\(\int \frac {(c+d x)^{3/2}}{(a+b x)^{7/2}} \, dx\) [1477]

Optimal result

Integrand size = 19, antiderivative size = 32 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{7/2}} \, dx=-\frac {2 (c+d x)^{5/2}}{5 (b c-a d) (a+b x)^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {37} \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{7/2}} \, dx=-\frac {2 (c+d x)^{5/2}}{5 (a+b x)^{5/2} (b c-a d)} \]

[In]

Int[(c + d*x)^(3/2)/(a + b*x)^(7/2),x]

[Out]

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

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (c+d x)^{5/2}}{5 (b c-a d) (a+b x)^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{7/2}} \, dx=-\frac {2 (c+d x)^{5/2}}{5 (b c-a d) (a+b x)^{5/2}} \]

[In]

Integrate[(c + d*x)^(3/2)/(a + b*x)^(7/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {2 \left (d x +c \right )^{\frac {5}{2}}}{5 \left (b x +a \right )^{\frac {5}{2}} \left (a d -b c \right )}\)	\(27\)
default	\(-\frac {\left (d x +c \right )^{\frac {3}{2}}}{b \left (b x +a \right )^{\frac {5}{2}}}+\frac {3 \left (a d -b c \right ) \left (-\frac {\sqrt {d x +c}}{2 b \left (b x +a \right )^{\frac {5}{2}}}+\frac {\left (a d -b c \right ) \left (-\frac {2 \sqrt {d x +c}}{5 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {5}{2}}}-\frac {4 d \left (-\frac {2 \sqrt {d x +c}}{3 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {3}{2}}}+\frac {4 d \sqrt {d x +c}}{3 \left (-a d +b c \right )^{2} \sqrt {b x +a}}\right )}{5 \left (-a d +b c \right )}\right )}{4 b}\right )}{2 b}\)	\(161\)

[In]

int((d*x+c)^(3/2)/(b*x+a)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (26) = 52\).

Time = 0.38 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.25 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{7/2}} \, dx=-\frac {2 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{5 \, {\left (a^{3} b c - a^{4} d + {\left (b^{4} c - a b^{3} d\right )} x^{3} + 3 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} x^{2} + 3 \, {\left (a^{2} b^{2} c - a^{3} b d\right )} x\right )}} \]

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{7/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {3}{2}}}{\left (a + b x\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate((d*x+c)**(3/2)/(b*x+a)**(7/2),x)

[Out]

Integral((c + d*x)**(3/2)/(a + b*x)**(7/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{7/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 374 vs. \(2 (26) = 52\).

Time = 0.44 (sec) , antiderivative size = 374, normalized size of antiderivative = 11.69 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{7/2}} \, dx=-\frac {4 \, {\left (\sqrt {b d} b^{8} c^{4} d^{2} {\left | b \right |} - 4 \, \sqrt {b d} a b^{7} c^{3} d^{3} {\left | b \right |} + 6 \, \sqrt {b d} a^{2} b^{6} c^{2} d^{4} {\left | b \right |} - 4 \, \sqrt {b d} a^{3} b^{5} c d^{5} {\left | b \right |} + \sqrt {b d} a^{4} b^{4} d^{6} {\left | b \right |} + 10 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} b^{4} c^{2} d^{2} {\left | b \right |} - 20 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a b^{3} c d^{3} {\left | b \right |} + 10 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a^{2} b^{2} d^{4} {\left | b \right |} + 5 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{8} d^{2} {\left | b \right |}\right )}}{5 \, {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{5} b^{3}} \]

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^(7/2),x, algorithm="giac")

[Out]

-4/5*(sqrt(b*d)*b^8*c^4*d^2*abs(b) - 4*sqrt(b*d)*a*b^7*c^3*d^3*abs(b) + 6*sqrt(b*d)*a^2*b^6*c^2*d^4*abs(b) - 4
*sqrt(b*d)*a^3*b^5*c*d^5*abs(b) + sqrt(b*d)*a^4*b^4*d^6*abs(b) + 10*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(
b^2*c + (b*x + a)*b*d - a*b*d))^4*b^4*c^2*d^2*abs(b) - 20*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b
*x + a)*b*d - a*b*d))^4*a*b^3*c*d^3*abs(b) + 10*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*
d - a*b*d))^4*a^2*b^2*d^4*abs(b) + 5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))
^8*d^2*abs(b))/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)^5*b^3)

Mupad [B] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^{7/2}} \, dx=\frac {2\,{\left (c+d\,x\right )}^{5/2}}{\left (5\,a\,d-5\,b\,c\right )\,{\left (a+b\,x\right )}^{5/2}} \]

[In]

int((c + d*x)^(3/2)/(a + b*x)^(7/2),x)

[Out]

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