3.582 \(\int \frac{1}{(d+e x) (a+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=154 \[ \frac{3 a^2 e^3+c d x \left (5 a e^2+2 c d^2\right )}{3 a^2 \sqrt{a+c x^2} \left (a e^2+c d^2\right )^2}+\frac{a e+c d x}{3 a \left (a+c x^2\right )^{3/2} \left (a e^2+c d^2\right )}-\frac{e^4 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{5/2}} \]

[Out]

(a*e + c*d*x)/(3*a*(c*d^2 + a*e^2)*(a + c*x^2)^(3/2)) + (3*a^2*e^3 + c*d*(2*c*d^2 + 5*a*e^2)*x)/(3*a^2*(c*d^2
+ a*e^2)^2*Sqrt[a + c*x^2]) - (e^4*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(c*d^2 + a*e^
2)^(5/2)

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Rubi [A]  time = 0.126313, antiderivative size = 154, normalized size of antiderivative = 1., 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 = {741, 823, 12, 725, 206} \[ \frac{3 a^2 e^3+c d x \left (5 a e^2+2 c d^2\right )}{3 a^2 \sqrt{a+c x^2} \left (a e^2+c d^2\right )^2}+\frac{a e+c d x}{3 a \left (a+c x^2\right )^{3/2} \left (a e^2+c d^2\right )}-\frac{e^4 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(a + c*x^2)^(5/2)),x]

[Out]

(a*e + c*d*x)/(3*a*(c*d^2 + a*e^2)*(a + c*x^2)^(3/2)) + (3*a^2*e^3 + c*d*(2*c*d^2 + 5*a*e^2)*x)/(3*a^2*(c*d^2
+ a*e^2)^2*Sqrt[a + c*x^2]) - (e^4*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(c*d^2 + a*e^
2)^(5/2)

Rule 741

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

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.149545, size = 153, normalized size = 0.99 \[ \frac{a^2 c e \left (d^2+6 d e x+3 e^2 x^2\right )+4 a^3 e^3+a c^2 d x \left (3 d^2+5 e^2 x^2\right )+2 c^3 d^3 x^3}{3 a^2 \left (a+c x^2\right )^{3/2} \left (a e^2+c d^2\right )^2}-\frac{e^4 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)*(a + c*x^2)^(5/2)),x]

[Out]

(4*a^3*e^3 + 2*c^3*d^3*x^3 + a^2*c*e*(d^2 + 6*d*e*x + 3*e^2*x^2) + a*c^2*d*x*(3*d^2 + 5*e^2*x^2))/(3*a^2*(c*d^
2 + a*e^2)^2*(a + c*x^2)^(3/2)) - (e^4*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(c*d^2 +
a*e^2)^(5/2)

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Maple [B]  time = 0.192, size = 454, normalized size = 3. \begin{align*}{\frac{e}{3\,a{e}^{2}+3\,c{d}^{2}} \left ( c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}} \right ) ^{-{\frac{3}{2}}}}+{\frac{cdx}{ \left ( 3\,a{e}^{2}+3\,c{d}^{2} \right ) a} \left ( c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}} \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,cdx}{ \left ( 3\,a{e}^{2}+3\,c{d}^{2} \right ){a}^{2}}{\frac{1}{\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}+{\frac{{e}^{3}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}{\frac{1}{\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}+{\frac{d{e}^{2}xc}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}a}{\frac{1}{\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}-{\frac{{e}^{3}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}\ln \left ({ \left ( 2\,{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)/(c*x^2+a)^(5/2),x)

[Out]

1/3*e/(a*e^2+c*d^2)/(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)+1/3*c*d/(a*e^2+c*d^2)/a/(c*(d/e+x)^2
-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)*x+2/3*c*d/(a*e^2+c*d^2)/a^2/(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^
2)/e^2)^(1/2)*x+e^3/(a*e^2+c*d^2)^2/(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)+e^2/(a*e^2+c*d^2)^2*
d/a/(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*x*c-e^3/(a*e^2+c*d^2)^2/((a*e^2+c*d^2)/e^2)^(1/2)*ln
((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(d/e+x)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e
^2)^(1/2))/(d/e+x))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.05014, size = 1690, normalized size = 10.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^2*c^2*e^4*x^4 + 2*a^3*c*e^4*x^2 + a^4*e^4)*sqrt(c*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e
^2 - (2*c^2*d^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2
)) + 2*(a^2*c^2*d^4*e + 5*a^3*c*d^2*e^3 + 4*a^4*e^5 + (2*c^4*d^5 + 7*a*c^3*d^3*e^2 + 5*a^2*c^2*d*e^4)*x^3 + 3*
(a^2*c^2*d^2*e^3 + a^3*c*e^5)*x^2 + 3*(a*c^3*d^5 + 3*a^2*c^2*d^3*e^2 + 2*a^3*c*d*e^4)*x)*sqrt(c*x^2 + a))/(a^4
*c^3*d^6 + 3*a^5*c^2*d^4*e^2 + 3*a^6*c*d^2*e^4 + a^7*e^6 + (a^2*c^5*d^6 + 3*a^3*c^4*d^4*e^2 + 3*a^4*c^3*d^2*e^
4 + a^5*c^2*e^6)*x^4 + 2*(a^3*c^4*d^6 + 3*a^4*c^3*d^4*e^2 + 3*a^5*c^2*d^2*e^4 + a^6*c*e^6)*x^2), -1/3*(3*(a^2*
c^2*e^4*x^4 + 2*a^3*c*e^4*x^2 + a^4*e^4)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c
*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - (a^2*c^2*d^4*e + 5*a^3*c*d^2*e^3 + 4*a^4*e^5 + (2*c
^4*d^5 + 7*a*c^3*d^3*e^2 + 5*a^2*c^2*d*e^4)*x^3 + 3*(a^2*c^2*d^2*e^3 + a^3*c*e^5)*x^2 + 3*(a*c^3*d^5 + 3*a^2*c
^2*d^3*e^2 + 2*a^3*c*d*e^4)*x)*sqrt(c*x^2 + a))/(a^4*c^3*d^6 + 3*a^5*c^2*d^4*e^2 + 3*a^6*c*d^2*e^4 + a^7*e^6 +
(a^2*c^5*d^6 + 3*a^3*c^4*d^4*e^2 + 3*a^4*c^3*d^2*e^4 + a^5*c^2*e^6)*x^4 + 2*(a^3*c^4*d^6 + 3*a^4*c^3*d^4*e^2
+ 3*a^5*c^2*d^2*e^4 + a^6*c*e^6)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + c x^{2}\right )^{\frac{5}{2}} \left (d + e x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x**2+a)**(5/2),x)

[Out]

Integral(1/((a + c*x**2)**(5/2)*(d + e*x)), x)

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Giac [B]  time = 1.35178, size = 1262, normalized size = 8.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+a)^(5/2),x, algorithm="giac")

[Out]

-2*arctan(((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^4/((c^2*d^4 + 2*a*c*d^2*e^2 +
a^2*e^4)*sqrt(-c*d^2 - a*e^2)) + 1/3*((((2*c^10*d^15 + 17*a*c^9*d^13*e^2 + 60*a^2*c^8*d^11*e^4 + 115*a^3*c^7*d
^9*e^6 + 130*a^4*c^6*d^7*e^8 + 87*a^5*c^5*d^5*e^10 + 32*a^6*c^4*d^3*e^12 + 5*a^7*c^3*d*e^14)*x/(a^2*c^9*d^16 +
8*a^3*c^8*d^14*e^2 + 28*a^4*c^7*d^12*e^4 + 56*a^5*c^6*d^10*e^6 + 70*a^6*c^5*d^8*e^8 + 56*a^7*c^4*d^6*e^10 + 2
8*a^8*c^3*d^4*e^12 + 8*a^9*c^2*d^2*e^14 + a^10*c*e^16) + 3*(a^2*c^8*d^12*e^3 + 6*a^3*c^7*d^10*e^5 + 15*a^4*c^6
*d^8*e^7 + 20*a^5*c^5*d^6*e^9 + 15*a^6*c^4*d^4*e^11 + 6*a^7*c^3*d^2*e^13 + a^8*c^2*e^15)/(a^2*c^9*d^16 + 8*a^3
*c^8*d^14*e^2 + 28*a^4*c^7*d^12*e^4 + 56*a^5*c^6*d^10*e^6 + 70*a^6*c^5*d^8*e^8 + 56*a^7*c^4*d^6*e^10 + 28*a^8*
c^3*d^4*e^12 + 8*a^9*c^2*d^2*e^14 + a^10*c*e^16))*x + 3*(a*c^9*d^15 + 8*a^2*c^8*d^13*e^2 + 27*a^3*c^7*d^11*e^4
+ 50*a^4*c^6*d^9*e^6 + 55*a^5*c^5*d^7*e^8 + 36*a^6*c^4*d^5*e^10 + 13*a^7*c^3*d^3*e^12 + 2*a^8*c^2*d*e^14)/(a^
2*c^9*d^16 + 8*a^3*c^8*d^14*e^2 + 28*a^4*c^7*d^12*e^4 + 56*a^5*c^6*d^10*e^6 + 70*a^6*c^5*d^8*e^8 + 56*a^7*c^4*
d^6*e^10 + 28*a^8*c^3*d^4*e^12 + 8*a^9*c^2*d^2*e^14 + a^10*c*e^16))*x + (a^2*c^8*d^14*e + 10*a^3*c^7*d^12*e^3
+ 39*a^4*c^6*d^10*e^5 + 80*a^5*c^5*d^8*e^7 + 95*a^6*c^4*d^6*e^9 + 66*a^7*c^3*d^4*e^11 + 25*a^8*c^2*d^2*e^13 +
4*a^9*c*e^15)/(a^2*c^9*d^16 + 8*a^3*c^8*d^14*e^2 + 28*a^4*c^7*d^12*e^4 + 56*a^5*c^6*d^10*e^6 + 70*a^6*c^5*d^8*
e^8 + 56*a^7*c^4*d^6*e^10 + 28*a^8*c^3*d^4*e^12 + 8*a^9*c^2*d^2*e^14 + a^10*c*e^16))/(c*x^2 + a)^(3/2)