3.4.43 \(\int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\) [343]

Optimal. Leaf size=108 \[ \frac {2 B \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 b B \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \]

[Out]

-2*b*B*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)+2*B*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/(a^2-b^2)/d/((a+b*cos(d*x+c))/(a+b))^(
1/2)

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Rubi [A]
time = 0.05, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {21, 2743, 2734, 2732} \begin {gather*} \frac {2 B \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 b B \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*B + b*B*Cos[c + d*x])/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(2*B*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/((a^2 - b^2)*d*Sqrt[(a + b*Cos[c + d*x])/
(a + b)]) - (2*b*B*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2732

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

Rule 2734

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

Rule 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

Rubi steps

\begin {align*} \int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx &=B \int \frac {1}{(a+b \cos (c+d x))^{3/2}} \, dx\\ &=-\frac {2 b B \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {(2 B) \int \frac {-\frac {a}{2}-\frac {1}{2} b \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{a^2-b^2}\\ &=-\frac {2 b B \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {B \int \sqrt {a+b \cos (c+d x)} \, dx}{a^2-b^2}\\ &=-\frac {2 b B \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {\left (B \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{\left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\\ &=\frac {2 B \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 b B \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 84, normalized size = 0.78 \begin {gather*} \frac {B \left (2 (a+b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-2 b \sin (c+d x)\right )}{(a-b) (a+b) d \sqrt {a+b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*B + b*B*Cos[c + d*x])/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(B*(2*(a + b)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - 2*b*Sin[c + d*x]))/((
a - b)*(a + b)*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [A]
time = 0.32, size = 218, normalized size = 2.02

method	result	size
default	\(-\frac {2 B \left (2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a -\sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) b \right )}{\left (a -b \right ) \left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b}\, d}\)	\(218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2*B*(2*b*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)
/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b)/(a-b)/(a+b)/sin
(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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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((a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*b*cos(d*x + c) + B*a)/(b*cos(d*x + c) + a)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 491, normalized size = 4.55 \begin {gather*} -\frac {6 \, \sqrt {b \cos \left (d x + c\right ) + a} B b^{2} \sin \left (d x + c\right ) + {\left (i \, \sqrt {2} B a b \cos \left (d x + c\right ) + i \, \sqrt {2} B a^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) + {\left (-i \, \sqrt {2} B a b \cos \left (d x + c\right ) - i \, \sqrt {2} B a^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) - 3 \, {\left (i \, \sqrt {2} B b^{2} \cos \left (d x + c\right ) + i \, \sqrt {2} B a b\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) - 3 \, {\left (-i \, \sqrt {2} B b^{2} \cos \left (d x + c\right ) - i \, \sqrt {2} B a b\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right )}{3 \, {\left ({\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b - a b^{3}\right )} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/3*(6*sqrt(b*cos(d*x + c) + a)*B*b^2*sin(d*x + c) + (I*sqrt(2)*B*a*b*cos(d*x + c) + I*sqrt(2)*B*a^2)*sqrt(b)
*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d
*x + c) + 2*a)/b) + (-I*sqrt(2)*B*a*b*cos(d*x + c) - I*sqrt(2)*B*a^2)*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 -
3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b) - 3*(I*sqrt(2)*
B*b^2*cos(d*x + c) + I*sqrt(2)*B*a*b)*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)
/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*
sin(d*x + c) + 2*a)/b)) - 3*(-I*sqrt(2)*B*b^2*cos(d*x + c) - I*sqrt(2)*B*a*b)*sqrt(b)*weierstrassZeta(4/3*(4*a
^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*
b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b)))/((a^2*b^2 - b^4)*d*cos(d*x + c) + (a^3*b - a*
b^3)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))**(5/2),x)

[Out]

B*Integral(1/(a*sqrt(a + b*cos(c + d*x)) + b*sqrt(a + b*cos(c + d*x))*cos(c + d*x)), 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((a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*b*cos(d*x + c) + B*a)/(b*cos(d*x + c) + a)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {B\,a+B\,b\,\cos \left (c+d\,x\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*a + B*b*cos(c + d*x))/(a + b*cos(c + d*x))^(5/2),x)

[Out]

int((B*a + B*b*cos(c + d*x))/(a + b*cos(c + d*x))^(5/2), x)

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