3.1689 \(\int \frac {(c+d x)^{5/4}}{(a+b x)^{7/4}} \, dx\)

Optimal. Leaf size=325 \[ \frac {5 d^{3/4} (b c-a d)^{3/2} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{3 \sqrt {2} b^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {10 d \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {4 (c+d x)^{5/4}}{3 b (a+b x)^{3/4}} \]

[Out]

10/3*d*(b*x+a)^(1/4)*(d*x+c)^(1/4)/b^2-4/3*(d*x+c)^(5/4)/b/(b*x+a)^(3/4)+5/6*d^(3/4)*(-a*d+b*c)^(3/2)*((b*x+a)
*(d*x+c))^(3/4)*(cos(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))^2)^(1/2)/cos(
2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*d^(
1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2))),1/2*2^(1/2))*(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1
/2)/(-a*d+b*c))*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d+b*(2*d*x+c))^2/(-a*d+b*c)^2/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d
*x+c))^(1/2)/(-a*d+b*c))^2)^(1/2)/b^(9/4)/(b*x+a)^(3/4)/(d*x+c)^(3/4)/(2*b*d*x+a*d+b*c)*2^(1/2)/((a*d+b*(2*d*x
+c))^2)^(1/2)

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Rubi [A]  time = 0.28, antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {47, 50, 62, 623, 220} \[ \frac {5 d^{3/4} (b c-a d)^{3/2} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{3 \sqrt {2} b^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {10 d \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {4 (c+d x)^{5/4}}{3 b (a+b x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/4)/(a + b*x)^(7/4),x]

[Out]

(10*d*(a + b*x)^(1/4)*(c + d*x)^(1/4))/(3*b^2) - (4*(c + d*x)^(5/4))/(3*b*(a + b*x)^(3/4)) + (5*d^(3/4)*(b*c -
a*d)^(3/2)*((a + b*x)*(c + d*x))^(3/4)*Sqrt[(b*c + a*d + 2*b*d*x)^2]*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(
c + d*x)])/(b*c - a*d))*Sqrt[(a*d + b*(c + 2*d*x))^2/((b*c - a*d)^2*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c
+ d*x)])/(b*c - a*d))^2)]*EllipticF[2*ArcTan[(Sqrt[2]*b^(1/4)*d^(1/4)*((a + b*x)*(c + d*x))^(1/4))/Sqrt[b*c -
a*d]], 1/2])/(3*Sqrt[2]*b^(9/4)*(a + b*x)^(3/4)*(c + d*x)^(3/4)*(b*c + a*d + 2*b*d*x)*Sqrt[(a*d + b*(c + 2*d*x
))^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 62

Int[((a_.) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Dist[((a + b*x)^m*(c + d*x)^m)/((a + b*x)
*(c + d*x))^m, Int[(a*c + (b*c + a*d)*x + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] &&
LtQ[-1, m, 0] && LeQ[3, Denominator[m], 4]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 623

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[(d*Sqrt[(b + 2*c*x)
^2])/(b + 2*c*x), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rubi steps

\begin {align*} \int \frac {(c+d x)^{5/4}}{(a+b x)^{7/4}} \, dx &=-\frac {4 (c+d x)^{5/4}}{3 b (a+b x)^{3/4}}+\frac {(5 d) \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{3/4}} \, dx}{3 b}\\ &=\frac {10 d \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {4 (c+d x)^{5/4}}{3 b (a+b x)^{3/4}}+\frac {(5 d (b c-a d)) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx}{6 b^2}\\ &=\frac {10 d \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {4 (c+d x)^{5/4}}{3 b (a+b x)^{3/4}}+\frac {\left (5 d (b c-a d) ((a+b x) (c+d x))^{3/4}\right ) \int \frac {1}{\left (a c+(b c+a d) x+b d x^2\right )^{3/4}} \, dx}{6 b^2 (a+b x)^{3/4} (c+d x)^{3/4}}\\ &=\frac {10 d \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {4 (c+d x)^{5/4}}{3 b (a+b x)^{3/4}}+\frac {\left (10 d (b c-a d) ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{3 b^2 (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x)}\\ &=\frac {10 d \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {4 (c+d x)^{5/4}}{3 b (a+b x)^{3/4}}+\frac {5 d^{3/4} (b c-a d)^{3/2} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{3 \sqrt {2} b^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 73, normalized size = 0.22 \[ -\frac {4 (c+d x)^{5/4} \, _2F_1\left (-\frac {5}{4},-\frac {3}{4};\frac {1}{4};\frac {d (a+b x)}{a d-b c}\right )}{3 b (a+b x)^{3/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/4)/(a + b*x)^(7/4),x]

[Out]

(-4*(c + d*x)^(5/4)*Hypergeometric2F1[-5/4, -3/4, 1/4, (d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(a + b*x)^(3/4)*((b
*(c + d*x))/(b*c - a*d))^(5/4))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {5}{4}}}{b^{2} x^{2} + 2 \, a b x + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(7/4),x, algorithm="fricas")

[Out]

integral((b*x + a)^(1/4)*(d*x + c)^(5/4)/(b^2*x^2 + 2*a*b*x + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {7}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(7/4),x, algorithm="giac")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(7/4), x)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {7}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/4)/(b*x+a)^(7/4),x)

[Out]

int((d*x+c)^(5/4)/(b*x+a)^(7/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {7}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(7/4),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(7/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{7/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/4)/(a + b*x)^(7/4),x)

[Out]

int((c + d*x)^(5/4)/(a + b*x)^(7/4), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {7}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/4)/(b*x+a)**(7/4),x)

[Out]

Integral((c + d*x)**(5/4)/(a + b*x)**(7/4), x)

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