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

Optimal. Leaf size=193 \[ -\frac {4 a^3 (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 a^2 (A+B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}-\frac {a (A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.46, antiderivative size = 193, 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 = {2973, 2740, 2737, 2667, 31} \[ -\frac {2 a^2 (A+B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}-\frac {4 a^3 (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a (A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2973

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*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(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)] && NeQ[m + n + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 1.50, size = 177, normalized size = 0.92 \[ -\frac {(a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((36 A+51 B) \sin (e+f x)-3 (A+3 B) \cos (2 (e+f x))+96 A \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-B \sin (3 (e+f x))+96 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )}{12 f \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(5/2)*(-3*(A + 3*B)*Cos[2*(e + f*x)] + 96*
A*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 96*B*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + (36*A + 51*B)*Sin
[e + f*x] - B*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*Sqrt[c - c*Sin[e + f*x]])

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fricas [F]  time = 4.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left ({\left (A + 2 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + B\right )} a^{2} + {\left (B a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{c \sin \left (f x + e\right ) - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(((A + 2*B)*a^2*cos(f*x + e)^2 - 2*(A + B)*a^2 + (B*a^2*cos(f*x + e)^2 - 2*(A + B)*a^2)*sin(f*x + e))*
sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c*sin(f*x + e) - c), 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+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.72, size = 590, normalized size = 3.06 \[ \frac {\left (15 A \sin \left (f x +e \right )-18 A \sin \left (f x +e \right ) \cos \left (f x +e \right )+7 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-48 A \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-15 A -17 B +24 A \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-48 A \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+24 A \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-26 B \sin \left (f x +e \right ) \cos \left (f x +e \right )-48 B \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+24 B \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-48 B \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+24 B \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+19 B \left (\cos ^{2}\left (f x +e \right )\right )+2 B \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+17 B \sin \left (f x +e \right )+3 A \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 A \cos \left (f x +e \right )-2 B \left (\cos ^{4}\left (f x +e \right )\right )-9 B \cos \left (f x +e \right )+3 A \left (\cos ^{3}\left (f x +e \right )\right )+9 B \left (\cos ^{3}\left (f x +e \right )\right )+48 A \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-24 A \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+48 B \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-24 B \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+15 A \left (\cos ^{2}\left (f x +e \right )\right )\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}}{6 f \left (\cos ^{3}\left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-2 \sin \left (f x +e \right ) \cos \left (f x +e \right )-2 \cos \left (f x +e \right )+4 \sin \left (f x +e \right )+4\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x)

[Out]

1/6/f*(2*B*cos(f*x+e)^3*sin(f*x+e)+15*A*sin(f*x+e)-18*A*sin(f*x+e)*cos(f*x+e)-48*A*sin(f*x+e)*ln(-(-1+cos(f*x+
e)+sin(f*x+e))/sin(f*x+e))+3*A*cos(f*x+e)^2*sin(f*x+e)+7*B*cos(f*x+e)^2*sin(f*x+e)-2*B*cos(f*x+e)^4-15*A-17*B+
3*A*cos(f*x+e)^3+24*A*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-48*A*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e
))+24*A*cos(f*x+e)*ln(2/(cos(f*x+e)+1))-26*B*sin(f*x+e)*cos(f*x+e)-48*B*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+
e))/sin(f*x+e))+24*B*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-48*B*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e)
)+24*B*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+19*B*cos(f*x+e)^2+9*B*cos(f*x+e)^3+17*B*sin(f*x+e)-3*A*cos(f*x+e)-9*B*c
os(f*x+e)+15*A*cos(f*x+e)^2+48*A*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-24*A*ln(2/(cos(f*x+e)+1))+48*B*ln(
-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-24*B*ln(2/(cos(f*x+e)+1)))*(a*(1+sin(f*x+e)))^(5/2)/(cos(f*x+e)^3-cos(
f*x+e)^2*sin(f*x+e)-3*cos(f*x+e)^2-2*sin(f*x+e)*cos(f*x+e)-2*cos(f*x+e)+4*sin(f*x+e)+4)/(-c*(sin(f*x+e)-1))^(1
/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2))/(c - c*sin(e + f*x))^(1/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2))/(c - c*sin(e + f*x))^(1/2), 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+a*sin(f*x+e))**(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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