\(\int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx\) [1709]

Optimal result

Integrand size = 19, antiderivative size = 66 \[ \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx=-\frac {4 \sqrt [4]{c+d x}}{5 (b c-a d) (a+b x)^{5/4}}+\frac {16 d \sqrt [4]{c+d x}}{5 (b c-a d)^2 \sqrt [4]{a+b x}} \]

[Out]

-4/5*(d*x+c)^(1/4)/(-a*d+b*c)/(b*x+a)^(5/4)+16/5*d*(d*x+c)^(1/4)/(-a*d+b*c)^2/(b*x+a)^(1/4)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {47, 37} \[ \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx=\frac {16 d \sqrt [4]{c+d x}}{5 \sqrt [4]{a+b x} (b c-a d)^2}-\frac {4 \sqrt [4]{c+d x}}{5 (a+b x)^{5/4} (b c-a d)} \]

[In]

Int[1/((a + b*x)^(9/4)*(c + d*x)^(3/4)),x]

[Out]

(-4*(c + d*x)^(1/4))/(5*(b*c - a*d)*(a + b*x)^(5/4)) + (16*d*(c + d*x)^(1/4))/(5*(b*c - a*d)^2*(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 47

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*} \text {integral}& = -\frac {4 \sqrt [4]{c+d x}}{5 (b c-a d) (a+b x)^{5/4}}-\frac {(4 d) \int \frac {1}{(a+b x)^{5/4} (c+d x)^{3/4}} \, dx}{5 (b c-a d)} \\ & = -\frac {4 \sqrt [4]{c+d x}}{5 (b c-a d) (a+b x)^{5/4}}+\frac {16 d \sqrt [4]{c+d x}}{5 (b c-a d)^2 \sqrt [4]{a+b x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx=-\frac {4 \sqrt [4]{c+d x} (b c-5 a d-4 b d x)}{5 (b c-a d)^2 (a+b x)^{5/4}} \]

[In]

Integrate[1/((a + b*x)^(9/4)*(c + d*x)^(3/4)),x]

[Out]

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

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {4 \left (d x +c \right )^{\frac {1}{4}} \left (4 b d x +5 a d -b c \right )}{5 \left (b x +a \right )^{\frac {5}{4}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(54\)

[In]

int(1/(b*x+a)^(9/4)/(d*x+c)^(3/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (54) = 108\).

Time = 0.23 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.79 \[ \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx=\frac {4 \, {\left (4 \, b d x - b c + 5 \, a d\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{5 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x\right )}} \]

[In]

integrate(1/(b*x+a)^(9/4)/(d*x+c)^(3/4),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {9}{4}} \left (c + d x\right )^{\frac {3}{4}}}\, dx \]

[In]

integrate(1/(b*x+a)**(9/4)/(d*x+c)**(3/4),x)

[Out]

Integral(1/((a + b*x)**(9/4)*(c + d*x)**(3/4)), x)

Maxima [F]

\[ \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {9}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(9/4)/(d*x+c)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(9/4)*(d*x + c)^(3/4)), x)

Giac [F]

\[ \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {9}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(9/4)/(d*x+c)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(9/4)*(d*x + c)^(3/4)), x)

Mupad [B] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx=\frac {\left (\frac {16\,d\,x}{5\,{\left (a\,d-b\,c\right )}^2}+\frac {20\,a\,d-4\,b\,c}{5\,b\,{\left (a\,d-b\,c\right )}^2}\right )\,{\left (c+d\,x\right )}^{1/4}}{x\,{\left (a+b\,x\right )}^{1/4}+\frac {a\,{\left (a+b\,x\right )}^{1/4}}{b}} \]

[In]

int(1/((a + b*x)^(9/4)*(c + d*x)^(3/4)),x)

[Out]

(((16*d*x)/(5*(a*d - b*c)^2) + (20*a*d - 4*b*c)/(5*b*(a*d - b*c)^2))*(c + d*x)^(1/4))/(x*(a + b*x)^(1/4) + (a*
(a + b*x)^(1/4))/b)