3.1698 \(\int \frac {1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=101 \[ -\frac {128 d^2 (c+d x)^{3/4}}{231 (a+b x)^{3/4} (b c-a d)^3}+\frac {32 d (c+d x)^{3/4}}{77 (a+b x)^{7/4} (b c-a d)^2}-\frac {4 (c+d x)^{3/4}}{11 (a+b x)^{11/4} (b c-a d)} \]

[Out]

-4/11*(d*x+c)^(3/4)/(-a*d+b*c)/(b*x+a)^(11/4)+32/77*d*(d*x+c)^(3/4)/(-a*d+b*c)^2/(b*x+a)^(7/4)-128/231*d^2*(d*
x+c)^(3/4)/(-a*d+b*c)^3/(b*x+a)^(3/4)

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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} \[ -\frac {128 d^2 (c+d x)^{3/4}}{231 (a+b x)^{3/4} (b c-a d)^3}+\frac {32 d (c+d x)^{3/4}}{77 (a+b x)^{7/4} (b c-a d)^2}-\frac {4 (c+d x)^{3/4}}{11 (a+b x)^{11/4} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(c + d*x)^(3/4))/(11*(b*c - a*d)*(a + b*x)^(11/4)) + (32*d*(c + d*x)^(3/4))/(77*(b*c - a*d)^2*(a + b*x)^(7
/4)) - (128*d^2*(c + d*x)^(3/4))/(231*(b*c - a*d)^3*(a + b*x)^(3/4))

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} \sqrt [4]{c+d x}} \, dx &=-\frac {4 (c+d x)^{3/4}}{11 (b c-a d) (a+b x)^{11/4}}-\frac {(8 d) \int \frac {1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx}{11 (b c-a d)}\\ &=-\frac {4 (c+d x)^{3/4}}{11 (b c-a d) (a+b x)^{11/4}}+\frac {32 d (c+d x)^{3/4}}{77 (b c-a d)^2 (a+b x)^{7/4}}+\frac {\left (32 d^2\right ) \int \frac {1}{(a+b x)^{7/4} \sqrt [4]{c+d x}} \, dx}{77 (b c-a d)^2}\\ &=-\frac {4 (c+d x)^{3/4}}{11 (b c-a d) (a+b x)^{11/4}}+\frac {32 d (c+d x)^{3/4}}{77 (b c-a d)^2 (a+b x)^{7/4}}-\frac {128 d^2 (c+d x)^{3/4}}{231 (b c-a d)^3 (a+b x)^{3/4}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 77, normalized size = 0.76 \[ -\frac {4 (c+d x)^{3/4} \left (77 a^2 d^2+22 a b d (4 d x-3 c)+b^2 \left (21 c^2-24 c d x+32 d^2 x^2\right )\right )}{231 (a+b x)^{11/4} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(c + d*x)^(3/4)*(77*a^2*d^2 + 22*a*b*d*(-3*c + 4*d*x) + b^2*(21*c^2 - 24*c*d*x + 32*d^2*x^2)))/(231*(b*c -
a*d)^3*(a + b*x)^(11/4))

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fricas [B]  time = 1.03, size = 252, normalized size = 2.50 \[ -\frac {4 \, {\left (32 \, b^{2} d^{2} x^{2} + 21 \, b^{2} c^{2} - 66 \, a b c d + 77 \, a^{2} d^{2} - 8 \, {\left (3 \, b^{2} c d - 11 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{231 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{3} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/231*(32*b^2*d^2*x^2 + 21*b^2*c^2 - 66*a*b*c*d + 77*a^2*d^2 - 8*(3*b^2*c*d - 11*a*b*d^2)*x)*(b*x + a)^(1/4)*
(d*x + c)^(3/4)/(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 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {15}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 105, normalized size = 1.04 \[ \frac {4 \left (d x +c \right )^{\frac {3}{4}} \left (32 b^{2} x^{2} d^{2}+88 a b \,d^{2} x -24 b^{2} c d x +77 a^{2} d^{2}-66 a b c d +21 b^{2} c^{2}\right )}{231 \left (b x +a \right )^{\frac {11}{4}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/231*(d*x+c)^(3/4)*(32*b^2*d^2*x^2+88*a*b*d^2*x-24*b^2*c*d*x+77*a^2*d^2-66*a*b*c*d+21*b^2*c^2)/(b*x+a)^(11/4)
/(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 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {15}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,x\right )}^{15/4}\,{\left (c+d\,x\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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