3.1665 \(\int \frac{(a+b x)^{7/2}}{(c+d x)^{7/4}} \, dx\)

Optimal. Leaf size=207 \[ -\frac{320 b^{3/4} (b c-a d)^{13/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{33 d^5 \sqrt{a+b x}}+\frac{56 b (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 d^2}-\frac{80 b (a+b x)^{3/2} \sqrt [4]{c+d x} (b c-a d)}{33 d^3}+\frac{160 b \sqrt{a+b x} \sqrt [4]{c+d x} (b c-a d)^2}{33 d^4}-\frac{4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}} \]

[Out]

(-4*(a + b*x)^(7/2))/(3*d*(c + d*x)^(3/4)) + (160*b*(b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(1/4))/(33*d^4) - (8
0*b*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(1/4))/(33*d^3) + (56*b*(a + b*x)^(5/2)*(c + d*x)^(1/4))/(33*d^2) -
(320*b^(3/4)*(b*c - a*d)^(13/4)*Sqrt[-((d*(a + b*x))/(b*c - a*d))]*EllipticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/
(b*c - a*d)^(1/4)], -1])/(33*d^5*Sqrt[a + b*x])

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Rubi [A]  time = 0.136313, antiderivative size = 207, 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, 224, 221} \[ -\frac{320 b^{3/4} (b c-a d)^{13/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{33 d^5 \sqrt{a+b x}}+\frac{56 b (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 d^2}-\frac{80 b (a+b x)^{3/2} \sqrt [4]{c+d x} (b c-a d)}{33 d^3}+\frac{160 b \sqrt{a+b x} \sqrt [4]{c+d x} (b c-a d)^2}{33 d^4}-\frac{4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(7/2)/(c + d*x)^(7/4),x]

[Out]

(-4*(a + b*x)^(7/2))/(3*d*(c + d*x)^(3/4)) + (160*b*(b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(1/4))/(33*d^4) - (8
0*b*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(1/4))/(33*d^3) + (56*b*(a + b*x)^(5/2)*(c + d*x)^(1/4))/(33*d^2) -
(320*b^(3/4)*(b*c - a*d)^(13/4)*Sqrt[-((d*(a + b*x))/(b*c - a*d))]*EllipticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/
(b*c - a*d)^(1/4)], -1])/(33*d^5*Sqrt[a + b*x])

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{(a+b x)^{7/2}}{(c+d x)^{7/4}} \, dx &=-\frac{4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}+\frac{(14 b) \int \frac{(a+b x)^{5/2}}{(c+d x)^{3/4}} \, dx}{3 d}\\ &=-\frac{4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}+\frac{56 b (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 d^2}-\frac{(140 b (b c-a d)) \int \frac{(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx}{33 d^2}\\ &=-\frac{4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}-\frac{80 b (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{33 d^3}+\frac{56 b (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 d^2}+\frac{\left (40 b (b c-a d)^2\right ) \int \frac{\sqrt{a+b x}}{(c+d x)^{3/4}} \, dx}{11 d^3}\\ &=-\frac{4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}+\frac{160 b (b c-a d)^2 \sqrt{a+b x} \sqrt [4]{c+d x}}{33 d^4}-\frac{80 b (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{33 d^3}+\frac{56 b (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 d^2}-\frac{\left (80 b (b c-a d)^3\right ) \int \frac{1}{\sqrt{a+b x} (c+d x)^{3/4}} \, dx}{33 d^4}\\ &=-\frac{4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}+\frac{160 b (b c-a d)^2 \sqrt{a+b x} \sqrt [4]{c+d x}}{33 d^4}-\frac{80 b (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{33 d^3}+\frac{56 b (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 d^2}-\frac{\left (320 b (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-\frac{b c}{d}+\frac{b x^4}{d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{33 d^5}\\ &=-\frac{4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}+\frac{160 b (b c-a d)^2 \sqrt{a+b x} \sqrt [4]{c+d x}}{33 d^4}-\frac{80 b (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{33 d^3}+\frac{56 b (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 d^2}-\frac{\left (320 b (b c-a d)^3 \sqrt{\frac{d (a+b x)}{-b c+a d}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{b x^4}{\left (a-\frac{b c}{d}\right ) d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{33 d^5 \sqrt{a+b x}}\\ &=-\frac{4 (a+b x)^{7/2}}{3 d (c+d x)^{3/4}}+\frac{160 b (b c-a d)^2 \sqrt{a+b x} \sqrt [4]{c+d x}}{33 d^4}-\frac{80 b (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{33 d^3}+\frac{56 b (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 d^2}-\frac{320 b^{3/4} (b c-a d)^{13/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{33 d^5 \sqrt{a+b x}}\\ \end{align*}

Mathematica [C]  time = 0.0705959, size = 73, normalized size = 0.35 \[ \frac{2 (a+b x)^{9/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{7/4} \, _2F_1\left (\frac{7}{4},\frac{9}{2};\frac{11}{2};\frac{d (a+b x)}{a d-b c}\right )}{9 b (c+d x)^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(7/2)/(c + d*x)^(7/4),x]

[Out]

(2*(a + b*x)^(9/2)*((b*(c + d*x))/(b*c - a*d))^(7/4)*Hypergeometric2F1[7/4, 9/2, 11/2, (d*(a + b*x))/(-(b*c) +
a*d)])/(9*b*(c + d*x)^(7/4))

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Maple [F]  time = 0.053, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{{\frac{7}{2}}} \left ( dx+c \right ) ^{-{\frac{7}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(7/2)/(d*x+c)^(7/4),x)

[Out]

int((b*x+a)^(7/2)/(d*x+c)^(7/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{7}{2}}}{{\left (d x + c\right )}^{\frac{7}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/2)/(d*x+c)^(7/4),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(7/2)/(d*x + c)^(7/4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \sqrt{b x + a}{\left (d x + c\right )}^{\frac{1}{4}}}{d^{2} x^{2} + 2 \, c d x + c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/2)/(d*x+c)^(7/4),x, algorithm="fricas")

[Out]

integral((b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(b*x + a)*(d*x + c)^(1/4)/(d^2*x^2 + 2*c*d*x + c^2), 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)**(7/2)/(d*x+c)**(7/4),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{7}{2}}}{{\left (d x + c\right )}^{\frac{7}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/2)/(d*x+c)^(7/4),x, algorithm="giac")

[Out]

integrate((b*x + a)^(7/2)/(d*x + c)^(7/4), x)