3.2217 \(\int \frac{(a+b x)^{3/2} (A+B x)}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=138 \[ -\frac{2 (a+b x)^{5/2} (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}+\frac{2 b^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{e^{7/2}}-\frac{2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac{2 b B \sqrt{a+b x}}{e^3 \sqrt{d+e x}} \]

[Out]

(-2*(B*d - A*e)*(a + b*x)^(5/2))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) - (2*B*(a + b*x)^(3/2))/(3*e^2*(d + e*x)^(3
/2)) - (2*b*B*Sqrt[a + b*x])/(e^3*Sqrt[d + e*x]) + (2*b^(3/2)*B*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[
d + e*x])])/e^(7/2)

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Rubi [A]  time = 0.0765241, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {78, 47, 63, 217, 206} \[ -\frac{2 (a+b x)^{5/2} (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}+\frac{2 b^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{e^{7/2}}-\frac{2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac{2 b B \sqrt{a+b x}}{e^3 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(B*d - A*e)*(a + b*x)^(5/2))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) - (2*B*(a + b*x)^(3/2))/(3*e^2*(d + e*x)^(3
/2)) - (2*b*B*Sqrt[a + b*x])/(e^3*Sqrt[d + e*x]) + (2*b^(3/2)*B*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[
d + e*x])])/e^(7/2)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 47

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 63

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 217

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]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{(a+b x)^{3/2} (A+B x)}{(d+e x)^{7/2}} \, dx &=-\frac{2 (B d-A e) (a+b x)^{5/2}}{5 e (b d-a e) (d+e x)^{5/2}}+\frac{B \int \frac{(a+b x)^{3/2}}{(d+e x)^{5/2}} \, dx}{e}\\ &=-\frac{2 (B d-A e) (a+b x)^{5/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac{2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}+\frac{(b B) \int \frac{\sqrt{a+b x}}{(d+e x)^{3/2}} \, dx}{e^2}\\ &=-\frac{2 (B d-A e) (a+b x)^{5/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac{2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac{2 b B \sqrt{a+b x}}{e^3 \sqrt{d+e x}}+\frac{\left (b^2 B\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{d+e x}} \, dx}{e^3}\\ &=-\frac{2 (B d-A e) (a+b x)^{5/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac{2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac{2 b B \sqrt{a+b x}}{e^3 \sqrt{d+e x}}+\frac{(2 b B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d-\frac{a e}{b}+\frac{e x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{e^3}\\ &=-\frac{2 (B d-A e) (a+b x)^{5/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac{2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac{2 b B \sqrt{a+b x}}{e^3 \sqrt{d+e x}}+\frac{(2 b B) \operatorname{Subst}\left (\int \frac{1}{1-\frac{e x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{d+e x}}\right )}{e^3}\\ &=-\frac{2 (B d-A e) (a+b x)^{5/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac{2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac{2 b B \sqrt{a+b x}}{e^3 \sqrt{d+e x}}+\frac{2 b^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{e^{7/2}}\\ \end{align*}

Mathematica [A]  time = 1.17342, size = 173, normalized size = 1.25 \[ \frac{2 \left (e^3 (a+b x)^3 (A e-B d)+\frac{5}{3} B e (a+b x) (d+e x) (a e-b d) (a e+3 b d+4 b e x)+\frac{5 B \sqrt{e} \sqrt{a+b x} (b d-a e)^{7/2} \left (\frac{b (d+e x)}{b d-a e}\right )^{5/2} \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )}{b}\right )}{5 e^4 \sqrt{a+b x} (d+e x)^{5/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e^3*(-(B*d) + A*e)*(a + b*x)^3 + (5*B*e*(-(b*d) + a*e)*(a + b*x)*(d + e*x)*(3*b*d + a*e + 4*b*e*x))/3 + (5
*B*Sqrt[e]*(b*d - a*e)^(7/2)*Sqrt[a + b*x]*((b*(d + e*x))/(b*d - a*e))^(5/2)*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/S
qrt[b*d - a*e]])/b))/(5*e^4*(b*d - a*e)*Sqrt[a + b*x]*(d + e*x)^(5/2))

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Maple [B]  time = 0.024, size = 780, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(7/2),x)

[Out]

-1/15*(b*x+a)^(1/2)*(-15*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^3*a*b
^2*e^4+15*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^3*b^3*d*e^3-45*B*ln(
1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^2*a*b^2*d*e^3+45*B*ln(1/2*(2*b*x*e+
2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^2*b^3*d^2*e^2+6*A*x^2*b^2*e^3*((b*x+a)*(e*x+d))^
(1/2)*(b*e)^(1/2)-45*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*a*b^2*d^2
*e^2+45*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*b^3*d^3*e+40*B*x^2*a*b
*e^3*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-46*B*x^2*b^2*d*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+12*A*x*a*b*e^3
*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-15*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)
^(1/2))*a*b^2*d^3*e+15*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^3*d^4+1
0*B*x*a^2*e^3*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+48*B*x*a*b*d*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-70*B*x*
b^2*d^2*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+6*A*a^2*e^3*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+4*B*a^2*d*e^2*((
b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+20*B*a*b*d^2*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-30*B*b^2*d^3*((b*x+a)*(e*
x+d))^(1/2)*(b*e)^(1/2))/((b*x+a)*(e*x+d))^(1/2)/(a*e-b*d)/(b*e)^(1/2)/(e*x+d)^(5/2)/e^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 26.6314, size = 1625, normalized size = 11.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*(B*b^2*d^4 - B*a*b*d^3*e + (B*b^2*d*e^3 - B*a*b*e^4)*x^3 + 3*(B*b^2*d^2*e^2 - B*a*b*d*e^3)*x^2 + 3*(
B*b^2*d^3*e - B*a*b*d^2*e^2)*x)*sqrt(b/e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b*e^2*x + b
*d*e + a*e^2)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(b/e) + 8*(b^2*d*e + a*b*e^2)*x) - 4*(15*B*b^2*d^3 - 10*B*a*b*d^
2*e - 2*B*a^2*d*e^2 - 3*A*a^2*e^3 + (23*B*b^2*d*e^2 - (20*B*a*b + 3*A*b^2)*e^3)*x^2 + (35*B*b^2*d^2*e - 24*B*a
*b*d*e^2 - (5*B*a^2 + 6*A*a*b)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b*d^4*e^3 - a*d^3*e^4 + (b*d*e^6 - a*e^7)
*x^3 + 3*(b*d^2*e^5 - a*d*e^6)*x^2 + 3*(b*d^3*e^4 - a*d^2*e^5)*x), -1/15*(15*(B*b^2*d^4 - B*a*b*d^3*e + (B*b^2
*d*e^3 - B*a*b*e^4)*x^3 + 3*(B*b^2*d^2*e^2 - B*a*b*d*e^3)*x^2 + 3*(B*b^2*d^3*e - B*a*b*d^2*e^2)*x)*sqrt(-b/e)*
arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(-b/e)/(b^2*e*x^2 + a*b*d + (b^2*d + a*b*e)*x
)) + 2*(15*B*b^2*d^3 - 10*B*a*b*d^2*e - 2*B*a^2*d*e^2 - 3*A*a^2*e^3 + (23*B*b^2*d*e^2 - (20*B*a*b + 3*A*b^2)*e
^3)*x^2 + (35*B*b^2*d^2*e - 24*B*a*b*d*e^2 - (5*B*a^2 + 6*A*a*b)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b*d^4*e
^3 - a*d^3*e^4 + (b*d*e^6 - a*e^7)*x^3 + 3*(b*d^2*e^5 - a*d*e^6)*x^2 + 3*(b*d^3*e^4 - a*d^2*e^5)*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(B*x+A)/(e*x+d)**(7/2),x)

[Out]

Timed out

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Giac [B]  time = 2.07236, size = 587, normalized size = 4.25 \begin{align*} \frac{B \sqrt{b}{\left | b \right |} e^{\frac{1}{2}} \log \left ({\left | -\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} + \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \right |}\right )}{64 \,{\left (b^{8} d e^{5} - a b^{7} e^{6}\right )}} + \frac{{\left ({\left (b x + a\right )}{\left (\frac{{\left (23 \, B b^{7} d^{2}{\left | b \right |} e^{4} - 43 \, B a b^{6} d{\left | b \right |} e^{5} - 3 \, A b^{7} d{\left | b \right |} e^{5} + 20 \, B a^{2} b^{5}{\left | b \right |} e^{6} + 3 \, A a b^{6}{\left | b \right |} e^{6}\right )}{\left (b x + a\right )}}{b^{12} d^{3} e^{6} - 3 \, a b^{11} d^{2} e^{7} + 3 \, a^{2} b^{10} d e^{8} - a^{3} b^{9} e^{9}} + \frac{35 \,{\left (B b^{8} d^{3}{\left | b \right |} e^{3} - 3 \, B a b^{7} d^{2}{\left | b \right |} e^{4} + 3 \, B a^{2} b^{6} d{\left | b \right |} e^{5} - B a^{3} b^{5}{\left | b \right |} e^{6}\right )}}{b^{12} d^{3} e^{6} - 3 \, a b^{11} d^{2} e^{7} + 3 \, a^{2} b^{10} d e^{8} - a^{3} b^{9} e^{9}}\right )} + \frac{15 \,{\left (B b^{9} d^{4}{\left | b \right |} e^{2} - 4 \, B a b^{8} d^{3}{\left | b \right |} e^{3} + 6 \, B a^{2} b^{7} d^{2}{\left | b \right |} e^{4} - 4 \, B a^{3} b^{6} d{\left | b \right |} e^{5} + B a^{4} b^{5}{\left | b \right |} e^{6}\right )}}{b^{12} d^{3} e^{6} - 3 \, a b^{11} d^{2} e^{7} + 3 \, a^{2} b^{10} d e^{8} - a^{3} b^{9} e^{9}}\right )} \sqrt{b x + a}}{960 \,{\left (b^{2} d +{\left (b x + a\right )} b e - a b e\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*B*sqrt(b)*abs(b)*e^(1/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/(
b^8*d*e^5 - a*b^7*e^6) + 1/960*((b*x + a)*((23*B*b^7*d^2*abs(b)*e^4 - 43*B*a*b^6*d*abs(b)*e^5 - 3*A*b^7*d*abs(
b)*e^5 + 20*B*a^2*b^5*abs(b)*e^6 + 3*A*a*b^6*abs(b)*e^6)*(b*x + a)/(b^12*d^3*e^6 - 3*a*b^11*d^2*e^7 + 3*a^2*b^
10*d*e^8 - a^3*b^9*e^9) + 35*(B*b^8*d^3*abs(b)*e^3 - 3*B*a*b^7*d^2*abs(b)*e^4 + 3*B*a^2*b^6*d*abs(b)*e^5 - B*a
^3*b^5*abs(b)*e^6)/(b^12*d^3*e^6 - 3*a*b^11*d^2*e^7 + 3*a^2*b^10*d*e^8 - a^3*b^9*e^9)) + 15*(B*b^9*d^4*abs(b)*
e^2 - 4*B*a*b^8*d^3*abs(b)*e^3 + 6*B*a^2*b^7*d^2*abs(b)*e^4 - 4*B*a^3*b^6*d*abs(b)*e^5 + B*a^4*b^5*abs(b)*e^6)
/(b^12*d^3*e^6 - 3*a*b^11*d^2*e^7 + 3*a^2*b^10*d*e^8 - a^3*b^9*e^9))*sqrt(b*x + a)/(b^2*d + (b*x + a)*b*e - a*
b*e)^(5/2)