3.393 \(\int \frac {a B+b B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=80 \[ -\frac {2 b B \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a d (a+b)}-\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {2 B \sin (c+d x)}{a d \sqrt {\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {21, 2802, 3059, 2639, 12, 2805} \[ -\frac {2 b B \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a d (a+b)}-\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {2 B \sin (c+d x)}{a d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 2805

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

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

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

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Mathematica [B]  time = 2.68, size = 196, normalized size = 2.45 \[ -\frac {B \left (\frac {2 \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt {\sin ^2(c+d x)}}+\frac {6 b \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {2 a \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{b}-\frac {4 \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(B*((6*b*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (2*a*(2*EllipticF[(c + d*x)/2, 2] - (2*a*El
lipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)))/b - (4*Sin[c + d*x])/Sqrt[Cos[c + d*x]] + (2*(-2*a*b*Ellipt
icE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*E
llipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2])))/(a*d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B b \cos \left (d x + c\right ) + B a}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*b*cos(d*x + c) + B*a)/((b*cos(d*x + c) + a)^2*cos(d*x + c)^(3/2)), x)

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maple [B]  time = 1.48, size = 355, normalized size = 4.44 \[ -\frac {2 B \left (-2 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (a -b \right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a -\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b \right )}{a \left (a -b \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*B+b*B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^2,x)

[Out]

-2*B*(-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(a-b)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-b*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)
)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a-(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*b)/a/(a-b)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1
/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B b \cos \left (d x + c\right ) + B a}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*b*cos(d*x + c) + B*a)/((b*cos(d*x + c) + a)^2*cos(d*x + c)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {B\,a+B\,b\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*a + B*b*cos(c + d*x))/(cos(c + d*x)^(3/2)*(a + b*cos(c + d*x))^2),x)

[Out]

int((B*a + B*b*cos(c + d*x))/(cos(c + d*x)^(3/2)*(a + b*cos(c + d*x))^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*cos(d*x+c))/cos(d*x+c)**(3/2)/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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