\(\int \frac {(a+b x)^{7/3}}{(c+d x)^{4/3}} \, dx\) [1615]

Optimal result

Integrand size = 19, antiderivative size = 241 \[ \int \frac {(a+b x)^{7/3}}{(c+d x)^{4/3}} \, dx=-\frac {3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}}-\frac {14 b (b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 d^3}+\frac {7 b (a+b x)^{4/3} (c+d x)^{2/3}}{2 d^2}-\frac {14 \sqrt [3]{b} (b c-a d)^2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} d^{10/3}}-\frac {7 \sqrt [3]{b} (b c-a d)^2 \log (a+b x)}{9 d^{10/3}}-\frac {7 \sqrt [3]{b} (b c-a d)^2 \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{3 d^{10/3}} \]

[Out]

-3*(b*x+a)^(7/3)/d/(d*x+c)^(1/3)-14/3*b*(-a*d+b*c)*(b*x+a)^(1/3)*(d*x+c)^(2/3)/d^3+7/2*b*(b*x+a)^(4/3)*(d*x+c)
^(2/3)/d^2-7/9*b^(1/3)*(-a*d+b*c)^2*ln(b*x+a)/d^(10/3)-7/3*b^(1/3)*(-a*d+b*c)^2*ln(-1+b^(1/3)*(d*x+c)^(1/3)/d^
(1/3)/(b*x+a)^(1/3))/d^(10/3)-14/9*b^(1/3)*(-a*d+b*c)^2*arctan(1/3*3^(1/2)+2/3*b^(1/3)*(d*x+c)^(1/3)/d^(1/3)/(
b*x+a)^(1/3)*3^(1/2))/d^(10/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {49, 52, 61} \[ \int \frac {(a+b x)^{7/3}}{(c+d x)^{4/3}} \, dx=-\frac {14 \sqrt [3]{b} (b c-a d)^2 \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3} d^{10/3}}-\frac {7 \sqrt [3]{b} (b c-a d)^2 \log (a+b x)}{9 d^{10/3}}-\frac {7 \sqrt [3]{b} (b c-a d)^2 \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{3 d^{10/3}}-\frac {14 b \sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d)}{3 d^3}+\frac {7 b (a+b x)^{4/3} (c+d x)^{2/3}}{2 d^2}-\frac {3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}} \]

[In]

Int[(a + b*x)^(7/3)/(c + d*x)^(4/3),x]

[Out]

(-3*(a + b*x)^(7/3))/(d*(c + d*x)^(1/3)) - (14*b*(b*c - a*d)*(a + b*x)^(1/3)*(c + d*x)^(2/3))/(3*d^3) + (7*b*(
a + b*x)^(4/3)*(c + d*x)^(2/3))/(2*d^2) - (14*b^(1/3)*(b*c - a*d)^2*ArcTan[1/Sqrt[3] + (2*b^(1/3)*(c + d*x)^(1
/3))/(Sqrt[3]*d^(1/3)*(a + b*x)^(1/3))])/(3*Sqrt[3]*d^(10/3)) - (7*b^(1/3)*(b*c - a*d)^2*Log[a + b*x])/(9*d^(1
0/3)) - (7*b^(1/3)*(b*c - a*d)^2*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))/(d^(1/3)*(a + b*x)^(1/3))])/(3*d^(10/3))

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 61

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, Simp[(-Sqrt
[3])*(q/d)*ArcTan[2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3))) + 1/Sqrt[3]], x] + (-Simp[3*(q/(2*d))*Log[q*
((a + b*x)^(1/3)/(c + d*x)^(1/3)) - 1], x] - Simp[(q/(2*d))*Log[c + d*x], x])] /; FreeQ[{a, b, c, d}, x] && Ne
Q[b*c - a*d, 0] && PosQ[d/b]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}}+\frac {(7 b) \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx}{d} \\ & = -\frac {3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}}+\frac {7 b (a+b x)^{4/3} (c+d x)^{2/3}}{2 d^2}-\frac {(14 b (b c-a d)) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx}{3 d^2} \\ & = -\frac {3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}}-\frac {14 b (b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 d^3}+\frac {7 b (a+b x)^{4/3} (c+d x)^{2/3}}{2 d^2}+\frac {\left (14 b (b c-a d)^2\right ) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{9 d^3} \\ & = -\frac {3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}}-\frac {14 b (b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 d^3}+\frac {7 b (a+b x)^{4/3} (c+d x)^{2/3}}{2 d^2}-\frac {14 \sqrt [3]{b} (b c-a d)^2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} d^{10/3}}-\frac {7 \sqrt [3]{b} (b c-a d)^2 \log (a+b x)}{9 d^{10/3}}-\frac {7 \sqrt [3]{b} (b c-a d)^2 \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{3 d^{10/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.18 \[ \int \frac {(a+b x)^{7/3}}{(c+d x)^{4/3}} \, dx=\frac {-\frac {3 \sqrt [3]{d} \sqrt [3]{a+b x} \left (18 a^2 d^2-a b d (49 c+13 d x)+b^2 \left (28 c^2+7 c d x-3 d^2 x^2\right )\right )}{\sqrt [3]{c+d x}}-28 \sqrt {3} \sqrt [3]{b} (b c-a d)^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}{2 \sqrt [3]{d} \sqrt [3]{a+b x}+\sqrt [3]{b} \sqrt [3]{c+d x}}\right )-28 \sqrt [3]{b} (b c-a d)^2 \log \left (\sqrt [3]{d} \sqrt [3]{a+b x}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )+14 \sqrt [3]{b} (b c-a d)^2 \log \left (d^{2/3} (a+b x)^{2/3}+\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{a+b x} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}\right )}{18 d^{10/3}} \]

[In]

Integrate[(a + b*x)^(7/3)/(c + d*x)^(4/3),x]

[Out]

((-3*d^(1/3)*(a + b*x)^(1/3)*(18*a^2*d^2 - a*b*d*(49*c + 13*d*x) + b^2*(28*c^2 + 7*c*d*x - 3*d^2*x^2)))/(c + d
*x)^(1/3) - 28*Sqrt[3]*b^(1/3)*(b*c - a*d)^2*ArcTan[(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))/(2*d^(1/3)*(a + b*x)^(1/
3) + b^(1/3)*(c + d*x)^(1/3))] - 28*b^(1/3)*(b*c - a*d)^2*Log[d^(1/3)*(a + b*x)^(1/3) - b^(1/3)*(c + d*x)^(1/3
)] + 14*b^(1/3)*(b*c - a*d)^2*Log[d^(2/3)*(a + b*x)^(2/3) + b^(1/3)*d^(1/3)*(a + b*x)^(1/3)*(c + d*x)^(1/3) +
b^(2/3)*(c + d*x)^(2/3)])/(18*d^(10/3))

Maple [F]

\[\int \frac {\left (b x +a \right )^{\frac {7}{3}}}{\left (d x +c \right )^{\frac {4}{3}}}d x\]

[In]

int((b*x+a)^(7/3)/(d*x+c)^(4/3),x)

[Out]

int((b*x+a)^(7/3)/(d*x+c)^(4/3),x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (187) = 374\).

Time = 0.24 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.76 \[ \int \frac {(a+b x)^{7/3}}{(c+d x)^{4/3}} \, dx=-\frac {28 \, \sqrt {3} {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} d \left (-\frac {b}{d}\right )^{\frac {2}{3}} + \sqrt {3} {\left (b d x + b c\right )}}{3 \, {\left (b d x + b c\right )}}\right ) + 14 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \log \left (\frac {{\left (d x + c\right )} \left (-\frac {b}{d}\right )^{\frac {2}{3}} - {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} \left (-\frac {b}{d}\right )^{\frac {1}{3}} + {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{d x + c}\right ) - 28 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \log \left (\frac {{\left (d x + c\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{d x + c}\right ) - 3 \, {\left (3 \, b^{2} d^{2} x^{2} - 28 \, b^{2} c^{2} + 49 \, a b c d - 18 \, a^{2} d^{2} - {\left (7 \, b^{2} c d - 13 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{18 \, {\left (d^{4} x + c d^{3}\right )}} \]

[In]

integrate((b*x+a)^(7/3)/(d*x+c)^(4/3),x, algorithm="fricas")

[Out]

-1/18*(28*sqrt(3)*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2 + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x)*(-b/d)^(1/3)*arc
tan(1/3*(2*sqrt(3)*(b*x + a)^(1/3)*(d*x + c)^(2/3)*d*(-b/d)^(2/3) + sqrt(3)*(b*d*x + b*c))/(b*d*x + b*c)) + 14
*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2 + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x)*(-b/d)^(1/3)*log(((d*x + c)*(-b/d
)^(2/3) - (b*x + a)^(1/3)*(d*x + c)^(2/3)*(-b/d)^(1/3) + (b*x + a)^(2/3)*(d*x + c)^(1/3))/(d*x + c)) - 28*(b^2
*c^3 - 2*a*b*c^2*d + a^2*c*d^2 + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x)*(-b/d)^(1/3)*log(((d*x + c)*(-b/d)^(1/
3) + (b*x + a)^(1/3)*(d*x + c)^(2/3))/(d*x + c)) - 3*(3*b^2*d^2*x^2 - 28*b^2*c^2 + 49*a*b*c*d - 18*a^2*d^2 - (
7*b^2*c*d - 13*a*b*d^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(d^4*x + c*d^3)

Sympy [F]

\[ \int \frac {(a+b x)^{7/3}}{(c+d x)^{4/3}} \, dx=\int \frac {\left (a + b x\right )^{\frac {7}{3}}}{\left (c + d x\right )^{\frac {4}{3}}}\, dx \]

[In]

integrate((b*x+a)**(7/3)/(d*x+c)**(4/3),x)

[Out]

Integral((a + b*x)**(7/3)/(c + d*x)**(4/3), x)

Maxima [F]

\[ \int \frac {(a+b x)^{7/3}}{(c+d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {7}{3}}}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \]

[In]

integrate((b*x+a)^(7/3)/(d*x+c)^(4/3),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(7/3)/(d*x + c)^(4/3), x)

Giac [F]

\[ \int \frac {(a+b x)^{7/3}}{(c+d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {7}{3}}}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \]

[In]

integrate((b*x+a)^(7/3)/(d*x+c)^(4/3),x, algorithm="giac")

[Out]

integrate((b*x + a)^(7/3)/(d*x + c)^(4/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{7/3}}{(c+d x)^{4/3}} \, dx=\int \frac {{\left (a+b\,x\right )}^{7/3}}{{\left (c+d\,x\right )}^{4/3}} \,d x \]

[In]

int((a + b*x)^(7/3)/(c + d*x)^(4/3),x)

[Out]

int((a + b*x)^(7/3)/(c + d*x)^(4/3), x)