3.1700 \(\int \frac {(a+b x)^{7/4}}{\sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=751 \[ \frac {7 (b c-a d)^{7/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{20 \sqrt {2} b^{3/4} d^{11/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {7 (b c-a d)^{7/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{10 \sqrt {2} b^{3/4} d^{11/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {7 (b c-a d) \sqrt {(a+b x) (c+d x)} \sqrt {(a d+b c+2 b d x)^2} \sqrt {(a d+b (c+2 d x))^2}}{10 \sqrt {b} d^{5/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )}-\frac {7 (a+b x)^{3/4} (c+d x)^{3/4} (b c-a d)}{15 d^2}+\frac {2 (a+b x)^{7/4} (c+d x)^{3/4}}{5 d} \]

[Out]

-7/15*(-a*d+b*c)*(b*x+a)^(3/4)*(d*x+c)^(3/4)/d^2+2/5*(b*x+a)^(7/4)*(d*x+c)^(3/4)/d+7/10*(-a*d+b*c)*((b*x+a)*(d
*x+c))^(1/2)*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d+b*(2*d*x+c))^2)^(1/2)/d^(5/2)/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*
d*x+a*d+b*c)/b^(1/2)/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))-7/20*(-a*d+b*c)^(7/2)*((b*x+a)*(
d*x+c))^(1/4)*(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)))*EllipticE(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^(3/4)/d^(11/4)/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*d*x+a*d+b*c)*2^(1/2)/((a*d+b
*(2*d*x+c))^2)^(1/2)+7/40*(-a*d+b*c)^(7/2)*((b*x+a)*(d*x+c))^(1/4)*(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^(3/4)/d^(11/4)/(b*x+a)^
(1/4)/(d*x+c)^(1/4)/(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 = 0.79, antiderivative size = 751, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {50, 62, 623, 305, 220, 1196} \[ \frac {7 (b c-a d)^{7/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{20 \sqrt {2} b^{3/4} d^{11/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {7 (b c-a d)^{7/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{10 \sqrt {2} b^{3/4} d^{11/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {7 (a+b x)^{3/4} (c+d x)^{3/4} (b c-a d)}{15 d^2}+\frac {7 (b c-a d) \sqrt {(a+b x) (c+d x)} \sqrt {(a d+b c+2 b d x)^2} \sqrt {(a d+b (c+2 d x))^2}}{10 \sqrt {b} d^{5/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )}+\frac {2 (a+b x)^{7/4} (c+d x)^{3/4}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(7/4)/(c + d*x)^(1/4),x]

[Out]

(-7*(b*c - a*d)*(a + b*x)^(3/4)*(c + d*x)^(3/4))/(15*d^2) + (2*(a + b*x)^(7/4)*(c + d*x)^(3/4))/(5*d) + (7*(b*
c - a*d)*Sqrt[(a + b*x)*(c + d*x)]*Sqrt[(b*c + a*d + 2*b*d*x)^2]*Sqrt[(a*d + b*(c + 2*d*x))^2])/(10*Sqrt[b]*d^
(5/2)*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(b*c + a*d + 2*b*d*x)*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])
/(b*c - a*d))) - (7*(b*c - a*d)^(7/2)*((a + b*x)*(c + d*x))^(1/4)*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)]*EllipticE[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])/(10*Sqrt[2]*b^(3/4)*d^(11/4)*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(b*c + a*d
+ 2*b*d*x)*Sqrt[(a*d + b*(c + 2*d*x))^2]) + (7*(b*c - a*d)^(7/2)*((a + b*x)*(c + d*x))^(1/4)*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])/(20*Sqrt[2]*b^(3/4)*d^(11/4)*(a + b*x)^(1/4
)*(c + d*x)^(1/4)*(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 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], 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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{7/4}}{\sqrt [4]{c+d x}} \, dx &=\frac {2 (a+b x)^{7/4} (c+d x)^{3/4}}{5 d}-\frac {(7 (b c-a d)) \int \frac {(a+b x)^{3/4}}{\sqrt [4]{c+d x}} \, dx}{10 d}\\ &=-\frac {7 (b c-a d) (a+b x)^{3/4} (c+d x)^{3/4}}{15 d^2}+\frac {2 (a+b x)^{7/4} (c+d x)^{3/4}}{5 d}+\frac {\left (7 (b c-a d)^2\right ) \int \frac {1}{\sqrt [4]{a+b x} \sqrt [4]{c+d x}} \, dx}{20 d^2}\\ &=-\frac {7 (b c-a d) (a+b x)^{3/4} (c+d x)^{3/4}}{15 d^2}+\frac {2 (a+b x)^{7/4} (c+d x)^{3/4}}{5 d}+\frac {\left (7 (b c-a d)^2 \sqrt [4]{(a+b x) (c+d x)}\right ) \int \frac {1}{\sqrt [4]{a c+(b c+a d) x+b d x^2}} \, dx}{20 d^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}\\ &=-\frac {7 (b c-a d) (a+b x)^{3/4} (c+d x)^{3/4}}{15 d^2}+\frac {2 (a+b x)^{7/4} (c+d x)^{3/4}}{5 d}+\frac {\left (7 (b c-a d)^2 \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\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 )}{5 d^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}\\ &=-\frac {7 (b c-a d) (a+b x)^{3/4} (c+d x)^{3/4}}{15 d^2}+\frac {2 (a+b x)^{7/4} (c+d x)^{3/4}}{5 d}+\frac {\left (7 (b c-a d)^3 \sqrt [4]{(a+b x) (c+d x)} \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 )}{10 \sqrt {b} d^{5/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}-\frac {\left (7 (b c-a d)^3 \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {2 \sqrt {b} \sqrt {d} x^2}{b c-a d}}{\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 )}{10 \sqrt {b} d^{5/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}\\ &=-\frac {7 (b c-a d) (a+b x)^{3/4} (c+d x)^{3/4}}{15 d^2}+\frac {2 (a+b x)^{7/4} (c+d x)^{3/4}}{5 d}+\frac {7 (b c-a d) \sqrt {(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \sqrt {(a d+b (c+2 d x))^2}}{10 \sqrt {b} d^{5/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )}-\frac {7 (b c-a d)^{7/2} \sqrt [4]{(a+b x) (c+d x)} \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}} E\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 )}{10 \sqrt {2} b^{3/4} d^{11/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {7 (b c-a d)^{7/2} \sqrt [4]{(a+b x) (c+d x)} \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 )}{20 \sqrt {2} b^{3/4} d^{11/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (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 = 0.03, size = 73, normalized size = 0.10 \[ \frac {4 (a+b x)^{11/4} \sqrt [4]{\frac {b (c+d x)}{b c-a d}} \, _2F_1\left (\frac {1}{4},\frac {11}{4};\frac {15}{4};\frac {d (a+b x)}{a d-b c}\right )}{11 b \sqrt [4]{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(7/4)/(c + d*x)^(1/4),x]

[Out]

(4*(a + b*x)^(11/4)*((b*(c + d*x))/(b*c - a*d))^(1/4)*Hypergeometric2F1[1/4, 11/4, 15/4, (d*(a + b*x))/(-(b*c)
+ a*d)])/(11*b*(c + d*x)^(1/4))

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{\frac {7}{4}}}{{\left (d x + c\right )}^{\frac {1}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/4)/(d*x+c)^(1/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{\frac {7}{4}}}{{\left (d x + c\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/4)/(d*x+c)^(1/4),x, algorithm="giac")

[Out]

integrate((b*x + a)^(7/4)/(d*x + c)^(1/4), x)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{\frac {7}{4}}}{\left (d x +c \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(7/4)/(d*x+c)^(1/4),x)

[Out]

int((b*x+a)^(7/4)/(d*x+c)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{\frac {7}{4}}}{{\left (d x + c\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/4)/(d*x+c)^(1/4),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(7/4)/(d*x + c)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(7/4)/(c + d*x)^(1/4),x)

[Out]

int((a + b*x)^(7/4)/(c + d*x)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(7/4)/(d*x+c)**(1/4),x)

[Out]

Integral((a + b*x)**(7/4)/(c + d*x)**(1/4), x)

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