\(\int \frac {(d+e x)^{7/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx\) [879]

Optimal result

Integrand size = 29, antiderivative size = 160 \[ \int \frac {(d+e x)^{7/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {256 d^3 \sqrt {c d^2-c e^2 x^2}}{35 c e \sqrt {d+e x}}-\frac {64 d^2 \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}{35 c e}-\frac {24 d (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}{35 c e}-\frac {2 (d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}}{7 c e} \]

[Out]

-24/35*d*(e*x+d)^(3/2)*(-c*e^2*x^2+c*d^2)^(1/2)/c/e-2/7*(e*x+d)^(5/2)*(-c*e^2*x^2+c*d^2)^(1/2)/c/e-256/35*d^3*
(-c*e^2*x^2+c*d^2)^(1/2)/c/e/(e*x+d)^(1/2)-64/35*d^2*(e*x+d)^(1/2)*(-c*e^2*x^2+c*d^2)^(1/2)/c/e

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {671, 663} \[ \int \frac {(d+e x)^{7/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {64 d^2 \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}{35 c e}-\frac {24 d (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}{35 c e}-\frac {2 (d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}}{7 c e}-\frac {256 d^3 \sqrt {c d^2-c e^2 x^2}}{35 c e \sqrt {d+e x}} \]

[In]

Int[(d + e*x)^(7/2)/Sqrt[c*d^2 - c*e^2*x^2],x]

[Out]

(-256*d^3*Sqrt[c*d^2 - c*e^2*x^2])/(35*c*e*Sqrt[d + e*x]) - (64*d^2*Sqrt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2])/(35
*c*e) - (24*d*(d + e*x)^(3/2)*Sqrt[c*d^2 - c*e^2*x^2])/(35*c*e) - (2*(d + e*x)^(5/2)*Sqrt[c*d^2 - c*e^2*x^2])/
(7*c*e)

Rule 663

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

Rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[2*c*d*(Simplify[m + p]/(c*(m + 2*p + 1))), Int[(d + e*x)^(m - 1)*(a + c*x^
2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p]
, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}}{7 c e}+\frac {1}{7} (12 d) \int \frac {(d+e x)^{5/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx \\ & = -\frac {24 d (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}{35 c e}-\frac {2 (d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}}{7 c e}+\frac {1}{35} \left (96 d^2\right ) \int \frac {(d+e x)^{3/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx \\ & = -\frac {64 d^2 \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}{35 c e}-\frac {24 d (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}{35 c e}-\frac {2 (d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}}{7 c e}+\frac {1}{35} \left (128 d^3\right ) \int \frac {\sqrt {d+e x}}{\sqrt {c d^2-c e^2 x^2}} \, dx \\ & = -\frac {256 d^3 \sqrt {c d^2-c e^2 x^2}}{35 c e \sqrt {d+e x}}-\frac {64 d^2 \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}{35 c e}-\frac {24 d (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}{35 c e}-\frac {2 (d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}}{7 c e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.42 \[ \int \frac {(d+e x)^{7/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {2 \sqrt {c \left (d^2-e^2 x^2\right )} \left (177 d^3+71 d^2 e x+27 d e^2 x^2+5 e^3 x^3\right )}{35 c e \sqrt {d+e x}} \]

[In]

Integrate[(d + e*x)^(7/2)/Sqrt[c*d^2 - c*e^2*x^2],x]

[Out]

(-2*Sqrt[c*(d^2 - e^2*x^2)]*(177*d^3 + 71*d^2*e*x + 27*d*e^2*x^2 + 5*e^3*x^3))/(35*c*e*Sqrt[d + e*x])

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.39

method	result	size
default	\(-\frac {2 \sqrt {c \left (-x^{2} e^{2}+d^{2}\right )}\, \left (5 e^{3} x^{3}+27 d \,e^{2} x^{2}+71 d^{2} e x +177 d^{3}\right )}{35 \sqrt {e x +d}\, c e}\)	\(62\)
gosper	\(-\frac {2 \left (-e x +d \right ) \left (5 e^{3} x^{3}+27 d \,e^{2} x^{2}+71 d^{2} e x +177 d^{3}\right ) \sqrt {e x +d}}{35 e \sqrt {-c \,x^{2} e^{2}+c \,d^{2}}}\)	\(66\)
risch	\(-\frac {2 \sqrt {-\frac {c \left (x^{2} e^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}\, \left (5 e^{3} x^{3}+27 d \,e^{2} x^{2}+71 d^{2} e x +177 d^{3}\right ) \left (-e x +d \right )}{35 \sqrt {-c \left (x^{2} e^{2}-d^{2}\right )}\, e \sqrt {-c \left (e x -d \right )}}\)	\(104\)

[In]

int((e*x+d)^(7/2)/(-c*e^2*x^2+c*d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/35/(e*x+d)^(1/2)*(c*(-e^2*x^2+d^2))^(1/2)/c*(5*e^3*x^3+27*d*e^2*x^2+71*d^2*e*x+177*d^3)/e

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.43 \[ \int \frac {(d+e x)^{7/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {2 \, {\left (5 \, e^{3} x^{3} + 27 \, d e^{2} x^{2} + 71 \, d^{2} e x + 177 \, d^{3}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{35 \, {\left (c e^{2} x + c d e\right )}} \]

[In]

integrate((e*x+d)^(7/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

-2/35*(5*e^3*x^3 + 27*d*e^2*x^2 + 71*d^2*e*x + 177*d^3)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)/(c*e^2*x + c*d*
e)

Sympy [F]

\[ \int \frac {(d+e x)^{7/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {7}{2}}}{\sqrt {- c \left (- d + e x\right ) \left (d + e x\right )}}\, dx \]

[In]

integrate((e*x+d)**(7/2)/(-c*e**2*x**2+c*d**2)**(1/2),x)

[Out]

Integral((d + e*x)**(7/2)/sqrt(-c*(-d + e*x)*(d + e*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int \frac {(d+e x)^{7/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx=\frac {2 \, {\left (5 \, e^{4} x^{4} + 22 \, d e^{3} x^{3} + 44 \, d^{2} e^{2} x^{2} + 106 \, d^{3} e x - 177 \, d^{4}\right )}}{35 \, \sqrt {-e x + d} \sqrt {c} e} \]

[In]

integrate((e*x+d)^(7/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

2/35*(5*e^4*x^4 + 22*d*e^3*x^3 + 44*d^2*e^2*x^2 + 106*d^3*e*x - 177*d^4)/(sqrt(-e*x + d)*sqrt(c)*e)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x)^{7/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx=\frac {2 \, {\left (\frac {128 \, \sqrt {2} \sqrt {c d} d^{3}}{c} - \frac {280 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} d^{3}}{c} + \frac {140 \, {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c^{2} d^{2} - 42 \, {\left ({\left (e x + d\right )} c - 2 \, c d\right )}^{2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c d - 5 \, {\left ({\left (e x + d\right )} c - 2 \, c d\right )}^{3} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{c^{4}}\right )}}{35 \, e} \]

[In]

integrate((e*x+d)^(7/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="giac")

[Out]

2/35*(128*sqrt(2)*sqrt(c*d)*d^3/c - 280*sqrt(-(e*x + d)*c + 2*c*d)*d^3/c + (140*(-(e*x + d)*c + 2*c*d)^(3/2)*c
^2*d^2 - 42*((e*x + d)*c - 2*c*d)^2*sqrt(-(e*x + d)*c + 2*c*d)*c*d - 5*((e*x + d)*c - 2*c*d)^3*sqrt(-(e*x + d)
*c + 2*c*d))/c^4)/e

Mupad [B] (verification not implemented)

Time = 9.99 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.61 \[ \int \frac {(d+e x)^{7/2}}{\sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {\sqrt {c\,d^2-c\,e^2\,x^2}\,\left (\frac {354\,d^3\,\sqrt {d+e\,x}}{35\,c\,e^2}+\frac {54\,d\,x^2\,\sqrt {d+e\,x}}{35\,c}+\frac {2\,e\,x^3\,\sqrt {d+e\,x}}{7\,c}+\frac {142\,d^2\,x\,\sqrt {d+e\,x}}{35\,c\,e}\right )}{x+\frac {d}{e}} \]

[In]

int((d + e*x)^(7/2)/(c*d^2 - c*e^2*x^2)^(1/2),x)

[Out]

-((c*d^2 - c*e^2*x^2)^(1/2)*((354*d^3*(d + e*x)^(1/2))/(35*c*e^2) + (54*d*x^2*(d + e*x)^(1/2))/(35*c) + (2*e*x
^3*(d + e*x)^(1/2))/(7*c) + (142*d^2*x*(d + e*x)^(1/2))/(35*c*e)))/(x + d/e)