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

Optimal. Leaf size=101 \[ -\frac {16 d^2 \sqrt {c+d x}}{15 \sqrt {a+b x} (b c-a d)^3}+\frac {8 d \sqrt {c+d x}}{15 (a+b x)^{3/2} (b c-a d)^2}-\frac {2 \sqrt {c+d x}}{5 (a+b x)^{5/2} (b c-a d)} \]

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Rubi [A]  time = 0.02, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, 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 {16 d^2 \sqrt {c+d x}}{15 \sqrt {a+b x} (b c-a d)^3}+\frac {8 d \sqrt {c+d x}}{15 (a+b x)^{3/2} (b c-a d)^2}-\frac {2 \sqrt {c+d x}}{5 (a+b x)^{5/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 75, normalized size = 0.74 \begin {gather*} -\frac {2 \sqrt {c+d x} \left (15 a^2 d^2-10 a b d (c-2 d x)+b^2 \left (3 c^2-4 c d x+8 d^2 x^2\right )\right )}{15 (a+b x)^{5/2} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x]*(15*a^2*d^2 - 10*a*b*d*(c - 2*d*x) + b^2*(3*c^2 - 4*c*d*x + 8*d^2*x^2)))/(15*(b*c - a*d)^3*(
a + b*x)^(5/2))

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IntegrateAlgebraic [A]  time = 0.10, size = 83, normalized size = 0.82 \begin {gather*} -\frac {2 \left (\frac {3 b^2 (c+d x)^{5/2}}{(a+b x)^{5/2}}+\frac {15 d^2 \sqrt {c+d x}}{\sqrt {a+b x}}-\frac {10 b d (c+d x)^{3/2}}{(a+b x)^{3/2}}\right )}{15 (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*((15*d^2*Sqrt[c + d*x])/Sqrt[a + b*x] - (10*b*d*(c + d*x)^(3/2))/(a + b*x)^(3/2) + (3*b^2*(c + d*x)^(5/2))
/(a + b*x)^(5/2)))/(15*(b*c - a*d)^3)

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fricas [B]  time = 1.32, size = 251, normalized size = 2.49 \begin {gather*} -\frac {2 \, {\left (8 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2} - 4 \, {\left (b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{15 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\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}\right )} x^{3} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(8*b^2*d^2*x^2 + 3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2 - 4*(b^2*c*d - 5*a*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x
+ c)/(a^3*b^3*c^3 - 3*a^4*b^2*c^2*d + 3*a^5*b*c*d^2 - a^6*d^3 + (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)*x^3 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^4*b^2*d^3)*x^2 + 3*(a^2*b^4*c^3 - 3*a^
3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b*d^3)*x)

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giac [B]  time = 1.27, size = 227, normalized size = 2.25 \begin {gather*} -\frac {32 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 5 \, {\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^{2} c + 5 \, {\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 d + 10 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}\right )} \sqrt {b d} b^{3} d^{2}}{15 \, {\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 )}^{5} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-32/15*(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 5*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)
)^2*b^2*c + 5*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + 10*(sqrt(b*d)*sqrt(b*x
+ a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)*sqrt(b*d)*b^3*d^2/((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*abs(b))

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maple [A]  time = 0.01, size = 105, normalized size = 1.04 \begin {gather*} \frac {2 \sqrt {d x +c}\, \left (8 b^{2} x^{2} d^{2}+20 a b \,d^{2} x -4 b^{2} c d x +15 a^{2} d^{2}-10 a b c d +3 b^{2} c^{2}\right )}{15 \left (b x +a \right )^{\frac {5}{2}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(d*x+c)^(1/2)*(8*b^2*d^2*x^2+20*a*b*d^2*x-4*b^2*c*d*x+15*a^2*d^2-10*a*b*c*d+3*b^2*c^2)/(b*x+a)^(5/2)/(a^3
*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)

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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)^(7/2)/(d*x+c)^(1/2),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 = 1.01, size = 133, normalized size = 1.32 \begin {gather*} \frac {\sqrt {c+d\,x}\,\left (\frac {16\,d^2\,x^2}{15\,{\left (a\,d-b\,c\right )}^3}+\frac {30\,a^2\,d^2-20\,a\,b\,c\,d+6\,b^2\,c^2}{15\,b^2\,{\left (a\,d-b\,c\right )}^3}+\frac {8\,d\,x\,\left (5\,a\,d-b\,c\right )}{15\,b\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,\sqrt {a+b\,x}+\frac {a^2\,\sqrt {a+b\,x}}{b^2}+\frac {2\,a\,x\,\sqrt {a+b\,x}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c + d*x)^(1/2)*((16*d^2*x^2)/(15*(a*d - b*c)^3) + (30*a^2*d^2 + 6*b^2*c^2 - 20*a*b*c*d)/(15*b^2*(a*d - b*c)^
3) + (8*d*x*(5*a*d - b*c))/(15*b*(a*d - b*c)^3)))/(x^2*(a + b*x)^(1/2) + (a^2*(a + b*x)^(1/2))/b^2 + (2*a*x*(a
+ b*x)^(1/2))/b)

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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 {7}{2}} \sqrt {c + d x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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