3.1513 \(\int \frac{(a+b x)^{7/2}}{(c+d x)^{5/2}} \, dx\)

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

[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))

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Rubi [A]  time = 0.0788944, antiderivative size = 170, normalized size of antiderivative = 1., 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 = {47, 50, 63, 217, 206} \[ \frac{35 b^2 (a+b x)^{3/2} \sqrt{c+d x}}{6 d^3}-\frac{35 b^2 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)}{4 d^4}+\frac{35 b^{3/2} (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 d^{9/2}}-\frac{14 b (a+b x)^{5/2}}{3 d^2 \sqrt{c+d x}}-\frac{2 (a+b x)^{7/2}}{3 d (c+d x)^{3/2}} \]

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 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 50

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 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)^{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [C]  time = 0.0788387, size = 73, normalized size = 0.43 \[ \frac{2 (a+b x)^{9/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/2} \, _2F_1\left (\frac{5}{2},\frac{9}{2};\frac{11}{2};\frac{d (a+b x)}{a d-b c}\right )}{9 b (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(9/2)*((b*(c + d*x))/(b*c - a*d))^(5/2)*Hypergeometric2F1[5/2, 9/2, 11/2, (d*(a + b*x))/(-(b*c) +
a*d)])/(9*b*(c + d*x)^(5/2))

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{{\frac{7}{2}}} \left ( dx+c \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

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., 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)^(7/2)/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 9.68811, size = 1423, normalized size = 8.37 \begin{align*} \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{align*}

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(-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)**(7/2)/(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.22686, size = 513, normalized size = 3.02 \begin{align*} \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{align*}

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))