3.16.2 \(\int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx\)

Optimal. Leaf size=101 \[ -\frac {128 d^2 \sqrt [4]{c+d x}}{45 \sqrt [4]{a+b x} (b c-a d)^3}+\frac {32 d \sqrt [4]{c+d x}}{45 (a+b x)^{5/4} (b c-a d)^2}-\frac {4 \sqrt [4]{c+d x}}{9 (a+b x)^{9/4} (b c-a d)} \]

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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]{c+d x}}{45 \sqrt [4]{a+b x} (b c-a d)^3}+\frac {32 d \sqrt [4]{c+d x}}{45 (a+b x)^{5/4} (b c-a d)^2}-\frac {4 \sqrt [4]{c+d x}}{9 (a+b x)^{9/4} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(13/4)*(c + d*x)^(3/4)),x]

[Out]

(-4*(c + d*x)^(1/4))/(9*(b*c - a*d)*(a + b*x)^(9/4)) + (32*d*(c + d*x)^(1/4))/(45*(b*c - a*d)^2*(a + b*x)^(5/4
)) - (128*d^2*(c + d*x)^(1/4))/(45*(b*c - a*d)^3*(a + b*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)^{13/4} (c+d x)^{3/4}} \, dx &=-\frac {4 \sqrt [4]{c+d x}}{9 (b c-a d) (a+b x)^{9/4}}-\frac {(8 d) \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx}{9 (b c-a d)}\\ &=-\frac {4 \sqrt [4]{c+d x}}{9 (b c-a d) (a+b x)^{9/4}}+\frac {32 d \sqrt [4]{c+d x}}{45 (b c-a d)^2 (a+b x)^{5/4}}+\frac {\left (32 d^2\right ) \int \frac {1}{(a+b x)^{5/4} (c+d x)^{3/4}} \, dx}{45 (b c-a d)^2}\\ &=-\frac {4 \sqrt [4]{c+d x}}{9 (b c-a d) (a+b x)^{9/4}}+\frac {32 d \sqrt [4]{c+d x}}{45 (b c-a d)^2 (a+b x)^{5/4}}-\frac {128 d^2 \sqrt [4]{c+d x}}{45 (b c-a d)^3 \sqrt [4]{a+b x}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 75, normalized size = 0.74 \begin {gather*} -\frac {4 \sqrt [4]{c+d x} \left (45 a^2 d^2-18 a b d (c-4 d x)+b^2 \left (5 c^2-8 c d x+32 d^2 x^2\right )\right )}{45 (a+b x)^{9/4} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(13/4)*(c + d*x)^(3/4)),x]

[Out]

(-4*(c + d*x)^(1/4)*(45*a^2*d^2 - 18*a*b*d*(c - 4*d*x) + b^2*(5*c^2 - 8*c*d*x + 32*d^2*x^2)))/(45*(b*c - a*d)^
3*(a + b*x)^(9/4))

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IntegrateAlgebraic [A]  time = 0.12, size = 83, normalized size = 0.82 \begin {gather*} -\frac {4 \left (\frac {5 b^2 (c+d x)^{9/4}}{(a+b x)^{9/4}}+\frac {45 d^2 \sqrt [4]{c+d x}}{\sqrt [4]{a+b x}}-\frac {18 b d (c+d x)^{5/4}}{(a+b x)^{5/4}}\right )}{45 (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^(13/4)*(c + d*x)^(3/4)),x]

[Out]

(-4*((45*d^2*(c + d*x)^(1/4))/(a + b*x)^(1/4) - (18*b*d*(c + d*x)^(5/4))/(a + b*x)^(5/4) + (5*b^2*(c + d*x)^(9
/4))/(a + b*x)^(9/4)))/(45*(b*c - a*d)^3)

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fricas [B]  time = 0.88, size = 251, normalized size = 2.49 \begin {gather*} -\frac {4 \, {\left (32 \, b^{2} d^{2} x^{2} + 5 \, b^{2} c^{2} - 18 \, a b c d + 45 \, a^{2} d^{2} - 8 \, {\left (b^{2} c d - 9 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{45 \, {\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)^(13/4)/(d*x+c)^(3/4),x, algorithm="fricas")

[Out]

-4/45*(32*b^2*d^2*x^2 + 5*b^2*c^2 - 18*a*b*c*d + 45*a^2*d^2 - 8*(b^2*c*d - 9*a*b*d^2)*x)*(b*x + a)^(3/4)*(d*x
+ c)^(1/4)/(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 [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x + a\right )}^{\frac {13}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(13/4)/(d*x+c)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(13/4)*(d*x + c)^(3/4)), x)

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maple [A]  time = 0.01, size = 105, normalized size = 1.04 \begin {gather*} \frac {4 \left (d x +c \right )^{\frac {1}{4}} \left (32 b^{2} x^{2} d^{2}+72 a b \,d^{2} x -8 b^{2} c d x +45 a^{2} d^{2}-18 a b c d +5 b^{2} c^{2}\right )}{45 \left (b x +a \right )^{\frac {9}{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)^(13/4)/(d*x+c)^(3/4),x)

[Out]

4/45*(d*x+c)^(1/4)*(32*b^2*d^2*x^2+72*a*b*d^2*x-8*b^2*c*d*x+45*a^2*d^2-18*a*b*c*d+5*b^2*c^2)/(b*x+a)^(9/4)/(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 {13}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(13/4)/(d*x+c)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(13/4)*(d*x + c)^(3/4)), x)

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mupad [B]  time = 1.02, size = 133, normalized size = 1.32 \begin {gather*} \frac {{\left (c+d\,x\right )}^{1/4}\,\left (\frac {128\,d^2\,x^2}{45\,{\left (a\,d-b\,c\right )}^3}+\frac {180\,a^2\,d^2-72\,a\,b\,c\,d+20\,b^2\,c^2}{45\,b^2\,{\left (a\,d-b\,c\right )}^3}+\frac {32\,d\,x\,\left (9\,a\,d-b\,c\right )}{45\,b\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,{\left (a+b\,x\right )}^{1/4}+\frac {a^2\,{\left (a+b\,x\right )}^{1/4}}{b^2}+\frac {2\,a\,x\,{\left (a+b\,x\right )}^{1/4}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(13/4)*(c + d*x)^(3/4)),x)

[Out]

((c + d*x)^(1/4)*((128*d^2*x^2)/(45*(a*d - b*c)^3) + (180*a^2*d^2 + 20*b^2*c^2 - 72*a*b*c*d)/(45*b^2*(a*d - b*
c)^3) + (32*d*x*(9*a*d - b*c))/(45*b*(a*d - b*c)^3)))/(x^2*(a + b*x)^(1/4) + (a^2*(a + b*x)^(1/4))/b^2 + (2*a*
x*(a + b*x)^(1/4))/b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(13/4)/(d*x+c)**(3/4),x)

[Out]

Timed out

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