3.985 \(\int \frac {(c x)^{7/2}}{(a+b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=146 \[ -\frac {5 a c^{7/2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}-\frac {5 a c^{7/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}+\frac {5 a c^3 \sqrt {c x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {c (c x)^{5/2}}{2 b \sqrt [4]{a+b x^2}} \]

[Out]

1/2*c*(c*x)^(5/2)/b/(b*x^2+a)^(1/4)-5/4*a*c^(7/2)*arctan(b^(1/4)*(c*x)^(1/2)/(b*x^2+a)^(1/4)/c^(1/2))/b^(9/4)-
5/4*a*c^(7/2)*arctanh(b^(1/4)*(c*x)^(1/2)/(b*x^2+a)^(1/4)/c^(1/2))/b^(9/4)+5/2*a*c^3*(c*x)^(1/2)/b^2/(b*x^2+a)
^(1/4)

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Rubi [A]  time = 0.08, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {285, 288, 329, 240, 212, 208, 205} \[ \frac {5 a c^3 \sqrt {c x}}{2 b^2 \sqrt [4]{a+b x^2}}-\frac {5 a c^{7/2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}-\frac {5 a c^{7/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}+\frac {c (c x)^{5/2}}{2 b \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(7/2)/(a + b*x^2)^(5/4),x]

[Out]

(5*a*c^3*Sqrt[c*x])/(2*b^2*(a + b*x^2)^(1/4)) + (c*(c*x)^(5/2))/(2*b*(a + b*x^2)^(1/4)) - (5*a*c^(7/2)*ArcTan[
(b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/(4*b^(9/4)) - (5*a*c^(7/2)*ArcTanh[(b^(1/4)*Sqrt[c*x])/(Sqrt
[c]*(a + b*x^2)^(1/4))])/(4*b^(9/4))

Rule 205

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

Rule 208

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

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 285

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(2*c*(c*x)^(m - 1))/(b*(2*m - 3)*(a + b*x
^2)^(1/4)), x] - Dist[(2*a*c^2*(m - 1))/(b*(2*m - 3)), Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a
, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

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Mathematica [C]  time = 0.04, size = 63, normalized size = 0.43 \[ \frac {c (c x)^{5/2} \left (1-\sqrt [4]{\frac {b x^2}{a}+1} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};-\frac {b x^2}{a}\right )\right )}{2 b \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(7/2)/(a + b*x^2)^(5/4),x]

[Out]

(c*(c*x)^(5/2)*(1 - (1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[5/4, 5/4, 9/4, -((b*x^2)/a)]))/(2*b*(a + b*x^2)^(1
/4))

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fricas [B]  time = 0.87, size = 389, normalized size = 2.66 \[ \frac {4 \, {\left (b c^{3} x^{2} + 5 \, a c^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} + 20 \, \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )} \arctan \left (-\frac {\left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {3}{4}} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a b^{7} c^{3} - {\left (b^{8} x^{2} + a b^{7}\right )} \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {3}{4}} \sqrt {\frac {\sqrt {b x^{2} + a} a^{2} c^{7} x + \sqrt {\frac {a^{4} c^{14}}{b^{9}}} {\left (b^{5} x^{2} + a b^{4}\right )}}{b x^{2} + a}}}{a^{4} b c^{14} x^{2} + a^{5} c^{14}}\right ) - 5 \, \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )} \log \left (\frac {5 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a c^{3} + \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )}\right )}}{b x^{2} + a}\right ) + 5 \, \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )} \log \left (\frac {5 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a c^{3} - \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )}\right )}}{b x^{2} + a}\right )}{8 \, {\left (b^{3} x^{2} + a b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

1/8*(4*(b*c^3*x^2 + 5*a*c^3)*(b*x^2 + a)^(3/4)*sqrt(c*x) + 20*(a^4*c^14/b^9)^(1/4)*(b^3*x^2 + a*b^2)*arctan(-(
(a^4*c^14/b^9)^(3/4)*(b*x^2 + a)^(3/4)*sqrt(c*x)*a*b^7*c^3 - (b^8*x^2 + a*b^7)*(a^4*c^14/b^9)^(3/4)*sqrt((sqrt
(b*x^2 + a)*a^2*c^7*x + sqrt(a^4*c^14/b^9)*(b^5*x^2 + a*b^4))/(b*x^2 + a)))/(a^4*b*c^14*x^2 + a^5*c^14)) - 5*(
a^4*c^14/b^9)^(1/4)*(b^3*x^2 + a*b^2)*log(5*((b*x^2 + a)^(3/4)*sqrt(c*x)*a*c^3 + (a^4*c^14/b^9)^(1/4)*(b^3*x^2
+ a*b^2))/(b*x^2 + a)) + 5*(a^4*c^14/b^9)^(1/4)*(b^3*x^2 + a*b^2)*log(5*((b*x^2 + a)^(3/4)*sqrt(c*x)*a*c^3 -
(a^4*c^14/b^9)^(1/4)*(b^3*x^2 + a*b^2))/(b*x^2 + a)))/(b^3*x^2 + a*b^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((c*x)^(7/2)/(b*x^2 + a)^(5/4), x)

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x \right )^{\frac {7}{2}}}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(7/2)/(b*x^2+a)^(5/4),x)

[Out]

int((c*x)^(7/2)/(b*x^2+a)^(5/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((c*x)^(7/2)/(b*x^2 + a)^(5/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x\right )}^{7/2}}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(7/2)/(a + b*x^2)^(5/4),x)

[Out]

int((c*x)^(7/2)/(a + b*x^2)^(5/4), x)

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sympy [C]  time = 24.49, size = 44, normalized size = 0.30 \[ \frac {c^{\frac {7}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(7/2)/(b*x**2+a)**(5/4),x)

[Out]

c**(7/2)*x**(9/2)*gamma(9/4)*hyper((5/4, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(13/4))

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