3.181 \(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=210 \[ -\frac {4 c^3 (A-2 B) \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 c^2 (A-2 B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a \sin (e+f x)+a}}-\frac {c (A-2 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a \sin (e+f x)+a}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

-1/2*(A-B)*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f/(a+a*sin(f*x+e))^(3/2)-1/2*(A-2*B)*c*cos(f*x+e)*(c-c*sin(f*x+e)
)^(3/2)/a/f/(a+a*sin(f*x+e))^(1/2)-4*(A-2*B)*c^3*cos(f*x+e)*ln(1+sin(f*x+e))/a/f/(a+a*sin(f*x+e))^(1/2)/(c-c*s
in(f*x+e))^(1/2)-2*(A-2*B)*c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.48, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2972, 2740, 2737, 2667, 31} \[ -\frac {2 c^2 (A-2 B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a \sin (e+f x)+a}}-\frac {4 c^3 (A-2 B) \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {c (A-2 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a \sin (e+f x)+a}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(-4*(A - 2*B)*c^3*Cos[e + f*x]*Log[1 + Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])
- (2*(A - 2*B)*c^2*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]) - ((A - 2*B)*c*Cos[e
+ f*x]*(c - c*Sin[e + f*x])^(3/2))/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x
])^(5/2))/(2*f*(a + a*Sin[e + f*x])^(3/2))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2737

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

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 2972

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

Rubi steps

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

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Mathematica [A]  time = 1.60, size = 212, normalized size = 1.01 \[ -\frac {c^2 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (2 A-7 B) \cos (2 (e+f x))+\sin (e+f x) \left (64 (A-2 B) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )-8 A+31 B\right )+64 A \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )+28 A+B \sin (3 (e+f x))-128 B \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )-16 B\right )}{8 f (a (\sin (e+f x)+1))^{3/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

-1/8*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(28*A - 16*B + 2*(2*A - 7*B)*Cos[2*(e
+ f*x)] + 64*A*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] - 128*B*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (-
8*A + 31*B + 64*(A - 2*B)*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Sin[e + f*x] + B*Sin[3*(e + f*x)]))/(f*(Co
s[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(3/2))

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fricas [F]  time = 8.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left ({\left (A - 2 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A - B\right )} c^{2} + {\left (B c^{2} \cos \left (f x + e\right )^{2} + 2 \, {\left (A - B\right )} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(((A - 2*B)*c^2*cos(f*x + e)^2 - 2*(A - B)*c^2 + (B*c^2*cos(f*x + e)^2 + 2*(A - B)*c^2)*sin(f*x + e))*
sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)

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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+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.63, size = 853, normalized size = 4.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(3/2),x)

[Out]

-1/2/f*(B*cos(f*x+e)^3*sin(f*x+e)-32*A*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+64*B*ln((1-cos(f*x+e)+sin(f*x+
e))/sin(f*x+e))+12*A*sin(f*x+e)-10*A*sin(f*x+e)*cos(f*x+e)+16*B*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))-2*A*cos(f*x+
e)^2*sin(f*x+e)+6*B*cos(f*x+e)^2*sin(f*x+e)+B*cos(f*x+e)^4+12*A-22*B+2*A*cos(f*x+e)^3-8*A*cos(f*x+e)^2*ln(2/(c
os(f*x+e)+1))+16*A*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-8*A*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+15*B*sin(f*x+e)*cos(f*x
+e)-32*B*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+16*B*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+16*A*cos(f*x+e)^2*ln((1-cos(f*x+
e)+sin(f*x+e))/sin(f*x+e))-32*B*cos(f*x+e)^2*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-8*A*cos(f*x+e)*sin(f*x+e
)*ln(2/(cos(f*x+e)+1))+16*B*ln(2/(cos(f*x+e)+1))*sin(f*x+e)*cos(f*x+e)+16*A*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*
x+e))/sin(f*x+e))-32*B*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-32*A*sin(f*x+e)*ln((1-cos(f*x+e)+si
n(f*x+e))/sin(f*x+e))+64*B*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+21*B*cos(f*x+e)^2-7*B*cos(f*x+e
)^3-22*B*sin(f*x+e)+16*A*sin(f*x+e)*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-32*B*sin(f*x+e)*cos(f*
x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-2*A*cos(f*x+e)+7*B*cos(f*x+e)-12*A*cos(f*x+e)^2+16*A*ln(2/(cos(f
*x+e)+1))-32*B*ln(2/(cos(f*x+e)+1)))*(-c*(sin(f*x+e)-1))^(5/2)/(cos(f*x+e)^2*sin(f*x+e)+cos(f*x+e)^3+2*sin(f*x
+e)*cos(f*x+e)-3*cos(f*x+e)^2-4*sin(f*x+e)-2*cos(f*x+e)+4)/(a*(1+sin(f*x+e)))^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(-c*sin(f*x + e) + c)^(5/2)/(a*sin(f*x + e) + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2))/(a + a*sin(e + f*x))^(3/2),x)

[Out]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2))/(a + a*sin(e + f*x))^(3/2), 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*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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