3.1520 \(\int \frac {1}{(a+b x)^{7/2} (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=172 \[ -\frac {256 b d^3 \sqrt {a+b x}}{15 \sqrt {c+d x} (b c-a d)^5}-\frac {128 d^3 \sqrt {a+b x}}{15 (c+d x)^{3/2} (b c-a d)^4}-\frac {32 d^2}{5 \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)^3}+\frac {16 d}{15 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)^2}-\frac {2}{5 (a+b x)^{5/2} (c+d x)^{3/2} (b c-a d)} \]

[Out]

-2/5/(-a*d+b*c)/(b*x+a)^(5/2)/(d*x+c)^(3/2)+16/15*d/(-a*d+b*c)^2/(b*x+a)^(3/2)/(d*x+c)^(3/2)-32/5*d^2/(-a*d+b*
c)^3/(d*x+c)^(3/2)/(b*x+a)^(1/2)-128/15*d^3*(b*x+a)^(1/2)/(-a*d+b*c)^4/(d*x+c)^(3/2)-256/15*b*d^3*(b*x+a)^(1/2
)/(-a*d+b*c)^5/(d*x+c)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 172, 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} \[ -\frac {256 b d^3 \sqrt {a+b x}}{15 \sqrt {c+d x} (b c-a d)^5}-\frac {128 d^3 \sqrt {a+b x}}{15 (c+d x)^{3/2} (b c-a d)^4}-\frac {32 d^2}{5 \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)^3}+\frac {16 d}{15 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)^2}-\frac {2}{5 (a+b x)^{5/2} (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(7/2)*(c + d*x)^(5/2)),x]

[Out]

-2/(5*(b*c - a*d)*(a + b*x)^(5/2)*(c + d*x)^(3/2)) + (16*d)/(15*(b*c - a*d)^2*(a + b*x)^(3/2)*(c + d*x)^(3/2))
- (32*d^2)/(5*(b*c - a*d)^3*Sqrt[a + b*x]*(c + d*x)^(3/2)) - (128*d^3*Sqrt[a + b*x])/(15*(b*c - a*d)^4*(c + d
*x)^(3/2)) - (256*b*d^3*Sqrt[a + b*x])/(15*(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)^{7/2} (c+d x)^{5/2}} \, dx &=-\frac {2}{5 (b c-a d) (a+b x)^{5/2} (c+d x)^{3/2}}-\frac {(8 d) \int \frac {1}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx}{5 (b c-a d)}\\ &=-\frac {2}{5 (b c-a d) (a+b x)^{5/2} (c+d x)^{3/2}}+\frac {16 d}{15 (b c-a d)^2 (a+b x)^{3/2} (c+d x)^{3/2}}+\frac {\left (16 d^2\right ) \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx}{5 (b c-a d)^2}\\ &=-\frac {2}{5 (b c-a d) (a+b x)^{5/2} (c+d x)^{3/2}}+\frac {16 d}{15 (b c-a d)^2 (a+b x)^{3/2} (c+d x)^{3/2}}-\frac {32 d^2}{5 (b c-a d)^3 \sqrt {a+b x} (c+d x)^{3/2}}-\frac {\left (64 d^3\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/2}} \, dx}{5 (b c-a d)^3}\\ &=-\frac {2}{5 (b c-a d) (a+b x)^{5/2} (c+d x)^{3/2}}+\frac {16 d}{15 (b c-a d)^2 (a+b x)^{3/2} (c+d x)^{3/2}}-\frac {32 d^2}{5 (b c-a d)^3 \sqrt {a+b x} (c+d x)^{3/2}}-\frac {128 d^3 \sqrt {a+b x}}{15 (b c-a d)^4 (c+d x)^{3/2}}-\frac {\left (128 b d^3\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx}{15 (b c-a d)^4}\\ &=-\frac {2}{5 (b c-a d) (a+b x)^{5/2} (c+d x)^{3/2}}+\frac {16 d}{15 (b c-a d)^2 (a+b x)^{3/2} (c+d x)^{3/2}}-\frac {32 d^2}{5 (b c-a d)^3 \sqrt {a+b x} (c+d x)^{3/2}}-\frac {128 d^3 \sqrt {a+b x}}{15 (b c-a d)^4 (c+d x)^{3/2}}-\frac {256 b d^3 \sqrt {a+b x}}{15 (b c-a d)^5 \sqrt {c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 170, normalized size = 0.99 \[ -\frac {2 \left (-5 a^4 d^4+20 a^3 b d^3 (3 c+2 d x)+30 a^2 b^2 d^2 \left (3 c^2+12 c d x+8 d^2 x^2\right )+20 a b^3 d \left (-c^3+6 c^2 d x+24 c d^2 x^2+16 d^3 x^3\right )+b^4 \left (3 c^4-8 c^3 d x+48 c^2 d^2 x^2+192 c d^3 x^3+128 d^4 x^4\right )\right )}{15 (a+b x)^{5/2} (c+d x)^{3/2} (b c-a d)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(7/2)*(c + d*x)^(5/2)),x]

[Out]

(-2*(-5*a^4*d^4 + 20*a^3*b*d^3*(3*c + 2*d*x) + 30*a^2*b^2*d^2*(3*c^2 + 12*c*d*x + 8*d^2*x^2) + 20*a*b^3*d*(-c^
3 + 6*c^2*d*x + 24*c*d^2*x^2 + 16*d^3*x^3) + b^4*(3*c^4 - 8*c^3*d*x + 48*c^2*d^2*x^2 + 192*c*d^3*x^3 + 128*d^4
*x^4)))/(15*(b*c - a*d)^5*(a + b*x)^(5/2)*(c + d*x)^(3/2))

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fricas [B]  time = 3.43, size = 715, normalized size = 4.16 \[ -\frac {2 \, {\left (128 \, b^{4} d^{4} x^{4} + 3 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d + 90 \, a^{2} b^{2} c^{2} d^{2} + 60 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4} + 64 \, {\left (3 \, b^{4} c d^{3} + 5 \, a b^{3} d^{4}\right )} x^{3} + 48 \, {\left (b^{4} c^{2} d^{2} + 10 \, a b^{3} c d^{3} + 5 \, a^{2} b^{2} d^{4}\right )} x^{2} - 8 \, {\left (b^{4} c^{3} d - 15 \, a b^{3} c^{2} d^{2} - 45 \, a^{2} b^{2} c d^{3} - 5 \, a^{3} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{15 \, {\left (a^{3} b^{5} c^{7} - 5 \, a^{4} b^{4} c^{6} d + 10 \, a^{5} b^{3} c^{5} d^{2} - 10 \, a^{6} b^{2} c^{4} d^{3} + 5 \, a^{7} b c^{3} d^{4} - a^{8} c^{2} d^{5} + {\left (b^{8} c^{5} d^{2} - 5 \, a b^{7} c^{4} d^{3} + 10 \, a^{2} b^{6} c^{3} d^{4} - 10 \, a^{3} b^{5} c^{2} d^{5} + 5 \, a^{4} b^{4} c d^{6} - a^{5} b^{3} d^{7}\right )} x^{5} + {\left (2 \, b^{8} c^{6} d - 7 \, a b^{7} c^{5} d^{2} + 5 \, a^{2} b^{6} c^{4} d^{3} + 10 \, a^{3} b^{5} c^{3} d^{4} - 20 \, a^{4} b^{4} c^{2} d^{5} + 13 \, a^{5} b^{3} c d^{6} - 3 \, a^{6} b^{2} d^{7}\right )} x^{4} + {\left (b^{8} c^{7} + a b^{7} c^{6} d - 17 \, a^{2} b^{6} c^{5} d^{2} + 35 \, a^{3} b^{5} c^{4} d^{3} - 25 \, a^{4} b^{4} c^{3} d^{4} - a^{5} b^{3} c^{2} d^{5} + 9 \, a^{6} b^{2} c d^{6} - 3 \, a^{7} b d^{7}\right )} x^{3} + {\left (3 \, a b^{7} c^{7} - 9 \, a^{2} b^{6} c^{6} d + a^{3} b^{5} c^{5} d^{2} + 25 \, a^{4} b^{4} c^{4} d^{3} - 35 \, a^{5} b^{3} c^{3} d^{4} + 17 \, a^{6} b^{2} c^{2} d^{5} - a^{7} b c d^{6} - a^{8} d^{7}\right )} x^{2} + {\left (3 \, a^{2} b^{6} c^{7} - 13 \, a^{3} b^{5} c^{6} d + 20 \, a^{4} b^{4} c^{5} d^{2} - 10 \, a^{5} b^{3} c^{4} d^{3} - 5 \, a^{6} b^{2} c^{3} d^{4} + 7 \, a^{7} b c^{2} d^{5} - 2 \, a^{8} c d^{6}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(7/2)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

-2/15*(128*b^4*d^4*x^4 + 3*b^4*c^4 - 20*a*b^3*c^3*d + 90*a^2*b^2*c^2*d^2 + 60*a^3*b*c*d^3 - 5*a^4*d^4 + 64*(3*
b^4*c*d^3 + 5*a*b^3*d^4)*x^3 + 48*(b^4*c^2*d^2 + 10*a*b^3*c*d^3 + 5*a^2*b^2*d^4)*x^2 - 8*(b^4*c^3*d - 15*a*b^3
*c^2*d^2 - 45*a^2*b^2*c*d^3 - 5*a^3*b*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^3*b^5*c^7 - 5*a^4*b^4*c^6*d + 10*
a^5*b^3*c^5*d^2 - 10*a^6*b^2*c^4*d^3 + 5*a^7*b*c^3*d^4 - a^8*c^2*d^5 + (b^8*c^5*d^2 - 5*a*b^7*c^4*d^3 + 10*a^2
*b^6*c^3*d^4 - 10*a^3*b^5*c^2*d^5 + 5*a^4*b^4*c*d^6 - a^5*b^3*d^7)*x^5 + (2*b^8*c^6*d - 7*a*b^7*c^5*d^2 + 5*a^
2*b^6*c^4*d^3 + 10*a^3*b^5*c^3*d^4 - 20*a^4*b^4*c^2*d^5 + 13*a^5*b^3*c*d^6 - 3*a^6*b^2*d^7)*x^4 + (b^8*c^7 + a
*b^7*c^6*d - 17*a^2*b^6*c^5*d^2 + 35*a^3*b^5*c^4*d^3 - 25*a^4*b^4*c^3*d^4 - a^5*b^3*c^2*d^5 + 9*a^6*b^2*c*d^6
- 3*a^7*b*d^7)*x^3 + (3*a*b^7*c^7 - 9*a^2*b^6*c^6*d + a^3*b^5*c^5*d^2 + 25*a^4*b^4*c^4*d^3 - 35*a^5*b^3*c^3*d^
4 + 17*a^6*b^2*c^2*d^5 - a^7*b*c*d^6 - a^8*d^7)*x^2 + (3*a^2*b^6*c^7 - 13*a^3*b^5*c^6*d + 20*a^4*b^4*c^5*d^2 -
10*a^5*b^3*c^4*d^3 - 5*a^6*b^2*c^3*d^4 + 7*a^7*b*c^2*d^5 - 2*a^8*c*d^6)*x)

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giac [B]  time = 3.92, size = 1203, normalized size = 6.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(7/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

-2/3*sqrt(b*x + a)*(11*(b^8*c^4*d^5*abs(b) - 4*a*b^7*c^3*d^6*abs(b) + 6*a^2*b^6*c^2*d^7*abs(b) - 4*a^3*b^5*c*d
^8*abs(b) + a^4*b^4*d^9*abs(b))*(b*x + a)/(b^11*c^9*d - 9*a*b^10*c^8*d^2 + 36*a^2*b^9*c^7*d^3 - 84*a^3*b^8*c^6
*d^4 + 126*a^4*b^7*c^5*d^5 - 126*a^5*b^6*c^4*d^6 + 84*a^6*b^5*c^3*d^7 - 36*a^7*b^4*c^2*d^8 + 9*a^8*b^3*c*d^9 -
a^9*b^2*d^10) + 12*(b^9*c^5*d^4*abs(b) - 5*a*b^8*c^4*d^5*abs(b) + 10*a^2*b^7*c^3*d^6*abs(b) - 10*a^3*b^6*c^2*
d^7*abs(b) + 5*a^4*b^5*c*d^8*abs(b) - a^5*b^4*d^9*abs(b))/(b^11*c^9*d - 9*a*b^10*c^8*d^2 + 36*a^2*b^9*c^7*d^3
- 84*a^3*b^8*c^6*d^4 + 126*a^4*b^7*c^5*d^5 - 126*a^5*b^6*c^4*d^6 + 84*a^6*b^5*c^3*d^7 - 36*a^7*b^4*c^2*d^8 + 9
*a^8*b^3*c*d^9 - a^9*b^2*d^10))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 4/15*(73*sqrt(b*d)*b^11*c^4*d^2 - 292*
sqrt(b*d)*a*b^10*c^3*d^3 + 438*sqrt(b*d)*a^2*b^9*c^2*d^4 - 292*sqrt(b*d)*a^3*b^8*c*d^5 + 73*sqrt(b*d)*a^4*b^7*
d^6 - 320*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^9*c^3*d^2 + 960*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^8*c^2*d^3 - 960*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^7*c*d^4 + 320*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^6*d^5 + 490*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^4*b^7*c^2*d^2 - 980*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^6*c*d^3 + 490*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^
5*d^4 - 240*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^5*c*d^2 + 240*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^4*d^3 + 45*sqrt(b*d)*(sqrt(b*d)*sqrt
(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b^3*d^2)/((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)^5)

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maple [A]  time = 0.01, size = 256, normalized size = 1.49 \[ -\frac {2 \left (-128 b^{4} x^{4} d^{4}-320 a \,b^{3} d^{4} x^{3}-192 b^{4} c \,d^{3} x^{3}-240 a^{2} b^{2} d^{4} x^{2}-480 a \,b^{3} c \,d^{3} x^{2}-48 b^{4} c^{2} d^{2} x^{2}-40 a^{3} b \,d^{4} x -360 a^{2} b^{2} c \,d^{3} x -120 a \,b^{3} c^{2} d^{2} x +8 b^{4} c^{3} d x +5 a^{4} d^{4}-60 a^{3} b c \,d^{3}-90 a^{2} b^{2} c^{2} d^{2}+20 a \,b^{3} c^{3} d -3 b^{4} c^{4}\right )}{15 \left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {3}{2}} \left (a^{5} d^{5}-5 a^{4} b c \,d^{4}+10 a^{3} b^{2} c^{2} d^{3}-10 a^{2} b^{3} c^{3} d^{2}+5 a \,b^{4} c^{4} d -b^{5} c^{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(7/2)/(d*x+c)^(5/2),x)

[Out]

-2/15*(-128*b^4*d^4*x^4-320*a*b^3*d^4*x^3-192*b^4*c*d^3*x^3-240*a^2*b^2*d^4*x^2-480*a*b^3*c*d^3*x^2-48*b^4*c^2
*d^2*x^2-40*a^3*b*d^4*x-360*a^2*b^2*c*d^3*x-120*a*b^3*c^2*d^2*x+8*b^4*c^3*d*x+5*a^4*d^4-60*a^3*b*c*d^3-90*a^2*
b^2*c^2*d^2+20*a*b^3*c^3*d-3*b^4*c^4)/(b*x+a)^(5/2)/(d*x+c)^(3/2)/(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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(7/2)/(d*x+c)^(5/2),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 = 1.53, size = 346, normalized size = 2.01 \[ \frac {\sqrt {c+d\,x}\,\left (\frac {32\,x^2\,\left (5\,a^2\,d^2+10\,a\,b\,c\,d+b^2\,c^2\right )}{5\,{\left (a\,d-b\,c\right )}^5}+\frac {256\,b^2\,d^2\,x^4}{15\,{\left (a\,d-b\,c\right )}^5}+\frac {-10\,a^4\,d^4+120\,a^3\,b\,c\,d^3+180\,a^2\,b^2\,c^2\,d^2-40\,a\,b^3\,c^3\,d+6\,b^4\,c^4}{15\,b^2\,d^2\,{\left (a\,d-b\,c\right )}^5}+\frac {x\,\left (80\,a^3\,b\,d^4+720\,a^2\,b^2\,c\,d^3+240\,a\,b^3\,c^2\,d^2-16\,b^4\,c^3\,d\right )}{15\,b^2\,d^2\,{\left (a\,d-b\,c\right )}^5}+\frac {128\,b\,d\,x^3\,\left (5\,a\,d+3\,b\,c\right )}{15\,{\left (a\,d-b\,c\right )}^5}\right )}{x^4\,\sqrt {a+b\,x}+\frac {x^2\,\sqrt {a+b\,x}\,\left (a^2\,d^2+4\,a\,b\,c\,d+b^2\,c^2\right )}{b^2\,d^2}+\frac {2\,x^3\,\left (a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b\,d}+\frac {a^2\,c^2\,\sqrt {a+b\,x}}{b^2\,d^2}+\frac {2\,a\,c\,x\,\left (a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b^2\,d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(7/2)*(c + d*x)^(5/2)),x)

[Out]

((c + d*x)^(1/2)*((32*x^2*(5*a^2*d^2 + b^2*c^2 + 10*a*b*c*d))/(5*(a*d - b*c)^5) + (256*b^2*d^2*x^4)/(15*(a*d -
b*c)^5) + (6*b^4*c^4 - 10*a^4*d^4 + 180*a^2*b^2*c^2*d^2 - 40*a*b^3*c^3*d + 120*a^3*b*c*d^3)/(15*b^2*d^2*(a*d
- b*c)^5) + (x*(80*a^3*b*d^4 - 16*b^4*c^3*d + 240*a*b^3*c^2*d^2 + 720*a^2*b^2*c*d^3))/(15*b^2*d^2*(a*d - b*c)^
5) + (128*b*d*x^3*(5*a*d + 3*b*c))/(15*(a*d - b*c)^5)))/(x^4*(a + b*x)^(1/2) + (x^2*(a + b*x)^(1/2)*(a^2*d^2 +
b^2*c^2 + 4*a*b*c*d))/(b^2*d^2) + (2*x^3*(a*d + b*c)*(a + b*x)^(1/2))/(b*d) + (a^2*c^2*(a + b*x)^(1/2))/(b^2*
d^2) + (2*a*c*x*(a*d + b*c)*(a + b*x)^(1/2))/(b^2*d^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x\right )^{\frac {7}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(7/2)/(d*x+c)**(5/2),x)

[Out]

Integral(1/((a + b*x)**(7/2)*(c + d*x)**(5/2)), x)

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