3.16.3 \(\int \frac {1}{(a+b x)^{17/4} (c+d x)^{3/4}} \, dx\)

Optimal. Leaf size=136 \[ \frac {512 d^3 \sqrt [4]{c+d x}}{195 \sqrt [4]{a+b x} (b c-a d)^4}-\frac {128 d^2 \sqrt [4]{c+d x}}{195 (a+b x)^{5/4} (b c-a d)^3}+\frac {16 d \sqrt [4]{c+d x}}{39 (a+b x)^{9/4} (b c-a d)^2}-\frac {4 \sqrt [4]{c+d x}}{13 (a+b x)^{13/4} (b c-a d)} \]

________________________________________________________________________________________

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]{c+d x}}{195 \sqrt [4]{a+b x} (b c-a d)^4}-\frac {128 d^2 \sqrt [4]{c+d x}}{195 (a+b x)^{5/4} (b c-a d)^3}+\frac {16 d \sqrt [4]{c+d x}}{39 (a+b x)^{9/4} (b c-a d)^2}-\frac {4 \sqrt [4]{c+d x}}{13 (a+b x)^{13/4} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(17/4)*(c + d*x)^(3/4)),x]

[Out]

(-4*(c + d*x)^(1/4))/(13*(b*c - a*d)*(a + b*x)^(13/4)) + (16*d*(c + d*x)^(1/4))/(39*(b*c - a*d)^2*(a + b*x)^(9
/4)) - (128*d^2*(c + d*x)^(1/4))/(195*(b*c - a*d)^3*(a + b*x)^(5/4)) + (512*d^3*(c + d*x)^(1/4))/(195*(b*c - a
*d)^4*(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)^{17/4} (c+d x)^{3/4}} \, dx &=-\frac {4 \sqrt [4]{c+d x}}{13 (b c-a d) (a+b x)^{13/4}}-\frac {(12 d) \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx}{13 (b c-a d)}\\ &=-\frac {4 \sqrt [4]{c+d x}}{13 (b c-a d) (a+b x)^{13/4}}+\frac {16 d \sqrt [4]{c+d x}}{39 (b c-a d)^2 (a+b x)^{9/4}}+\frac {\left (32 d^2\right ) \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx}{39 (b c-a d)^2}\\ &=-\frac {4 \sqrt [4]{c+d x}}{13 (b c-a d) (a+b x)^{13/4}}+\frac {16 d \sqrt [4]{c+d x}}{39 (b c-a d)^2 (a+b x)^{9/4}}-\frac {128 d^2 \sqrt [4]{c+d x}}{195 (b c-a d)^3 (a+b x)^{5/4}}-\frac {\left (128 d^3\right ) \int \frac {1}{(a+b x)^{5/4} (c+d x)^{3/4}} \, dx}{195 (b c-a d)^3}\\ &=-\frac {4 \sqrt [4]{c+d x}}{13 (b c-a d) (a+b x)^{13/4}}+\frac {16 d \sqrt [4]{c+d x}}{39 (b c-a d)^2 (a+b x)^{9/4}}-\frac {128 d^2 \sqrt [4]{c+d x}}{195 (b c-a d)^3 (a+b x)^{5/4}}+\frac {512 d^3 \sqrt [4]{c+d x}}{195 (b c-a d)^4 \sqrt [4]{a+b x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 116, normalized size = 0.85 \begin {gather*} \frac {4 \sqrt [4]{c+d x} \left (195 a^3 d^3-117 a^2 b d^2 (c-4 d x)+13 a b^2 d \left (5 c^2-8 c d x+32 d^2 x^2\right )+b^3 \left (-15 c^3+20 c^2 d x-32 c d^2 x^2+128 d^3 x^3\right )\right )}{195 (a+b x)^{13/4} (b c-a d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(17/4)*(c + d*x)^(3/4)),x]

[Out]

(4*(c + d*x)^(1/4)*(195*a^3*d^3 - 117*a^2*b*d^2*(c - 4*d*x) + 13*a*b^2*d*(5*c^2 - 8*c*d*x + 32*d^2*x^2) + b^3*
(-15*c^3 + 20*c^2*d*x - 32*c*d^2*x^2 + 128*d^3*x^3)))/(195*(b*c - a*d)^4*(a + b*x)^(13/4))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.13, size = 109, normalized size = 0.80 \begin {gather*} -\frac {4 \left (\frac {15 b^3 (c+d x)^{13/4}}{(a+b x)^{13/4}}-\frac {65 b^2 d (c+d x)^{9/4}}{(a+b x)^{9/4}}-\frac {195 d^3 \sqrt [4]{c+d x}}{\sqrt [4]{a+b x}}+\frac {117 b d^2 (c+d x)^{5/4}}{(a+b x)^{5/4}}\right )}{195 (b c-a d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^(17/4)*(c + d*x)^(3/4)),x]

[Out]

(-4*((-195*d^3*(c + d*x)^(1/4))/(a + b*x)^(1/4) + (117*b*d^2*(c + d*x)^(5/4))/(a + b*x)^(5/4) - (65*b^2*d*(c +
d*x)^(9/4))/(a + b*x)^(9/4) + (15*b^3*(c + d*x)^(13/4))/(a + b*x)^(13/4)))/(195*(b*c - a*d)^4)

________________________________________________________________________________________

fricas [B]  time = 1.32, size = 419, normalized size = 3.08 \begin {gather*} \frac {4 \, {\left (128 \, b^{3} d^{3} x^{3} - 15 \, b^{3} c^{3} + 65 \, a b^{2} c^{2} d - 117 \, a^{2} b c d^{2} + 195 \, a^{3} d^{3} - 32 \, {\left (b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x^{2} + 4 \, {\left (5 \, b^{3} c^{2} d - 26 \, a b^{2} c d^{2} + 117 \, a^{2} b d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{195 \, {\left (a^{4} b^{4} c^{4} - 4 \, a^{5} b^{3} c^{3} d + 6 \, a^{6} b^{2} c^{2} d^{2} - 4 \, a^{7} b c d^{3} + a^{8} d^{4} + {\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{4} + 4 \, {\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{3} + 6 \, {\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x^{2} + 4 \, {\left (a^{3} b^{5} c^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(17/4)/(d*x+c)^(3/4),x, algorithm="fricas")

[Out]

4/195*(128*b^3*d^3*x^3 - 15*b^3*c^3 + 65*a*b^2*c^2*d - 117*a^2*b*c*d^2 + 195*a^3*d^3 - 32*(b^3*c*d^2 - 13*a*b^
2*d^3)*x^2 + 4*(5*b^3*c^2*d - 26*a*b^2*c*d^2 + 117*a^2*b*d^3)*x)*(b*x + a)^(3/4)*(d*x + c)^(1/4)/(a^4*b^4*c^4
- 4*a^5*b^3*c^3*d + 6*a^6*b^2*c^2*d^2 - 4*a^7*b*c*d^3 + a^8*d^4 + (b^8*c^4 - 4*a*b^7*c^3*d + 6*a^2*b^6*c^2*d^2
- 4*a^3*b^5*c*d^3 + a^4*b^4*d^4)*x^4 + 4*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 - 4*a^4*b^4*c*d^3 +
a^5*b^3*d^4)*x^3 + 6*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4*a^5*b^3*c*d^3 + a^6*b^2*d^4)*x^2
+ 4*(a^3*b^5*c^4 - 4*a^4*b^4*c^3*d + 6*a^5*b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^4)*x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x + a\right )}^{\frac {17}{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)^(17/4)/(d*x+c)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(17/4)*(d*x + c)^(3/4)), x)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 171, normalized size = 1.26 \begin {gather*} \frac {4 \left (d x +c \right )^{\frac {1}{4}} \left (128 b^{3} d^{3} x^{3}+416 a \,b^{2} d^{3} x^{2}-32 b^{3} c \,d^{2} x^{2}+468 a^{2} b \,d^{3} x -104 a \,b^{2} c \,d^{2} x +20 b^{3} c^{2} d x +195 a^{3} d^{3}-117 a^{2} b c \,d^{2}+65 a \,b^{2} c^{2} d -15 b^{3} c^{3}\right )}{195 \left (b x +a \right )^{\frac {13}{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)^(17/4)/(d*x+c)^(3/4),x)

[Out]

4/195*(d*x+c)^(1/4)*(128*b^3*d^3*x^3+416*a*b^2*d^3*x^2-32*b^3*c*d^2*x^2+468*a^2*b*d^3*x-104*a*b^2*c*d^2*x+20*b
^3*c^2*d*x+195*a^3*d^3-117*a^2*b*c*d^2+65*a*b^2*c^2*d-15*b^3*c^3)/(b*x+a)^(13/4)/(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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x + a\right )}^{\frac {17}{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)^(17/4)/(d*x+c)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(17/4)*(d*x + c)^(3/4)), x)

________________________________________________________________________________________

mupad [B]  time = 1.26, size = 209, normalized size = 1.54 \begin {gather*} \frac {{\left (c+d\,x\right )}^{1/4}\,\left (\frac {512\,d^3\,x^3}{195\,{\left (a\,d-b\,c\right )}^4}+\frac {780\,a^3\,d^3-468\,a^2\,b\,c\,d^2+260\,a\,b^2\,c^2\,d-60\,b^3\,c^3}{195\,b^3\,{\left (a\,d-b\,c\right )}^4}+\frac {16\,d\,x\,\left (117\,a^2\,d^2-26\,a\,b\,c\,d+5\,b^2\,c^2\right )}{195\,b^2\,{\left (a\,d-b\,c\right )}^4}+\frac {128\,d^2\,x^2\,\left (13\,a\,d-b\,c\right )}{195\,b\,{\left (a\,d-b\,c\right )}^4}\right )}{x^3\,{\left (a+b\,x\right )}^{1/4}+\frac {a^3\,{\left (a+b\,x\right )}^{1/4}}{b^3}+\frac {3\,a\,x^2\,{\left (a+b\,x\right )}^{1/4}}{b}+\frac {3\,a^2\,x\,{\left (a+b\,x\right )}^{1/4}}{b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(17/4)*(c + d*x)^(3/4)),x)

[Out]

((c + d*x)^(1/4)*((512*d^3*x^3)/(195*(a*d - b*c)^4) + (780*a^3*d^3 - 60*b^3*c^3 + 260*a*b^2*c^2*d - 468*a^2*b*
c*d^2)/(195*b^3*(a*d - b*c)^4) + (16*d*x*(117*a^2*d^2 + 5*b^2*c^2 - 26*a*b*c*d))/(195*b^2*(a*d - b*c)^4) + (12
8*d^2*x^2*(13*a*d - b*c))/(195*b*(a*d - b*c)^4)))/(x^3*(a + b*x)^(1/4) + (a^3*(a + b*x)^(1/4))/b^3 + (3*a*x^2*
(a + b*x)^(1/4))/b + (3*a^2*x*(a + b*x)^(1/4))/b^2)

________________________________________________________________________________________

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)**(17/4)/(d*x+c)**(3/4),x)

[Out]

Timed out

________________________________________________________________________________________