3.179 \(\int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=134 \[ \frac {2 a^2 (6 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (18 A+25 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]

[Out]

2/15*a^2*(6*A+5*B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+2/15*a^2*(18*A+25*B)*sin(d*x+c)/d/cos(
d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)+2/5*a*A*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)

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Rubi [A]  time = 0.34, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2975, 2980, 2771} \[ \frac {2 a^2 (6 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (18 A+25 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(7/2),x]

[Out]

(2*a^2*(6*A + 5*B)*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(18*A + 25*B)*Sin
[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) + (2*a*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/
(5*d*Cos[c + d*x]^(5/2))

Rule 2771

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2975

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

Rule 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{2} a (6 A+5 B)+\frac {1}{2} a (2 A+5 B) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (6 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{15} (a (18 A+25 B)) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (6 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (18 A+25 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 80, normalized size = 0.60 \[ \frac {a \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} (2 (9 A+5 B) \cos (c+d x)+(18 A+25 B) \cos (2 (c+d x))+24 A+25 B)}{15 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(7/2),x]

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(24*A + 25*B + 2*(9*A + 5*B)*Cos[c + d*x] + (18*A + 25*B)*Cos[2*(c + d*x)])*Tan[
(c + d*x)/2])/(15*d*Cos[c + d*x]^(5/2))

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fricas [A]  time = 1.35, size = 88, normalized size = 0.66 \[ \frac {2 \, {\left ({\left (18 \, A + 25 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (9 \, A + 5 \, B\right )} a \cos \left (d x + c\right ) + 3 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/15*((18*A + 25*B)*a*cos(d*x + c)^2 + (9*A + 5*B)*a*cos(d*x + c) + 3*A*a)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d
*x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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giac [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+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.23, size = 87, normalized size = 0.65 \[ -\frac {2 a \left (-1+\cos \left (d x +c \right )\right ) \left (18 A \left (\cos ^{2}\left (d x +c \right )\right )+25 B \left (\cos ^{2}\left (d x +c \right )\right )+9 A \cos \left (d x +c \right )+5 B \cos \left (d x +c \right )+3 A \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{15 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x)

[Out]

-2/15/d*a*(-1+cos(d*x+c))*(18*A*cos(d*x+c)^2+25*B*cos(d*x+c)^2+9*A*cos(d*x+c)+5*B*cos(d*x+c)+3*A)*(a*(1+cos(d*
x+c)))^(1/2)/sin(d*x+c)/cos(d*x+c)^(5/2)

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maxima [B]  time = 0.87, size = 344, normalized size = 2.57 \[ \frac {4 \, {\left (\frac {5 \, {\left (\frac {3 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )} B}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}}} + \frac {3 \, {\left (\frac {5 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {7 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {2 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )} A {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}}\right )}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

4/15*(5*(3*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) +
1)^3 + 2*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)*B/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(
-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)) + 3*(5*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sqrt(
2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 7*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 2*sqr
t(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)*A*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((sin(d*x + c)
/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(2*sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)))/d

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mupad [B]  time = 3.09, size = 195, normalized size = 1.46 \[ \frac {2\,a\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\left (48\,A\,\sin \left (c+d\,x\right )+50\,B\,\sin \left (c+d\,x\right )+36\,A\,\sin \left (2\,c+2\,d\,x\right )+66\,A\,\sin \left (3\,c+3\,d\,x\right )+18\,A\,\sin \left (4\,c+4\,d\,x\right )+18\,A\,\sin \left (5\,c+5\,d\,x\right )+20\,B\,\sin \left (2\,c+2\,d\,x\right )+75\,B\,\sin \left (3\,c+3\,d\,x\right )+10\,B\,\sin \left (4\,c+4\,d\,x\right )+25\,B\,\sin \left (5\,c+5\,d\,x\right )\right )}{15\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (10\,\cos \left (c+d\,x\right )+8\,\cos \left (2\,c+2\,d\,x\right )+5\,\cos \left (3\,c+3\,d\,x\right )+2\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (5\,c+5\,d\,x\right )+6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(3/2))/cos(c + d*x)^(7/2),x)

[Out]

(2*a*(a*(cos(c + d*x) + 1))^(1/2)*(48*A*sin(c + d*x) + 50*B*sin(c + d*x) + 36*A*sin(2*c + 2*d*x) + 66*A*sin(3*
c + 3*d*x) + 18*A*sin(4*c + 4*d*x) + 18*A*sin(5*c + 5*d*x) + 20*B*sin(2*c + 2*d*x) + 75*B*sin(3*c + 3*d*x) + 1
0*B*sin(4*c + 4*d*x) + 25*B*sin(5*c + 5*d*x)))/(15*d*cos(c + d*x)^(1/2)*(10*cos(c + d*x) + 8*cos(2*c + 2*d*x)
+ 5*cos(3*c + 3*d*x) + 2*cos(4*c + 4*d*x) + cos(5*c + 5*d*x) + 6))

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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+a*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(7/2),x)

[Out]

Timed out

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