3.335 \(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=269 \[ \frac {d (A-B) \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} F_1\left (n+1;\frac {1}{2},2;n+2;\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right )}{f (n+1) (c-d)^2 (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}-\frac {B \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} F_1\left (n+1;\frac {1}{2},1;n+2;\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right )}{a f (n+1) (c-d) (1-\sin (e+f x)) \sqrt {a \sin (e+f x)+a}} \]

[Out]

-B*AppellF1(1+n,1,1/2,2+n,(c+d*sin(f*x+e))/(c-d),(c+d*sin(f*x+e))/(c+d))*cos(f*x+e)*(c+d*sin(f*x+e))^(1+n)*(d*
(1-sin(f*x+e))/(c+d))^(1/2)/a/(c-d)/f/(1+n)/(1-sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)+(A-B)*d*AppellF1(1+n,2,1/2,2
+n,(c+d*sin(f*x+e))/(c-d),(c+d*sin(f*x+e))/(c+d))*cos(f*x+e)*(c+d*sin(f*x+e))^(1+n)*(d*(1-sin(f*x+e))/(c+d))^(
1/2)/(c-d)^2/f/(1+n)/(a-a*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.47, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {2987, 2788, 137, 136} \[ \frac {d (A-B) \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} F_1\left (n+1;\frac {1}{2},2;n+2;\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right )}{f (n+1) (c-d)^2 (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}-\frac {B \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} F_1\left (n+1;\frac {1}{2},1;n+2;\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right )}{a f (n+1) (c-d) (1-\sin (e+f x)) \sqrt {a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*AppellF1[1 + n, 1/2, 1, 2 + n, (c + d*Sin[e + f*x])/(c + d), (c + d*Sin[e + f*x])/(c - d)]*Cos[e + f*x]*S
qrt[(d*(1 - Sin[e + f*x]))/(c + d)]*(c + d*Sin[e + f*x])^(1 + n))/(a*(c - d)*f*(1 + n)*(1 - Sin[e + f*x])*Sqrt
[a + a*Sin[e + f*x]])) + ((A - B)*d*AppellF1[1 + n, 1/2, 2, 2 + n, (c + d*Sin[e + f*x])/(c + d), (c + d*Sin[e
+ f*x])/(c - d)]*Cos[e + f*x]*Sqrt[(d*(1 - Sin[e + f*x]))/(c + d)]*(c + d*Sin[e + f*x])^(1 + n))/((c - d)^2*f*
(1 + n)*(a - a*Sin[e + f*x])*Sqrt[a + a*Sin[e + f*x]])

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 137

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 2788

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

Rule 2987

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b - a*B)/b, Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x
], x] + Dist[B/b, 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] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A*b + a*B, 0]

Rubi steps

\begin {align*} \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx &=(A-B) \int \frac {(c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx+\frac {B \int \frac {(c+d \sin (e+f x))^n}{\sqrt {a+a \sin (e+f x)}} \, dx}{a}\\ &=\frac {\left (a^2 (A-B) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^n}{\sqrt {a-a x} (a+a x)^2} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {(a B \cos (e+f x)) \operatorname {Subst}\left (\int \frac {(c+d x)^n}{\sqrt {a-a x} (a+a x)} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\left (a^2 (A-B) \cos (e+f x) \sqrt {\frac {d (a-a \sin (e+f x))}{a c+a d}}\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^n}{(a+a x)^2 \sqrt {\frac {a d}{a c+a d}-\frac {a d x}{a c+a d}}} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}+\frac {\left (a B \cos (e+f x) \sqrt {\frac {d (a-a \sin (e+f x))}{a c+a d}}\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^n}{(a+a x) \sqrt {\frac {a d}{a c+a d}-\frac {a d x}{a c+a d}}} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=-\frac {B F_1\left (1+n;\frac {1}{2},1;2+n;\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right ) \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{1+n}}{(c-d) f (1+n) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}+\frac {(A-B) d F_1\left (1+n;\frac {1}{2},2;2+n;\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right ) \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{1+n}}{(c-d)^2 f (1+n) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

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Mathematica [B]  time = 17.23, size = 603, normalized size = 2.24 \[ \frac {\sec (e+f x) (c+d \sin (e+f x))^n \left (a A (\sin (e+f x)+1) \left (a \sqrt {2-2 \sin (e+f x)} (\sin (e+f x)+1) \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (1;\frac {1}{2},-n;2;\frac {1}{2} (\sin (e+f x)+1),\frac {d (\sin (e+f x)+1)}{d-c}\right )-\frac {4 \sqrt {\frac {\sin (e+f x)-1}{\sin (e+f x)+1}} \left (\frac {c-d}{d \sin (e+f x)+d}+1\right )^{-n} \left (2 a (2 n+1) F_1\left (\frac {1}{2}-n;-\frac {1}{2},-n;\frac {3}{2}-n;\frac {2}{\sin (e+f x)+1},\frac {d-c}{\sin (e+f x) d+d}\right )+a (2 n-1) (\sin (e+f x)+1) F_1\left (-n-\frac {1}{2};-\frac {1}{2},-n;\frac {1}{2}-n;\frac {2}{\sin (e+f x)+1},\frac {d-c}{\sin (e+f x) d+d}\right )\right )}{4 n^2-1}\right )+a B (\sin (e+f x)+1) \left (a \sqrt {2-2 \sin (e+f x)} (\sin (e+f x)+1) \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (1;\frac {1}{2},-n;2;\frac {1}{2} (\sin (e+f x)+1),\frac {d (\sin (e+f x)+1)}{d-c}\right )-\frac {4 \sqrt {\frac {\sin (e+f x)-1}{\sin (e+f x)+1}} \left (\frac {c-d}{d \sin (e+f x)+d}+1\right )^{-n} \left (a (2 n-1) (\sin (e+f x)+1) F_1\left (-n-\frac {1}{2};-\frac {1}{2},-n;\frac {1}{2}-n;\frac {2}{\sin (e+f x)+1},\frac {d-c}{\sin (e+f x) d+d}\right )-2 a (2 n+1) F_1\left (\frac {1}{2}-n;-\frac {1}{2},-n;\frac {3}{2}-n;\frac {2}{\sin (e+f x)+1},\frac {d-c}{\sin (e+f x) d+d}\right )\right )}{4 n^2-1}\right )\right )}{8 a^3 f \sqrt {a (\sin (e+f x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sec[e + f*x]*(c + d*Sin[e + f*x])^n*(a*B*(1 + Sin[e + f*x])*((a*AppellF1[1, 1/2, -n, 2, (1 + Sin[e + f*x])/2,
(d*(1 + Sin[e + f*x]))/(-c + d)]*Sqrt[2 - 2*Sin[e + f*x]]*(1 + Sin[e + f*x]))/((c + d*Sin[e + f*x])/(c - d))^
n - (4*Sqrt[(-1 + Sin[e + f*x])/(1 + Sin[e + f*x])]*(-2*a*(1 + 2*n)*AppellF1[1/2 - n, -1/2, -n, 3/2 - n, 2/(1
+ Sin[e + f*x]), (-c + d)/(d + d*Sin[e + f*x])] + a*(-1 + 2*n)*AppellF1[-1/2 - n, -1/2, -n, 1/2 - n, 2/(1 + Si
n[e + f*x]), (-c + d)/(d + d*Sin[e + f*x])]*(1 + Sin[e + f*x])))/((-1 + 4*n^2)*(1 + (c - d)/(d + d*Sin[e + f*x
]))^n)) + a*A*(1 + Sin[e + f*x])*((a*AppellF1[1, 1/2, -n, 2, (1 + Sin[e + f*x])/2, (d*(1 + Sin[e + f*x]))/(-c
+ d)]*Sqrt[2 - 2*Sin[e + f*x]]*(1 + Sin[e + f*x]))/((c + d*Sin[e + f*x])/(c - d))^n - (4*Sqrt[(-1 + Sin[e + f*
x])/(1 + Sin[e + f*x])]*(2*a*(1 + 2*n)*AppellF1[1/2 - n, -1/2, -n, 3/2 - n, 2/(1 + Sin[e + f*x]), (-c + d)/(d
+ d*Sin[e + f*x])] + a*(-1 + 2*n)*AppellF1[-1/2 - n, -1/2, -n, 1/2 - n, 2/(1 + Sin[e + f*x]), (-c + d)/(d + d*
Sin[e + f*x])]*(1 + Sin[e + f*x])))/((-1 + 4*n^2)*(1 + (c - d)/(d + d*Sin[e + f*x]))^n))))/(8*a^3*f*Sqrt[a*(1
+ Sin[e + f*x])])

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.98, size = 0, normalized size = 0.00 \[ \int \frac {\left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{n}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))**n/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Integral((A + B*sin(e + f*x))*(c + d*sin(e + f*x))**n/(a*(sin(e + f*x) + 1))**(3/2), x)

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