3.16.8 \(\int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx\)

Optimal. Leaf size=101 \[ \frac {128 d^2 \sqrt [4]{a+b x}}{21 \sqrt [4]{c+d x} (b c-a d)^3}+\frac {32 d}{21 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)^2}-\frac {4}{7 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)} \]

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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 {128 d^2 \sqrt [4]{a+b x}}{21 \sqrt [4]{c+d x} (b c-a d)^3}+\frac {32 d}{21 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)^2}-\frac {4}{7 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(11/4)*(c + d*x)^(5/4)),x]

[Out]

-4/(7*(b*c - a*d)*(a + b*x)^(7/4)*(c + d*x)^(1/4)) + (32*d)/(21*(b*c - a*d)^2*(a + b*x)^(3/4)*(c + d*x)^(1/4))
+ (128*d^2*(a + b*x)^(1/4))/(21*(b*c - a*d)^3*(c + d*x)^(1/4))

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)^{11/4} (c+d x)^{5/4}} \, dx &=-\frac {4}{7 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}-\frac {(8 d) \int \frac {1}{(a+b x)^{7/4} (c+d x)^{5/4}} \, dx}{7 (b c-a d)}\\ &=-\frac {4}{7 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}+\frac {32 d}{21 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}+\frac {\left (32 d^2\right ) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{5/4}} \, dx}{21 (b c-a d)^2}\\ &=-\frac {4}{7 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}+\frac {32 d}{21 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}+\frac {128 d^2 \sqrt [4]{a+b x}}{21 (b c-a d)^3 \sqrt [4]{c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 76, normalized size = 0.75 \begin {gather*} \frac {84 a^2 d^2+56 a b d (c+4 d x)+4 b^2 \left (-3 c^2+8 c d x+32 d^2 x^2\right )}{21 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(11/4)*(c + d*x)^(5/4)),x]

[Out]

(84*a^2*d^2 + 56*a*b*d*(c + 4*d*x) + 4*b^2*(-3*c^2 + 8*c*d*x + 32*d^2*x^2))/(21*(b*c - a*d)^3*(a + b*x)^(7/4)*
(c + d*x)^(1/4))

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IntegrateAlgebraic [A]  time = 0.13, size = 73, normalized size = 0.72 \begin {gather*} \frac {4 (c+d x)^{7/4} \left (\frac {21 d^2 (a+b x)^2}{(c+d x)^2}+\frac {14 b d (a+b x)}{c+d x}-3 b^2\right )}{21 (a+b x)^{7/4} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^(11/4)*(c + d*x)^(5/4)),x]

[Out]

(4*(c + d*x)^(7/4)*(-3*b^2 + (21*d^2*(a + b*x)^2)/(c + d*x)^2 + (14*b*d*(a + b*x))/(c + d*x)))/(21*(b*c - a*d)
^3*(a + b*x)^(7/4))

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fricas [B]  time = 1.39, size = 273, normalized size = 2.70 \begin {gather*} \frac {4 \, {\left (32 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c^{2} + 14 \, a b c d + 21 \, a^{2} d^{2} + 8 \, {\left (b^{2} c d + 7 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{21 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(11/4)/(d*x+c)^(5/4),x, algorithm="fricas")

[Out]

4/21*(32*b^2*d^2*x^2 - 3*b^2*c^2 + 14*a*b*c*d + 21*a^2*d^2 + 8*(b^2*c*d + 7*a*b*d^2)*x)*(b*x + a)^(1/4)*(d*x +
c)^(3/4)/(a^2*b^3*c^4 - 3*a^3*b^2*c^3*d + 3*a^4*b*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*
b^3*c*d^3 - a^3*b^2*d^4)*x^3 + (b^5*c^4 - a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^2
+ (2*a*b^4*c^4 - 5*a^2*b^3*c^3*d + 3*a^3*b^2*c^2*d^2 + a^4*b*c*d^3 - a^5*d^4)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x + a\right )}^{\frac {11}{4}} {\left (d x + c\right )}^{\frac {5}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(11/4)/(d*x+c)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(11/4)*(d*x + c)^(5/4)), x)

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maple [A]  time = 0.01, size = 105, normalized size = 1.04 \begin {gather*} -\frac {4 \left (32 b^{2} x^{2} d^{2}+56 a b \,d^{2} x +8 b^{2} c d x +21 a^{2} d^{2}+14 a b c d -3 b^{2} c^{2}\right )}{21 \left (b x +a \right )^{\frac {7}{4}} \left (d x +c \right )^{\frac {1}{4}} \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)^(11/4)/(d*x+c)^(5/4),x)

[Out]

-4/21*(32*b^2*d^2*x^2+56*a*b*d^2*x+8*b^2*c*d*x+21*a^2*d^2+14*a*b*c*d-3*b^2*c^2)/(b*x+a)^(7/4)/(d*x+c)^(1/4)/(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]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x + a\right )}^{\frac {11}{4}} {\left (d x + c\right )}^{\frac {5}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(11/4)/(d*x+c)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(11/4)*(d*x + c)^(5/4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+b\,x\right )}^{11/4}\,{\left (c+d\,x\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(11/4)*(c + d*x)^(5/4)),x)

[Out]

int(1/((a + b*x)^(11/4)*(c + d*x)^(5/4)), x)

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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 {11}{4}} \left (c + d x\right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(11/4)/(d*x+c)**(5/4),x)

[Out]

Integral(1/((a + b*x)**(11/4)*(c + d*x)**(5/4)), x)

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