3.16.9 \(\int \frac {1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx\)

Optimal. Leaf size=136 \[ -\frac {512 d^3 \sqrt [4]{a+b x}}{77 \sqrt [4]{c+d x} (b c-a d)^4}-\frac {128 d^2}{77 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)^3}+\frac {48 d}{77 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)^2}-\frac {4}{11 (a+b x)^{11/4} \sqrt [4]{c+d x} (b c-a d)} \]

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Rubi [A]  time = 0.03, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {512 d^3 \sqrt [4]{a+b x}}{77 \sqrt [4]{c+d x} (b c-a d)^4}-\frac {128 d^2}{77 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)^3}+\frac {48 d}{77 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)^2}-\frac {4}{11 (a+b x)^{11/4} \sqrt [4]{c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(15/4)*(c + d*x)^(5/4)),x]

[Out]

-4/(11*(b*c - a*d)*(a + b*x)^(11/4)*(c + d*x)^(1/4)) + (48*d)/(77*(b*c - a*d)^2*(a + b*x)^(7/4)*(c + d*x)^(1/4
)) - (128*d^2)/(77*(b*c - a*d)^3*(a + b*x)^(3/4)*(c + d*x)^(1/4)) - (512*d^3*(a + b*x)^(1/4))/(77*(b*c - a*d)^
4*(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)^{15/4} (c+d x)^{5/4}} \, dx &=-\frac {4}{11 (b c-a d) (a+b x)^{11/4} \sqrt [4]{c+d x}}-\frac {(12 d) \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx}{11 (b c-a d)}\\ &=-\frac {4}{11 (b c-a d) (a+b x)^{11/4} \sqrt [4]{c+d x}}+\frac {48 d}{77 (b c-a d)^2 (a+b x)^{7/4} \sqrt [4]{c+d x}}+\frac {\left (96 d^2\right ) \int \frac {1}{(a+b x)^{7/4} (c+d x)^{5/4}} \, dx}{77 (b c-a d)^2}\\ &=-\frac {4}{11 (b c-a d) (a+b x)^{11/4} \sqrt [4]{c+d x}}+\frac {48 d}{77 (b c-a d)^2 (a+b x)^{7/4} \sqrt [4]{c+d x}}-\frac {128 d^2}{77 (b c-a d)^3 (a+b x)^{3/4} \sqrt [4]{c+d x}}-\frac {\left (128 d^3\right ) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{5/4}} \, dx}{77 (b c-a d)^3}\\ &=-\frac {4}{11 (b c-a d) (a+b x)^{11/4} \sqrt [4]{c+d x}}+\frac {48 d}{77 (b c-a d)^2 (a+b x)^{7/4} \sqrt [4]{c+d x}}-\frac {128 d^2}{77 (b c-a d)^3 (a+b x)^{3/4} \sqrt [4]{c+d x}}-\frac {512 d^3 \sqrt [4]{a+b x}}{77 (b c-a d)^4 \sqrt [4]{c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 116, normalized size = 0.85 \begin {gather*} -\frac {4 \left (77 a^3 d^3+77 a^2 b d^2 (c+4 d x)+11 a b^2 d \left (-3 c^2+8 c d x+32 d^2 x^2\right )+b^3 \left (7 c^3-12 c^2 d x+32 c d^2 x^2+128 d^3 x^3\right )\right )}{77 (a+b x)^{11/4} \sqrt [4]{c+d x} (b c-a d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(15/4)*(c + d*x)^(5/4)),x]

[Out]

(-4*(77*a^3*d^3 + 77*a^2*b*d^2*(c + 4*d*x) + 11*a*b^2*d*(-3*c^2 + 8*c*d*x + 32*d^2*x^2) + b^3*(7*c^3 - 12*c^2*
d*x + 32*c*d^2*x^2 + 128*d^3*x^3)))/(77*(b*c - a*d)^4*(a + b*x)^(11/4)*(c + d*x)^(1/4))

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IntegrateAlgebraic [A]  time = 0.13, size = 95, normalized size = 0.70 \begin {gather*} -\frac {4 (c+d x)^{11/4} \left (-\frac {33 b^2 d (a+b x)}{c+d x}+\frac {77 d^3 (a+b x)^3}{(c+d x)^3}+\frac {77 b d^2 (a+b x)^2}{(c+d x)^2}+7 b^3\right )}{77 (a+b x)^{11/4} (b c-a d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^(15/4)*(c + d*x)^(5/4)),x]

[Out]

(-4*(c + d*x)^(11/4)*(7*b^3 + (77*d^3*(a + b*x)^3)/(c + d*x)^3 + (77*b*d^2*(a + b*x)^2)/(c + d*x)^2 - (33*b^2*
d*(a + b*x))/(c + d*x)))/(77*(b*c - a*d)^4*(a + b*x)^(11/4))

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fricas [B]  time = 2.58, size = 457, normalized size = 3.36 \begin {gather*} -\frac {4 \, {\left (128 \, b^{3} d^{3} x^{3} + 7 \, b^{3} c^{3} - 33 \, a b^{2} c^{2} d + 77 \, a^{2} b c d^{2} + 77 \, a^{3} d^{3} + 32 \, {\left (b^{3} c d^{2} + 11 \, a b^{2} d^{3}\right )} x^{2} - 4 \, {\left (3 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} - 77 \, a^{2} b d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{77 \, {\left (a^{3} b^{4} c^{5} - 4 \, a^{4} b^{3} c^{4} d + 6 \, a^{5} b^{2} c^{3} d^{2} - 4 \, a^{6} b c^{2} d^{3} + a^{7} c d^{4} + {\left (b^{7} c^{4} d - 4 \, a b^{6} c^{3} d^{2} + 6 \, a^{2} b^{5} c^{2} d^{3} - 4 \, a^{3} b^{4} c d^{4} + a^{4} b^{3} d^{5}\right )} x^{4} + {\left (b^{7} c^{5} - a b^{6} c^{4} d - 6 \, a^{2} b^{5} c^{3} d^{2} + 14 \, a^{3} b^{4} c^{2} d^{3} - 11 \, a^{4} b^{3} c d^{4} + 3 \, a^{5} b^{2} d^{5}\right )} x^{3} + 3 \, {\left (a b^{6} c^{5} - 3 \, a^{2} b^{5} c^{4} d + 2 \, a^{3} b^{4} c^{3} d^{2} + 2 \, a^{4} b^{3} c^{2} d^{3} - 3 \, a^{5} b^{2} c d^{4} + a^{6} b d^{5}\right )} x^{2} + {\left (3 \, a^{2} b^{5} c^{5} - 11 \, a^{3} b^{4} c^{4} d + 14 \, a^{4} b^{3} c^{3} d^{2} - 6 \, a^{5} b^{2} c^{2} d^{3} - a^{6} b c d^{4} + a^{7} d^{5}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(15/4)/(d*x+c)^(5/4),x, algorithm="fricas")

[Out]

-4/77*(128*b^3*d^3*x^3 + 7*b^3*c^3 - 33*a*b^2*c^2*d + 77*a^2*b*c*d^2 + 77*a^3*d^3 + 32*(b^3*c*d^2 + 11*a*b^2*d
^3)*x^2 - 4*(3*b^3*c^2*d - 22*a*b^2*c*d^2 - 77*a^2*b*d^3)*x)*(b*x + a)^(1/4)*(d*x + c)^(3/4)/(a^3*b^4*c^5 - 4*
a^4*b^3*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d - 4*a*b^6*c^3*d^2 + 6*a^2*b^5*c^2
*d^3 - 4*a^3*b^4*c*d^4 + a^4*b^3*d^5)*x^4 + (b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b^4*c^2*d^3 -
11*a^4*b^3*c*d^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3*b^4*c^3*d^2 + 2*a^4*b^3*c^2*d^3
- 3*a^5*b^2*c*d^4 + a^6*b*d^5)*x^2 + (3*a^2*b^5*c^5 - 11*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*a^5*b^2*c^2*d
^3 - a^6*b*c*d^4 + a^7*d^5)*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 {15}{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)^(15/4)/(d*x+c)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(15/4)*(d*x + c)^(5/4)), x)

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maple [A]  time = 0.01, size = 171, normalized size = 1.26 \begin {gather*} -\frac {4 \left (128 b^{3} d^{3} x^{3}+352 a \,b^{2} d^{3} x^{2}+32 b^{3} c \,d^{2} x^{2}+308 a^{2} b \,d^{3} x +88 a \,b^{2} c \,d^{2} x -12 b^{3} c^{2} d x +77 a^{3} d^{3}+77 a^{2} b c \,d^{2}-33 a \,b^{2} c^{2} d +7 b^{3} c^{3}\right )}{77 \left (b x +a \right )^{\frac {11}{4}} \left (d x +c \right )^{\frac {1}{4}} \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(15/4)/(d*x+c)^(5/4),x)

[Out]

-4/77*(128*b^3*d^3*x^3+352*a*b^2*d^3*x^2+32*b^3*c*d^2*x^2+308*a^2*b*d^3*x+88*a*b^2*c*d^2*x-12*b^3*c^2*d*x+77*a
^3*d^3+77*a^2*b*c*d^2-33*a*b^2*c^2*d+7*b^3*c^3)/(b*x+a)^(11/4)/(d*x+c)^(1/4)/(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^2*
c^2*d^2-4*a*b^3*c^3*d+b^4*c^4)

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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 {15}{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)^(15/4)/(d*x+c)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(15/4)*(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 )}^{15/4}\,{\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)^(15/4)*(c + d*x)^(5/4)),x)

[Out]

int(1/((a + b*x)^(15/4)*(c + d*x)^(5/4)), x)

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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)**(15/4)/(d*x+c)**(5/4),x)

[Out]

Timed out

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