3.1688 \(\int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx\)

Optimal. Leaf size=332 \[ \frac {5 (b c-a d)^{5/2} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} b^{9/4} \sqrt [4]{d} (a+b x)^{3/4} (c+d x)^{3/4} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {5 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c-a d)}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b} \]

[Out]

5/3*(-a*d+b*c)*(b*x+a)^(1/4)*(d*x+c)^(1/4)/b^2+2/3*(b*x+a)^(1/4)*(d*x+c)^(5/4)/b+5/12*(-a*d+b*c)^(5/2)*((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)/d^(1/4)/(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)

________________________________________________________________________________________

Rubi [A]  time = 0.29, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {50, 62, 623, 220} \[ \frac {5 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c-a d)}{3 b^2}+\frac {5 (b c-a d)^{5/2} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{6 \sqrt {2} b^{9/4} \sqrt [4]{d} (a+b x)^{3/4} (c+d x)^{3/4} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(b*c - a*d)*(a + b*x)^(1/4)*(c + d*x)^(1/4))/(3*b^2) + (2*(a + b*x)^(1/4)*(c + d*x)^(5/4))/(3*b) + (5*(b*c
- a*d)^(5/2)*((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])/(6*Sqrt[2]*b^(9/4)*d^(1/4)*(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 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)^{3/4}} \, dx &=\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {(5 (b c-a d)) \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{3/4}} \, dx}{6 b}\\ &=\frac {5 (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {\left (5 (b c-a d)^2\right ) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx}{12 b^2}\\ &=\frac {5 (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {\left (5 (b c-a d)^2 ((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}{12 b^2 (a+b x)^{3/4} (c+d x)^{3/4}}\\ &=\frac {5 (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {\left (5 (b c-a d)^2 ((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 {5 (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {5 (b c-a d)^{5/2} ((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 )}{6 \sqrt {2} b^{9/4} \sqrt [4]{d} (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*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 71, normalized size = 0.21 \[ \frac {4 \sqrt [4]{a+b x} (c+d x)^{5/4} \, _2F_1\left (-\frac {5}{4},\frac {1}{4};\frac {5}{4};\frac {d (a+b x)}{a d-b c}\right )}{b \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {3}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(3/4),x, algorithm="fricas")

[Out]

integral((d*x + c)^(5/4)/(b*x + a)^(3/4), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(3/4),x, algorithm="giac")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(3/4), x)

________________________________________________________________________________________

maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/4)/(b*x+a)^(3/4),x)

[Out]

int((d*x+c)^(5/4)/(b*x+a)^(3/4),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(3/4), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/4)/(a + b*x)^(3/4),x)

[Out]

int((c + d*x)^(5/4)/(a + b*x)^(3/4), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/4)/(b*x+a)**(3/4),x)

[Out]

Integral((c + d*x)**(5/4)/(a + b*x)**(3/4), x)

________________________________________________________________________________________