3.8.18 \(\int \frac {1}{(a+b x) (c+d x^2)^{3/4}} \, dx\) [718]

Optimal. Leaf size=268 \[ -\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )}{\left (b^2 c+a^2 d\right )^{3/4}}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )}{\left (b^2 c+a^2 d\right )^{3/4}}+\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (-\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{\left (b^2 c+a^2 d\right ) x}+\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{\left (b^2 c+a^2 d\right ) x} \]

[Out]

-arctan((d*x^2+c)^(1/4)*b^(1/2)/(a^2*d+b^2*c)^(1/4))*b^(1/2)/(a^2*d+b^2*c)^(3/4)-arctanh((d*x^2+c)^(1/4)*b^(1/
2)/(a^2*d+b^2*c)^(1/4))*b^(1/2)/(a^2*d+b^2*c)^(3/4)+a*c^(1/4)*EllipticPi((d*x^2+c)^(1/4)/c^(1/4),-b*c^(1/2)/(a
^2*d+b^2*c)^(1/2),I)*(-d*x^2/c)^(1/2)/(a^2*d+b^2*c)/x+a*c^(1/4)*EllipticPi((d*x^2+c)^(1/4)/c^(1/4),b*c^(1/2)/(
a^2*d+b^2*c)^(1/2),I)*(-d*x^2/c)^(1/2)/(a^2*d+b^2*c)/x

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Rubi [A]
time = 0.18, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {761, 410, 109, 418, 1232, 455, 65, 218, 214, 211} \begin {gather*} \frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (-\frac {b \sqrt {c}}{\sqrt {d a^2+b^2 c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{x \left (a^2 d+b^2 c\right )}+\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (\frac {b \sqrt {c}}{\sqrt {d a^2+b^2 c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{x \left (a^2 d+b^2 c\right )}-\frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\left (a^2 d+b^2 c\right )^{3/4}}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\left (a^2 d+b^2 c\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(c + d*x^2)^(3/4)),x]

[Out]

-((Sqrt[b]*ArcTan[(Sqrt[b]*(c + d*x^2)^(1/4))/(b^2*c + a^2*d)^(1/4)])/(b^2*c + a^2*d)^(3/4)) - (Sqrt[b]*ArcTan
h[(Sqrt[b]*(c + d*x^2)^(1/4))/(b^2*c + a^2*d)^(1/4)])/(b^2*c + a^2*d)^(3/4) + (a*c^(1/4)*Sqrt[-((d*x^2)/c)]*El
lipticPi[-((b*Sqrt[c])/Sqrt[b^2*c + a^2*d]), ArcSin[(c + d*x^2)^(1/4)/c^(1/4)], -1])/((b^2*c + a^2*d)*x) + (a*
c^(1/4)*Sqrt[-((d*x^2)/c)]*EllipticPi[(b*Sqrt[c])/Sqrt[b^2*c + a^2*d], ArcSin[(c + d*x^2)^(1/4)/c^(1/4)], -1])
/((b^2*c + a^2*d)*x)

Rule 65

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 109

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(3/4)), x_Symbol] :> Dist[-4, Subst[
Int[1/((b*e - a*f - b*x^4)*Sqrt[c - d*(e/f) + d*(x^4/f)]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e
, f}, x] && GtQ[-f/(d*e - c*f), 0]

Rule 211

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

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 410

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

Rule 418

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

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 761

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(3/4)), x_Symbol] :> Dist[d, Int[1/((d^2 - e^2*x^2)*(a + c*x^
2)^(3/4)), x], x] - Dist[e, Int[x/((d^2 - e^2*x^2)*(a + c*x^2)^(3/4)), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ
[c*d^2 + a*e^2, 0]

Rule 1232

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

Rubi steps

\begin {align*} \int \frac {1}{(a+b x) \left (c+d x^2\right )^{3/4}} \, dx &=a \int \frac {1}{\left (a^2-b^2 x^2\right ) \left (c+d x^2\right )^{3/4}} \, dx-b \int \frac {x}{\left (a^2-b^2 x^2\right ) \left (c+d x^2\right )^{3/4}} \, dx\\ &=-\left (\frac {1}{2} b \text {Subst}\left (\int \frac {1}{\left (a^2-b^2 x\right ) (c+d x)^{3/4}} \, dx,x,x^2\right )\right )+\frac {\left (a \sqrt {-\frac {d x^2}{c}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-\frac {d x}{c}} \left (a^2-b^2 x\right ) (c+d x)^{3/4}} \, dx,x,x^2\right )}{2 x}\\ &=-\frac {(2 b) \text {Subst}\left (\int \frac {1}{a^2+\frac {b^2 c}{d}-\frac {b^2 x^4}{d}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{d}-\frac {\left (2 a \sqrt {-\frac {d x^2}{c}}\right ) \text {Subst}\left (\int \frac {1}{\left (-b^2 c-a^2 d+b^2 x^4\right ) \sqrt {1-\frac {x^4}{c}}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{x}\\ &=-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {b^2 c+a^2 d}-b x^2} \, dx,x,\sqrt [4]{c+d x^2}\right )}{\sqrt {b^2 c+a^2 d}}-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {b^2 c+a^2 d}+b x^2} \, dx,x,\sqrt [4]{c+d x^2}\right )}{\sqrt {b^2 c+a^2 d}}+\frac {\left (a \sqrt {-\frac {d x^2}{c}}\right ) \text {Subst}\left (\int \frac {1}{\left (1-\frac {b x^2}{\sqrt {b^2 c+a^2 d}}\right ) \sqrt {1-\frac {x^4}{c}}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{\left (b^2 c+a^2 d\right ) x}+\frac {\left (a \sqrt {-\frac {d x^2}{c}}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{\sqrt {b^2 c+a^2 d}}\right ) \sqrt {1-\frac {x^4}{c}}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{\left (b^2 c+a^2 d\right ) x}\\ &=-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )}{\left (b^2 c+a^2 d\right )^{3/4}}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )}{\left (b^2 c+a^2 d\right )^{3/4}}+\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (-\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{\left (b^2 c+a^2 d\right ) x}+\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{\left (b^2 c+a^2 d\right ) x}\\ \end {align*}

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Mathematica [A]
time = 7.37, size = 233, normalized size = 0.87 \begin {gather*} \frac {-\sqrt {b} \sqrt [4]{b^2 c+a^2 d} x \left (\tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )+\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )\right )+a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (-\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )+a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{\left (b^2 c+a^2 d\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)*(c + d*x^2)^(3/4)),x]

[Out]

(-(Sqrt[b]*(b^2*c + a^2*d)^(1/4)*x*(ArcTan[(Sqrt[b]*(c + d*x^2)^(1/4))/(b^2*c + a^2*d)^(1/4)] + ArcTanh[(Sqrt[
b]*(c + d*x^2)^(1/4))/(b^2*c + a^2*d)^(1/4)])) + a*c^(1/4)*Sqrt[-((d*x^2)/c)]*EllipticPi[-((b*Sqrt[c])/Sqrt[b^
2*c + a^2*d]), ArcSin[(c + d*x^2)^(1/4)/c^(1/4)], -1] + a*c^(1/4)*Sqrt[-((d*x^2)/c)]*EllipticPi[(b*Sqrt[c])/Sq
rt[b^2*c + a^2*d], ArcSin[(c + d*x^2)^(1/4)/c^(1/4)], -1])/((b^2*c + a^2*d)*x)

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b x +a \right ) \left (d \,x^{2}+c \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x^2+c)^(3/4),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x^2+c)^(3/4),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right ) \left (c + d x^{2}\right )^{\frac {3}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x**2+c)**(3/4),x)

[Out]

Integral(1/((a + b*x)*(c + d*x**2)**(3/4)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x^2+c)^(3/4),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (d\,x^2+c\right )}^{3/4}\,\left (a+b\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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