frac {(8 d) \int \frac {1}{(a+b x)^{7/2} (c+d x)^{3/2}} \, dx}{7 (b c-a d)}\\ &=-\frac {2}{7 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}+\frac {16 d}{35 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}+\frac {\left (48 d^2\right ) \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx}{35 (b c-a d)^2}\\ &=-\frac {2}{7 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}+\frac {16 d}{35 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}-\frac {32 d^2}{35 (b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}-\frac {\left (64 d^3\right ) \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx}{35 (b c-a d)^3}\\ &=-\frac {2}{7 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}+\frac {16 d}{35 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}-\frac {32 d^2}{35 (b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}+\frac {128 d^3}{35 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}+\frac {\left (128 d^4\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx}{35 (b c-a d)^4}\\ &=-\frac {2}{7 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}+\frac {16 d}{35 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}-\frac {32 d^2}{35 (b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}+\frac {128 d^3}{35 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}+\frac {256 d^4 \sqrt {a+b x}}{35 (b c-a d)^5 \sqrt {c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 166, normalized size = 0.97 \begin {gather*} \frac {2 \left (35 a^4 d^4+140 a^3 b d^3 (c+2 d x)+70 a^2 b^2 d^2 \left (-c^2+4 c d x+8 d^2 x^2\right )+28 a b^3 d \left (c^3-2 c^2 d x+8 c d^2 x^2+16 d^3 x^3\right )+b^4 \left (-5 c^4+8 c^3 d x-16 c^2 d^2 x^2+64 c d^3 x^3+128 d^4 x^4\right )\right )}{35 (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(9/2)*(c + d*x)^(3/2)),x]

[Out]

(2*(35*a^4*d^4 + 140*a^3*b*d^3*(c + 2*d*x) + 70*a^2*b^2*d^2*(-c^2 + 4*c*d*x + 8*d^2*x^2) + 28*a*b^3*d*(c^3 - 2
*c^2*d*x + 8*c*d^2*x^2 + 16*d^3*x^3) + b^4*(-5*c^4 + 8*c^3*d*x - 16*c^2*d^2*x^2 + 64*c*d^3*x^3 + 128*d^4*x^4))
)/(35*(b*c - a*d)^5*(a + b*x)^(7/2)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.14, size = 117, normalized size = 0.68 \begin {gather*} \frac {2 (c+d x)^{7/2} \left (\frac {28 b^3 d (a+b x)}{c+d x}-\frac {70 b^2 d^2 (a+b x)^2}{(c+d x)^2}+\frac {35 d^4 (a+b x)^4}{(c+d x)^4}+\frac {140 b d^3 (a+b x)^3}{(c+d x)^3}-5 b^4\right )}{35 (a+b x)^{7/2} (b c-a d)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^(9/2)*(c + d*x)^(3/2)),x]

[Out]

(2*(c + d*x)^(7/2)*(-5*b^4 + (35*d^4*(a + b*x)^4)/(c + d*x)^4 + (140*b*d^3*(a + b*x)^3)/(c + d*x)^3 - (70*b^2*
d^2*(a + b*x)^2)/(c + d*x)^2 + (28*b^3*d*(a + b*x))/(c + d*x)))/(35*(b*c - a*d)^5*(a + b*x)^(7/2))

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fricas [B]  time = 8.19, size = 689, normalized size = 4.03 \begin {gather*} \frac {2 \, {\left (128 \, b^{4} d^{4} x^{4} - 5 \, b^{4} c^{4} + 28 \, a b^{3} c^{3} d - 70 \, a^{2} b^{2} c^{2} d^{2} + 140 \, a^{3} b c d^{3} + 35 \, a^{4} d^{4} + 64 \, {\left (b^{4} c d^{3} + 7 \, a b^{3} d^{4}\right )} x^{3} - 16 \, {\left (b^{4} c^{2} d^{2} - 14 \, a b^{3} c d^{3} - 35 \, a^{2} b^{2} d^{4}\right )} x^{2} + 8 \, {\left (b^{4} c^{3} d - 7 \, a b^{3} c^{2} d^{2} + 35 \, a^{2} b^{2} c d^{3} + 35 \, a^{3} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{35 \, {\left (a^{4} b^{5} c^{6} - 5 \, a^{5} b^{4} c^{5} d + 10 \, a^{6} b^{3} c^{4} d^{2} - 10 \, a^{7} b^{2} c^{3} d^{3} + 5 \, a^{8} b c^{2} d^{4} - a^{9} c d^{5} + {\left (b^{9} c^{5} d - 5 \, a b^{8} c^{4} d^{2} + 10 \, a^{2} b^{7} c^{3} d^{3} - 10 \, a^{3} b^{6} c^{2} d^{4} + 5 \, a^{4} b^{5} c d^{5} - a^{5} b^{4} d^{6}\right )} x^{5} + {\left (b^{9} c^{6} - a b^{8} c^{5} d - 10 \, a^{2} b^{7} c^{4} d^{2} + 30 \, a^{3} b^{6} c^{3} d^{3} - 35 \, a^{4} b^{5} c^{2} d^{4} + 19 \, a^{5} b^{4} c d^{5} - 4 \, a^{6} b^{3} d^{6}\right )} x^{4} + 2 \, {\left (2 \, a b^{8} c^{6} - 7 \, a^{2} b^{7} c^{5} d + 5 \, a^{3} b^{6} c^{4} d^{2} + 10 \, a^{4} b^{5} c^{3} d^{3} - 20 \, a^{5} b^{4} c^{2} d^{4} + 13 \, a^{6} b^{3} c d^{5} - 3 \, a^{7} b^{2} d^{6}\right )} x^{3} + 2 \, {\left (3 \, a^{2} b^{7} c^{6} - 13 \, a^{3} b^{6} c^{5} d + 20 \, a^{4} b^{5} c^{4} d^{2} - 10 \, a^{5} b^{4} c^{3} d^{3} - 5 \, a^{6} b^{3} c^{2} d^{4} + 7 \, a^{7} b^{2} c d^{5} - 2 \, a^{8} b d^{6}\right )} x^{2} + {\left (4 \, a^{3} b^{6} c^{6} - 19 \, a^{4} b^{5} c^{5} d + 35 \, a^{5} b^{4} c^{4} d^{2} - 30 \, a^{6} b^{3} c^{3} d^{3} + 10 \, a^{7} b^{2} c^{2} d^{4} + a^{8} b c d^{5} - a^{9} d^{6}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(9/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

2/35*(128*b^4*d^4*x^4 - 5*b^4*c^4 + 28*a*b^3*c^3*d - 70*a^2*b^2*c^2*d^2 + 140*a^3*b*c*d^3 + 35*a^4*d^4 + 64*(b
^4*c*d^3 + 7*a*b^3*d^4)*x^3 - 16*(b^4*c^2*d^2 - 14*a*b^3*c*d^3 - 35*a^2*b^2*d^4)*x^2 + 8*(b^4*c^3*d - 7*a*b^3*
c^2*d^2 + 35*a^2*b^2*c*d^3 + 35*a^3*b*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^4*b^5*c^6 - 5*a^5*b^4*c^5*d + 10*
a^6*b^3*c^4*d^2 - 10*a^7*b^2*c^3*d^3 + 5*a^8*b*c^2*d^4 - a^9*c*d^5 + (b^9*c^5*d - 5*a*b^8*c^4*d^2 + 10*a^2*b^7
*c^3*d^3 - 10*a^3*b^6*c^2*d^4 + 5*a^4*b^5*c*d^5 - a^5*b^4*d^6)*x^5 + (b^9*c^6 - a*b^8*c^5*d - 10*a^2*b^7*c^4*d
^2 + 30*a^3*b^6*c^3*d^3 - 35*a^4*b^5*c^2*d^4 + 19*a^5*b^4*c*d^5 - 4*a^6*b^3*d^6)*x^4 + 2*(2*a*b^8*c^6 - 7*a^2*
b^7*c^5*d + 5*a^3*b^6*c^4*d^2 + 10*a^4*b^5*c^3*d^3 - 20*a^5*b^4*c^2*d^4 + 13*a^6*b^3*c*d^5 - 3*a^7*b^2*d^6)*x^
3 + 2*(3*a^2*b^7*c^6 - 13*a^3*b^6*c^5*d + 20*a^4*b^5*c^4*d^2 - 10*a^5*b^4*c^3*d^3 - 5*a^6*b^3*c^2*d^4 + 7*a^7*
b^2*c*d^5 - 2*a^8*b*d^6)*x^2 + (4*a^3*b^6*c^6 - 19*a^4*b^5*c^5*d + 35*a^5*b^4*c^4*d^2 - 30*a^6*b^3*c^3*d^3 + 1
0*a^7*b^2*c^2*d^4 + a^8*b*c*d^5 - a^9*d^6)*x)

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giac [B]  time = 4.77, size = 1518, normalized size = 8.88

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(9/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

2*sqrt(b*x + a)*b^2*d^4/((b^5*c^5*abs(b) - 5*a*b^4*c^4*d*abs(b) + 10*a^2*b^3*c^3*d^2*abs(b) - 10*a^3*b^2*c^2*d
^3*abs(b) + 5*a^4*b*c*d^4*abs(b) - a^5*d^5*abs(b))*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) + 4/35*(93*sqrt(b*d)*b
^14*c^6*d^3 - 558*sqrt(b*d)*a*b^13*c^5*d^4 + 1395*sqrt(b*d)*a^2*b^12*c^4*d^5 - 1860*sqrt(b*d)*a^3*b^11*c^3*d^6
+ 1395*sqrt(b*d)*a^4*b^10*c^2*d^7 - 558*sqrt(b*d)*a^5*b^9*c*d^8 + 93*sqrt(b*d)*a^6*b^8*d^9 - 616*sqrt(b*d)*(s
qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^12*c^5*d^3 + 3080*sqrt(b*d)*(sqrt(b*d)*sqrt(
b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^11*c^4*d^4 - 6160*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - s
qrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^10*c^3*d^5 + 6160*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^9*c^2*d^6 - 3080*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)
*b*d - a*b*d))^2*a^4*b^8*c*d^7 + 616*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))
^2*a^5*b^7*d^8 + 1673*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^10*c^4*d^3
- 6692*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^9*c^3*d^4 + 10038*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^8*c^2*d^5 - 6692*sqrt(b*d)*(sqrt
(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^7*c*d^6 + 1673*sqrt(b*d)*(sqrt(b*d)*sqrt(b*
x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^4*b^6*d^7 - 2240*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b
^2*c + (b*x + a)*b*d - a*b*d))^6*b^8*c^3*d^3 + 6720*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^7*c^2*d^4 - 6720*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d
))^6*a^2*b^6*c*d^5 + 2240*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^3*b^5*
d^6 + 1015*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b^6*c^2*d^3 - 2030*sqrt
(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a*b^5*c*d^4 + 1015*sqrt(b*d)*(sqrt(b*d
)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^2*b^4*d^5 - 280*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^10*b^4*c*d^3 + 280*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b
*x + a)*b*d - a*b*d))^10*a*b^3*d^4 + 35*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*
d))^12*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*abs(b) + a^
4*d^4*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)^7)

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maple [A]  time = 0.01, size = 256, normalized size = 1.50 \begin {gather*} -\frac {2 \left (128 b^{4} x^{4} d^{4}+448 a \,b^{3} d^{4} x^{3}+64 b^{4} c \,d^{3} x^{3}+560 a^{2} b^{2} d^{4} x^{2}+224 a \,b^{3} c \,d^{3} x^{2}-16 b^{4} c^{2} d^{2} x^{2}+280 a^{3} b \,d^{4} x +280 a^{2} b^{2} c \,d^{3} x -56 a \,b^{3} c^{2} d^{2} x +8 b^{4} c^{3} d x +35 a^{4} d^{4}+140 a^{3} b c \,d^{3}-70 a^{2} b^{2} c^{2} d^{2}+28 a \,b^{3} c^{3} d -5 b^{4} c^{4}\right )}{35 \left (b x +a \right )^{\frac {7}{2}} \sqrt {d x +c}\, \left (a^{5} d^{5}-5 a^{4} b c \,d^{4}+10 a^{3} b^{2} c^{2} d^{3}-10 a^{2} b^{3} c^{3} d^{2}+5 a \,b^{4} c^{4} d -b^{5} c^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(9/2)/(d*x+c)^(3/2),x)

[Out]

-2/35*(128*b^4*d^4*x^4+448*a*b^3*d^4*x^3+64*b^4*c*d^3*x^3+560*a^2*b^2*d^4*x^2+224*a*b^3*c*d^3*x^2-16*b^4*c^2*d
^2*x^2+280*a^3*b*d^4*x+280*a^2*b^2*c*d^3*x-56*a*b^3*c^2*d^2*x+8*b^4*c^3*d*x+35*a^4*d^4+140*a^3*b*c*d^3-70*a^2*
b^2*c^2*d^2+28*a*b^3*c^3*d-5*b^4*c^4)/(b*x+a)^(7/2)/(d*x+c)^(1/2)/(a^5*d^5-5*a^4*b*c*d^4+10*a^3*b^2*c^2*d^3-10
*a^2*b^3*c^3*d^2+5*a*b^4*c^4*d-b^5*c^5)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(9/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [B]  time = 1.50, size = 337, normalized size = 1.97 \begin {gather*} -\frac {\sqrt {c+d\,x}\,\left (\frac {256\,b\,d^3\,x^4}{35\,{\left (a\,d-b\,c\right )}^5}+\frac {128\,d^2\,x^3\,\left (7\,a\,d+b\,c\right )}{35\,{\left (a\,d-b\,c\right )}^5}+\frac {70\,a^4\,d^4+280\,a^3\,b\,c\,d^3-140\,a^2\,b^2\,c^2\,d^2+56\,a\,b^3\,c^3\,d-10\,b^4\,c^4}{35\,b^3\,d\,{\left (a\,d-b\,c\right )}^5}+\frac {x\,\left (560\,a^3\,b\,d^4+560\,a^2\,b^2\,c\,d^3-112\,a\,b^3\,c^2\,d^2+16\,b^4\,c^3\,d\right )}{35\,b^3\,d\,{\left (a\,d-b\,c\right )}^5}+\frac {32\,d\,x^2\,\left (35\,a^2\,d^2+14\,a\,b\,c\,d-b^2\,c^2\right )}{35\,b\,{\left (a\,d-b\,c\right )}^5}\right )}{x^4\,\sqrt {a+b\,x}+\frac {a^3\,c\,\sqrt {a+b\,x}}{b^3\,d}+\frac {x^3\,\left (3\,a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b\,d}+\frac {3\,a\,x^2\,\left (a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b^2\,d}+\frac {a^2\,x\,\left (a\,d+3\,b\,c\right )\,\sqrt {a+b\,x}}{b^3\,d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(9/2)*(c + d*x)^(3/2)),x)

[Out]

-((c + d*x)^(1/2)*((256*b*d^3*x^4)/(35*(a*d - b*c)^5) + (128*d^2*x^3*(7*a*d + b*c))/(35*(a*d - b*c)^5) + (70*a
^4*d^4 - 10*b^4*c^4 - 140*a^2*b^2*c^2*d^2 + 56*a*b^3*c^3*d + 280*a^3*b*c*d^3)/(35*b^3*d*(a*d - b*c)^5) + (x*(5
60*a^3*b*d^4 + 16*b^4*c^3*d - 112*a*b^3*c^2*d^2 + 560*a^2*b^2*c*d^3))/(35*b^3*d*(a*d - b*c)^5) + (32*d*x^2*(35
*a^2*d^2 - b^2*c^2 + 14*a*b*c*d))/(35*b*(a*d - b*c)^5)))/(x^4*(a + b*x)^(1/2) + (a^3*c*(a + b*x)^(1/2))/(b^3*d
) + (x^3*(3*a*d + b*c)*(a + b*x)^(1/2))/(b*d) + (3*a*x^2*(a*d + b*c)*(a + b*x)^(1/2))/(b^2*d) + (a^2*x*(a*d +
3*b*c)*(a + b*x)^(1/2))/(b^3*d))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right )^{\frac {9}{2}} \left (c + d x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(9/2)/(d*x+c)**(3/2),x)

[Out]

Integral(1/((a + b*x)**(9/2)*(c + d*x)**(3/2)), x)

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