\(\int \frac {1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx\) [1697]

Optimal result

Integrand size = 19, antiderivative size = 66 \[ \int \frac {1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx=-\frac {4 (c+d x)^{3/4}}{7 (b c-a d) (a+b x)^{7/4}}+\frac {16 d (c+d x)^{3/4}}{21 (b c-a d)^2 (a+b x)^{3/4}} \]

[Out]

-4/7*(d*x+c)^(3/4)/(-a*d+b*c)/(b*x+a)^(7/4)+16/21*d*(d*x+c)^(3/4)/(-a*d+b*c)^2/(b*x+a)^(3/4)

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} \[ \int \frac {1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx=\frac {16 d (c+d x)^{3/4}}{21 (a+b x)^{3/4} (b c-a d)^2}-\frac {4 (c+d x)^{3/4}}{7 (a+b x)^{7/4} (b c-a d)} \]

[In]

Int[1/((a + b*x)^(11/4)*(c + d*x)^(1/4)),x]

[Out]

(-4*(c + d*x)^(3/4))/(7*(b*c - a*d)*(a + b*x)^(7/4)) + (16*d*(c + d*x)^(3/4))/(21*(b*c - a*d)^2*(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 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 (c+d x)^{3/4}}{7 (b c-a d) (a+b x)^{7/4}}-\frac {(4 d) \int \frac {1}{(a+b x)^{7/4} \sqrt [4]{c+d x}} \, dx}{7 (b c-a d)} \\ & = -\frac {4 (c+d x)^{3/4}}{7 (b c-a d) (a+b x)^{7/4}}+\frac {16 d (c+d x)^{3/4}}{21 (b c-a d)^2 (a+b x)^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx=-\frac {4 (c+d x)^{3/4} (3 b c-7 a d-4 b d x)}{21 (b c-a d)^2 (a+b x)^{7/4}} \]

[In]

Integrate[1/((a + b*x)^(11/4)*(c + d*x)^(1/4)),x]

[Out]

(-4*(c + d*x)^(3/4)*(3*b*c - 7*a*d - 4*b*d*x))/(21*(b*c - a*d)^2*(a + b*x)^(7/4))

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {4 \left (d x +c \right )^{\frac {3}{4}} \left (4 b d x +7 a d -3 b c \right )}{21 \left (b x +a \right )^{\frac {7}{4}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(54\)

[In]

int(1/(b*x+a)^(11/4)/(d*x+c)^(1/4),x,method=_RETURNVERBOSE)

[Out]

4/21*(d*x+c)^(3/4)*(4*b*d*x+7*a*d-3*b*c)/(b*x+a)^(7/4)/(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 = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.79 \[ \int \frac {1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx=\frac {4 \, {\left (4 \, b d x - 3 \, b c + 7 \, a d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{21 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x\right )}} \]

[In]

integrate(1/(b*x+a)^(11/4)/(d*x+c)^(1/4),x, algorithm="fricas")

[Out]

4/21*(4*b*d*x - 3*b*c + 7*a*d)*(b*x + a)^(1/4)*(d*x + c)^(3/4)/(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]

\[ \int \frac {1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {11}{4}} \sqrt [4]{c + d x}}\, dx \]

[In]

integrate(1/(b*x+a)**(11/4)/(d*x+c)**(1/4),x)

[Out]

Integral(1/((a + b*x)**(11/4)*(c + d*x)**(1/4)), x)

Maxima [F]

\[ \int \frac {1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {11}{4}} {\left (d x + c\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(11/4)/(d*x+c)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(11/4)*(d*x + c)^(1/4)), x)

Giac [F]

\[ \int \frac {1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {11}{4}} {\left (d x + c\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(11/4)/(d*x+c)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(11/4)*(d*x + c)^(1/4)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{11/4}\,{\left (c+d\,x\right )}^{1/4}} \,d x \]

[In]

int(1/((a + b*x)^(11/4)*(c + d*x)^(1/4)),x)

[Out]

int(1/((a + b*x)^(11/4)*(c + d*x)^(1/4)), x)