\(\int \frac {(a+b x)^{7/2}}{(c+d x)^{3/2}} \, dx\) [1502]

Optimal result

Integrand size = 19, antiderivative size = 174 \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{3/2}} \, dx=-\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}+\frac {35 b (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^4}-\frac {35 b (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^3}+\frac {7 b (a+b x)^{5/2} \sqrt {c+d x}}{3 d^2}-\frac {35 \sqrt {b} (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 d^{9/2}} \]

[Out]

-35/8*(-a*d+b*c)^3*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*b^(1/2)/d^(9/2)-2*(b*x+a)^(7/2)/d/(d*x
+c)^(1/2)-35/12*b*(-a*d+b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d^3+7/3*b*(b*x+a)^(5/2)*(d*x+c)^(1/2)/d^2+35/8*b*(-a*
d+b*c)^2*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^4

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {49, 52, 65, 223, 212} \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{3/2}} \, dx=-\frac {35 \sqrt {b} (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 d^{9/2}}+\frac {35 b \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}{8 d^4}-\frac {35 b (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{12 d^3}+\frac {7 b (a+b x)^{5/2} \sqrt {c+d x}}{3 d^2}-\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}} \]

[In]

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

[Out]

(-2*(a + b*x)^(7/2))/(d*Sqrt[c + d*x]) + (35*b*(b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*d^4) - (35*b*(b*c
- a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(12*d^3) + (7*b*(a + b*x)^(5/2)*Sqrt[c + d*x])/(3*d^2) - (35*Sqrt[b]*(b
*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*d^(9/2))

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 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}+\frac {(7 b) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx}{d} \\ & = -\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}+\frac {7 b (a+b x)^{5/2} \sqrt {c+d x}}{3 d^2}-\frac {(35 b (b c-a d)) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{6 d^2} \\ & = -\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}-\frac {35 b (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^3}+\frac {7 b (a+b x)^{5/2} \sqrt {c+d x}}{3 d^2}+\frac {\left (35 b (b c-a d)^2\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{8 d^3} \\ & = -\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}+\frac {35 b (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^4}-\frac {35 b (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^3}+\frac {7 b (a+b x)^{5/2} \sqrt {c+d x}}{3 d^2}-\frac {\left (35 b (b c-a d)^3\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 d^4} \\ & = -\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}+\frac {35 b (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^4}-\frac {35 b (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^3}+\frac {7 b (a+b x)^{5/2} \sqrt {c+d x}}{3 d^2}-\frac {\left (35 (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{8 d^4} \\ & = -\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}+\frac {35 b (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^4}-\frac {35 b (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^3}+\frac {7 b (a+b x)^{5/2} \sqrt {c+d x}}{3 d^2}-\frac {\left (35 (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{8 d^4} \\ & = -\frac {2 (a+b x)^{7/2}}{d \sqrt {c+d x}}+\frac {35 b (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^4}-\frac {35 b (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^3}+\frac {7 b (a+b x)^{5/2} \sqrt {c+d x}}{3 d^2}-\frac {35 \sqrt {b} (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 d^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {a+b x} \left (-48 a^3 d^3+3 a^2 b d^2 (77 c+29 d x)+2 a b^2 d \left (-140 c^2-49 c d x+19 d^2 x^2\right )+b^3 \left (105 c^3+35 c^2 d x-14 c d^2 x^2+8 d^3 x^3\right )\right )}{24 d^4 \sqrt {c+d x}}-\frac {35 \sqrt {b} (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 d^{9/2}} \]

[In]

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

[Out]

(Sqrt[a + b*x]*(-48*a^3*d^3 + 3*a^2*b*d^2*(77*c + 29*d*x) + 2*a*b^2*d*(-140*c^2 - 49*c*d*x + 19*d^2*x^2) + b^3
*(105*c^3 + 35*c^2*d*x - 14*c*d^2*x^2 + 8*d^3*x^3)))/(24*d^4*Sqrt[c + d*x]) - (35*Sqrt[b]*(b*c - a*d)^3*ArcTan
h[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(8*d^(9/2))

Maple [F]

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (138) = 276\).

Time = 0.38 (sec) , antiderivative size = 603, normalized size of antiderivative = 3.47 \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{3/2}} \, dx=\left [-\frac {105 \, {\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3} + {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (8 \, b^{3} d^{3} x^{3} + 105 \, b^{3} c^{3} - 280 \, a b^{2} c^{2} d + 231 \, a^{2} b c d^{2} - 48 \, a^{3} d^{3} - 2 \, {\left (7 \, b^{3} c d^{2} - 19 \, a b^{2} d^{3}\right )} x^{2} + {\left (35 \, b^{3} c^{2} d - 98 \, a b^{2} c d^{2} + 87 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (d^{5} x + c d^{4}\right )}}, \frac {105 \, {\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3} + {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{3} + 105 \, b^{3} c^{3} - 280 \, a b^{2} c^{2} d + 231 \, a^{2} b c d^{2} - 48 \, a^{3} d^{3} - 2 \, {\left (7 \, b^{3} c d^{2} - 19 \, a b^{2} d^{3}\right )} x^{2} + {\left (35 \, b^{3} c^{2} d - 98 \, a b^{2} c d^{2} + 87 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (d^{5} x + c d^{4}\right )}}\right ] \]

[In]

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

[Out]

[-1/96*(105*(b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 - a^3*c*d^3 + (b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*
d^3 - a^3*d^4)*x)*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d^2*x + b*c*d + a*d^2)*
sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(8*b^3*d^3*x^3 + 105*b^3*c^3 - 280*a*b^2*
c^2*d + 231*a^2*b*c*d^2 - 48*a^3*d^3 - 2*(7*b^3*c*d^2 - 19*a*b^2*d^3)*x^2 + (35*b^3*c^2*d - 98*a*b^2*c*d^2 + 8
7*a^2*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^5*x + c*d^4), 1/48*(105*(b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2
*d^2 - a^3*c*d^3 + (b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - a^3*d^4)*x)*sqrt(-b/d)*arctan(1/2*(2*b*d*x +
b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x)) + 2*(8*b^3*d^3*x^3
+ 105*b^3*c^3 - 280*a*b^2*c^2*d + 231*a^2*b*c*d^2 - 48*a^3*d^3 - 2*(7*b^3*c*d^2 - 19*a*b^2*d^3)*x^2 + (35*b^3
*c^2*d - 98*a*b^2*c*d^2 + 87*a^2*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^5*x + c*d^4)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((b*x+a)^(7/2)/(d*x+c)^(3/2),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. 279 vs. \(2 (138) = 276\).

Time = 0.40 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.60 \[ \int \frac {(a+b x)^{7/2}}{(c+d x)^{3/2}} \, dx=\frac {{\left ({\left (b x + a\right )} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} b^{2}}{d {\left | b \right |}} - \frac {7 \, {\left (b^{3} c d^{5} - a b^{2} d^{6}\right )}}{d^{7} {\left | b \right |}}\right )} + \frac {35 \, {\left (b^{4} c^{2} d^{4} - 2 \, a b^{3} c d^{5} + a^{2} b^{2} d^{6}\right )}}{d^{7} {\left | b \right |}}\right )} + \frac {105 \, {\left (b^{5} c^{3} d^{3} - 3 \, a b^{4} c^{2} d^{4} + 3 \, a^{2} b^{3} c d^{5} - a^{3} b^{2} d^{6}\right )}}{d^{7} {\left | b \right |}}\right )} \sqrt {b x + a}}{24 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {35 \, {\left (b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{8 \, \sqrt {b d} d^{4} {\left | b \right |}} \]

[In]

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

[Out]

1/24*((b*x + a)*(2*(b*x + a)*(4*(b*x + a)*b^2/(d*abs(b)) - 7*(b^3*c*d^5 - a*b^2*d^6)/(d^7*abs(b))) + 35*(b^4*c
^2*d^4 - 2*a*b^3*c*d^5 + a^2*b^2*d^6)/(d^7*abs(b))) + 105*(b^5*c^3*d^3 - 3*a*b^4*c^2*d^4 + 3*a^2*b^3*c*d^5 - a
^3*b^2*d^6)/(d^7*abs(b)))*sqrt(b*x + a)/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) + 35/8*(b^5*c^3 - 3*a*b^4*c^2*d +
3*a^2*b^3*c*d^2 - a^3*b^2*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(
b*d)*d^4*abs(b))

Mupad [F(-1)]

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

[In]

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

[Out]

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