3.1489 \(\int \frac{(c+d x)^{5/2}}{(a+b x)^{11/2}} \, dx\)
Optimal. Leaf size=66 \[ \frac{4 d (c+d x)^{7/2}}{63 (a+b x)^{7/2} (b c-a d)^2}-\frac{2 (c+d x)^{7/2}}{9 (a+b x)^{9/2} (b c-a d)} \]
[Out]
(-2*(c + d*x)^(7/2))/(9*(b*c - a*d)*(a + b*x)^(9/2)) + (4*d*(c + d*x)^(7/2))/(63*(b*c - a*d)^2*(a + b*x)^(7/2)
)
________________________________________________________________________________________
Rubi [A] time = 0.0090831, antiderivative size = 66, normalized size of antiderivative = 1.,
number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used =
{45, 37} \[ \frac{4 d (c+d x)^{7/2}}{63 (a+b x)^{7/2} (b c-a d)^2}-\frac{2 (c+d x)^{7/2}}{9 (a+b x)^{9/2} (b c-a d)} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)^(5/2)/(a + b*x)^(11/2),x]
[Out]
(-2*(c + d*x)^(7/2))/(9*(b*c - a*d)*(a + b*x)^(9/2)) + (4*d*(c + d*x)^(7/2))/(63*(b*c - a*d)^2*(a + b*x)^(7/2)
)
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{(c+d x)^{5/2}}{(a+b x)^{11/2}} \, dx &=-\frac{2 (c+d x)^{7/2}}{9 (b c-a d) (a+b x)^{9/2}}-\frac{(2 d) \int \frac{(c+d x)^{5/2}}{(a+b x)^{9/2}} \, dx}{9 (b c-a d)}\\ &=-\frac{2 (c+d x)^{7/2}}{9 (b c-a d) (a+b x)^{9/2}}+\frac{4 d (c+d x)^{7/2}}{63 (b c-a d)^2 (a+b x)^{7/2}}\\ \end{align*}
Mathematica [A] time = 0.0314782, size = 46, normalized size = 0.7 \[ \frac{2 (c+d x)^{7/2} (9 a d-7 b c+2 b d x)}{63 (a+b x)^{9/2} (b c-a d)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)^(5/2)/(a + b*x)^(11/2),x]
[Out]
(2*(c + d*x)^(7/2)*(-7*b*c + 9*a*d + 2*b*d*x))/(63*(b*c - a*d)^2*(a + b*x)^(9/2))
________________________________________________________________________________________
Maple [A] time = 0.005, size = 54, normalized size = 0.8 \begin{align*}{\frac{4\,bdx+18\,ad-14\,bc}{63\,{a}^{2}{d}^{2}-126\,abcd+63\,{b}^{2}{c}^{2}} \left ( dx+c \right ) ^{{\frac{7}{2}}} \left ( bx+a \right ) ^{-{\frac{9}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x+c)^(5/2)/(b*x+a)^(11/2),x)
[Out]
2/63*(d*x+c)^(7/2)*(2*b*d*x+9*a*d-7*b*c)/(b*x+a)^(9/2)/(a^2*d^2-2*a*b*c*d+b^2*c^2)
________________________________________________________________________________________
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((d*x+c)^(5/2)/(b*x+a)^(11/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [B] time = 44.933, size = 605, normalized size = 9.17 \begin{align*} \frac{2 \,{\left (2 \, b d^{4} x^{4} - 7 \, b c^{4} + 9 \, a c^{3} d -{\left (b c d^{3} - 9 \, a d^{4}\right )} x^{3} - 3 \,{\left (5 \, b c^{2} d^{2} - 9 \, a c d^{3}\right )} x^{2} -{\left (19 \, b c^{3} d - 27 \, a c^{2} d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{63 \,{\left (a^{5} b^{2} c^{2} - 2 \, a^{6} b c d + a^{7} d^{2} +{\left (b^{7} c^{2} - 2 \, a b^{6} c d + a^{2} b^{5} d^{2}\right )} x^{5} + 5 \,{\left (a b^{6} c^{2} - 2 \, a^{2} b^{5} c d + a^{3} b^{4} d^{2}\right )} x^{4} + 10 \,{\left (a^{2} b^{5} c^{2} - 2 \, a^{3} b^{4} c d + a^{4} b^{3} d^{2}\right )} x^{3} + 10 \,{\left (a^{3} b^{4} c^{2} - 2 \, a^{4} b^{3} c d + a^{5} b^{2} d^{2}\right )} x^{2} + 5 \,{\left (a^{4} b^{3} c^{2} - 2 \, a^{5} b^{2} c d + a^{6} b d^{2}\right )} x\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^(5/2)/(b*x+a)^(11/2),x, algorithm="fricas")
[Out]
2/63*(2*b*d^4*x^4 - 7*b*c^4 + 9*a*c^3*d - (b*c*d^3 - 9*a*d^4)*x^3 - 3*(5*b*c^2*d^2 - 9*a*c*d^3)*x^2 - (19*b*c^
3*d - 27*a*c^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^5*b^2*c^2 - 2*a^6*b*c*d + a^7*d^2 + (b^7*c^2 - 2*a*b^6*c
*d + a^2*b^5*d^2)*x^5 + 5*(a*b^6*c^2 - 2*a^2*b^5*c*d + a^3*b^4*d^2)*x^4 + 10*(a^2*b^5*c^2 - 2*a^3*b^4*c*d + a^
4*b^3*d^2)*x^3 + 10*(a^3*b^4*c^2 - 2*a^4*b^3*c*d + a^5*b^2*d^2)*x^2 + 5*(a^4*b^3*c^2 - 2*a^5*b^2*c*d + a^6*b*d
^2)*x)
________________________________________________________________________________________
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((d*x+c)**(5/2)/(b*x+a)**(11/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 2.15476, size = 2465, normalized size = 37.35 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^(5/2)/(b*x+a)^(11/2),x, algorithm="giac")
[Out]
8/63*(sqrt(b*d)*b^14*c^7*d^4*abs(b) - 7*sqrt(b*d)*a*b^13*c^6*d^5*abs(b) + 21*sqrt(b*d)*a^2*b^12*c^5*d^6*abs(b)
- 35*sqrt(b*d)*a^3*b^11*c^4*d^7*abs(b) + 35*sqrt(b*d)*a^4*b^10*c^3*d^8*abs(b) - 21*sqrt(b*d)*a^5*b^9*c^2*d^9*
abs(b) + 7*sqrt(b*d)*a^6*b^8*c*d^10*abs(b) - sqrt(b*d)*a^7*b^7*d^11*abs(b) - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x +
a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^12*c^6*d^4*abs(b) + 54*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr
t(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^11*c^5*d^5*abs(b) - 135*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^10*c^4*d^6*abs(b) + 180*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^3*d^7*abs(b) - 135*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^2*d^8*abs(b) + 54*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*c*d^9*abs(b) - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*
d))^2*a^6*b^6*d^10*abs(b) - 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^1
0*c^5*d^4*abs(b) + 135*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^4*d
^5*abs(b) - 270*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^3*d^6*ab
s(b) + 270*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^2*d^7*abs(b)
- 135*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*c*d^8*abs(b) + 27*sq
rt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^5*b^5*d^9*abs(b) - 189*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^4*d^4*abs(b) + 756*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^3*d^5*abs(b) - 1134*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^2*d^6*abs(b) + 756*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*c*d^7*abs(b) - 189*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - s
qrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^4*b^4*d^8*abs(b) - 189*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^3*d^4*abs(b) + 567*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^2*d^5*abs(b) - 567*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*c*d^6*abs(b) + 189*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*
d))^8*a^3*b^3*d^7*abs(b) - 315*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^2*d^4*abs(b) + 630*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*c*d^
5*abs(b) - 315*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^10*a^2*b^2*d^6*abs(b)
- 105*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*c*d^4*abs(b) + 105*sqr
t(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^12*a*b*d^5*abs(b) - 63*sqrt(b*d)*(sqrt(
b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^14*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)^9*b^3)