3.16.13 \(\int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx\) [1513]

Optimal. Leaf size=170 \[ -\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}-\frac {14 b (a+b x)^{5/2}}{3 d^2 \sqrt {c+d x}}-\frac {35 b^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4}+\frac {35 b^2 (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3}+\frac {35 b^{3/2} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{9/2}} \]

[Out]

-2/3*(b*x+a)^(7/2)/d/(d*x+c)^(3/2)+35/4*b^(3/2)*(-a*d+b*c)^2*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/
2))/d^(9/2)-14/3*b*(b*x+a)^(5/2)/d^2/(d*x+c)^(1/2)+35/6*b^2*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d^3-35/4*b^2*(-a*d+b*c
)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^4

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Rubi [A]
time = 0.06, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {49, 52, 65, 223, 212} \begin {gather*} \frac {35 b^{3/2} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{9/2}}-\frac {35 b^2 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 d^4}+\frac {35 b^2 (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3}-\frac {14 b (a+b x)^{5/2}}{3 d^2 \sqrt {c+d x}}-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(7/2)/(c + d*x)^(5/2),x]

[Out]

(-2*(a + b*x)^(7/2))/(3*d*(c + d*x)^(3/2)) - (14*b*(a + b*x)^(5/2))/(3*d^2*Sqrt[c + d*x]) - (35*b^2*(b*c - a*d
)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*d^4) + (35*b^2*(a + b*x)^(3/2)*Sqrt[c + d*x])/(6*d^3) + (35*b^(3/2)*(b*c - a
*d)^2*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*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*} \int \frac {(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx &=-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}+\frac {(7 b) \int \frac {(a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx}{3 d}\\ &=-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}-\frac {14 b (a+b x)^{5/2}}{3 d^2 \sqrt {c+d x}}+\frac {\left (35 b^2\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{3 d^2}\\ &=-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}-\frac {14 b (a+b x)^{5/2}}{3 d^2 \sqrt {c+d x}}+\frac {35 b^2 (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3}-\frac {\left (35 b^2 (b c-a d)\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{4 d^3}\\ &=-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}-\frac {14 b (a+b x)^{5/2}}{3 d^2 \sqrt {c+d x}}-\frac {35 b^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4}+\frac {35 b^2 (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3}+\frac {\left (35 b^2 (b c-a d)^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 d^4}\\ &=-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}-\frac {14 b (a+b x)^{5/2}}{3 d^2 \sqrt {c+d x}}-\frac {35 b^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4}+\frac {35 b^2 (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3}+\frac {\left (35 b (b c-a d)^2\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 )}{4 d^4}\\ &=-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}-\frac {14 b (a+b x)^{5/2}}{3 d^2 \sqrt {c+d x}}-\frac {35 b^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4}+\frac {35 b^2 (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3}+\frac {\left (35 b (b c-a d)^2\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 )}{4 d^4}\\ &=-\frac {2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}}-\frac {14 b (a+b x)^{5/2}}{3 d^2 \sqrt {c+d x}}-\frac {35 b^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4}+\frac {35 b^2 (a+b x)^{3/2} \sqrt {c+d x}}{6 d^3}+\frac {35 b^{3/2} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 166, normalized size = 0.98 \begin {gather*} -\frac {\sqrt {a+b x} \left (8 a^3 d^3+8 a^2 b d^2 (7 c+10 d x)-a b^2 d \left (175 c^2+238 c d x+39 d^2 x^2\right )+b^3 \left (105 c^3+140 c^2 d x+21 c d^2 x^2-6 d^3 x^3\right )\right )}{12 d^4 (c+d x)^{3/2}}+\frac {35 b^{3/2} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 d^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(7/2)/(c + d*x)^(5/2),x]

[Out]

-1/12*(Sqrt[a + b*x]*(8*a^3*d^3 + 8*a^2*b*d^2*(7*c + 10*d*x) - a*b^2*d*(175*c^2 + 238*c*d*x + 39*d^2*x^2) + b^
3*(105*c^3 + 140*c^2*d*x + 21*c*d^2*x^2 - 6*d^3*x^3)))/(d^4*(c + d*x)^(3/2)) + (35*b^(3/2)*(b*c - a*d)^2*ArcTa
nh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(4*d^(9/2))

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{\frac {7}{2}}}{\left (d x +c \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(7/2)/(d*x+c)^(5/2),x)

[Out]

int((b*x+a)^(7/2)/(d*x+c)^(5/2),x)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/2)/(d*x+c)^(5/2),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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (132) = 264\).
time = 1.54, size = 657, normalized size = 3.86 \begin {gather*} \left [\frac {105 \, {\left (b^{3} c^{4} - 2 \, a b^{2} c^{3} d + a^{2} b c^{2} d^{2} + {\left (b^{3} c^{2} d^{2} - 2 \, a b^{2} c d^{3} + a^{2} b d^{4}\right )} x^{2} + 2 \, {\left (b^{3} c^{3} d - 2 \, a b^{2} c^{2} d^{2} + a^{2} b c d^{3}\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 (6 \, b^{3} d^{3} x^{3} - 105 \, b^{3} c^{3} + 175 \, a b^{2} c^{2} d - 56 \, a^{2} b c d^{2} - 8 \, a^{3} d^{3} - 3 \, {\left (7 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x^{2} - 2 \, {\left (70 \, b^{3} c^{2} d - 119 \, a b^{2} c d^{2} + 40 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}}, -\frac {105 \, {\left (b^{3} c^{4} - 2 \, a b^{2} c^{3} d + a^{2} b c^{2} d^{2} + {\left (b^{3} c^{2} d^{2} - 2 \, a b^{2} c d^{3} + a^{2} b d^{4}\right )} x^{2} + 2 \, {\left (b^{3} c^{3} d - 2 \, a b^{2} c^{2} d^{2} + a^{2} b c d^{3}\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 (6 \, b^{3} d^{3} x^{3} - 105 \, b^{3} c^{3} + 175 \, a b^{2} c^{2} d - 56 \, a^{2} b c d^{2} - 8 \, a^{3} d^{3} - 3 \, {\left (7 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x^{2} - 2 \, {\left (70 \, b^{3} c^{2} d - 119 \, a b^{2} c d^{2} + 40 \, a^{2} b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{24 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/2)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(105*(b^3*c^4 - 2*a*b^2*c^3*d + a^2*b*c^2*d^2 + (b^3*c^2*d^2 - 2*a*b^2*c*d^3 + a^2*b*d^4)*x^2 + 2*(b^3*c
^3*d - 2*a*b^2*c^2*d^2 + a^2*b*c*d^3)*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*(6*b^3*d^3*x^3 - 1
05*b^3*c^3 + 175*a*b^2*c^2*d - 56*a^2*b*c*d^2 - 8*a^3*d^3 - 3*(7*b^3*c*d^2 - 13*a*b^2*d^3)*x^2 - 2*(70*b^3*c^2
*d - 119*a*b^2*c*d^2 + 40*a^2*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^6*x^2 + 2*c*d^5*x + c^2*d^4), -1/24*(1
05*(b^3*c^4 - 2*a*b^2*c^3*d + a^2*b*c^2*d^2 + (b^3*c^2*d^2 - 2*a*b^2*c*d^3 + a^2*b*d^4)*x^2 + 2*(b^3*c^3*d - 2
*a*b^2*c^2*d^2 + a^2*b*c*d^3)*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*(6*b^3*d^3*x^3 - 105*b^3*c^3 + 175*a*b^2*c^2*d - 56*a^2*b*c
*d^2 - 8*a^3*d^3 - 3*(7*b^3*c*d^2 - 13*a*b^2*d^3)*x^2 - 2*(70*b^3*c^2*d - 119*a*b^2*c*d^2 + 40*a^2*b*d^3)*x)*s
qrt(b*x + a)*sqrt(d*x + c))/(d^6*x^2 + 2*c*d^5*x + c^2*d^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{\frac {7}{2}}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(7/2)/(d*x+c)**(5/2),x)

[Out]

Integral((a + b*x)**(7/2)/(c + d*x)**(5/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (132) = 264\).
time = 0.80, size = 380, normalized size = 2.24 \begin {gather*} \frac {{\left ({\left (3 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (b^{6} c d^{6} - a b^{5} d^{7}\right )} {\left (b x + a\right )}}{b^{2} c d^{7} {\left | b \right |} - a b d^{8} {\left | b \right |}} - \frac {7 \, {\left (b^{7} c^{2} d^{5} - 2 \, a b^{6} c d^{6} + a^{2} b^{5} d^{7}\right )}}{b^{2} c d^{7} {\left | b \right |} - a b d^{8} {\left | b \right |}}\right )} - \frac {140 \, {\left (b^{8} c^{3} d^{4} - 3 \, a b^{7} c^{2} d^{5} + 3 \, a^{2} b^{6} c d^{6} - a^{3} b^{5} d^{7}\right )}}{b^{2} c d^{7} {\left | b \right |} - a b d^{8} {\left | b \right |}}\right )} {\left (b x + a\right )} - \frac {105 \, {\left (b^{9} c^{4} d^{3} - 4 \, a b^{8} c^{3} d^{4} + 6 \, a^{2} b^{7} c^{2} d^{5} - 4 \, a^{3} b^{6} c d^{6} + a^{4} b^{5} d^{7}\right )}}{b^{2} c d^{7} {\left | b \right |} - a b d^{8} {\left | b \right |}}\right )} \sqrt {b x + a}}{12 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {35 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\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 )}{4 \, \sqrt {b d} d^{4} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

1/12*((3*(b*x + a)*(2*(b^6*c*d^6 - a*b^5*d^7)*(b*x + a)/(b^2*c*d^7*abs(b) - a*b*d^8*abs(b)) - 7*(b^7*c^2*d^5 -
2*a*b^6*c*d^6 + a^2*b^5*d^7)/(b^2*c*d^7*abs(b) - a*b*d^8*abs(b))) - 140*(b^8*c^3*d^4 - 3*a*b^7*c^2*d^5 + 3*a^
2*b^6*c*d^6 - a^3*b^5*d^7)/(b^2*c*d^7*abs(b) - a*b*d^8*abs(b)))*(b*x + a) - 105*(b^9*c^4*d^3 - 4*a*b^8*c^3*d^4
+ 6*a^2*b^7*c^2*d^5 - 4*a^3*b^6*c*d^6 + a^4*b^5*d^7)/(b^2*c*d^7*abs(b) - a*b*d^8*abs(b)))*sqrt(b*x + a)/(b^2*
c + (b*x + a)*b*d - a*b*d)^(3/2) - 35/4*(b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*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))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{7/2}}{{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(7/2)/(c + d*x)^(5/2),x)

[Out]

int((a + b*x)^(7/2)/(c + d*x)^(5/2), x)

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