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

Optimal. Leaf size=229 \[ -\frac{2 a^2 \left (A \left (8 n^2+30 n+25\right )+2 B \left (4 n^2+13 n+9\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};1-\sin (e+f x)\right )}{f (2 n+3) (2 n+5) \sqrt{a \sin (e+f x)+a}}-\frac{2 a^2 (A (2 n+5)+2 B (n+3)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) (2 n+5) \sqrt{a \sin (e+f x)+a}}-\frac{2 a B \cos (e+f x) \sqrt{a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5)} \]

[Out]

(-2*a^2*(2*B*(9 + 13*n + 4*n^2) + A*(25 + 30*n + 8*n^2))*Cos[e + f*x]*Hypergeometric2F1[1/2, -n, 3/2, 1 - Sin[
e + f*x]]*(d*Sin[e + f*x])^n)/(f*(3 + 2*n)*(5 + 2*n)*Sin[e + f*x]^n*Sqrt[a + a*Sin[e + f*x]]) - (2*a^2*(2*B*(3
+ n) + A*(5 + 2*n))*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(d*f*(3 + 2*n)*(5 + 2*n)*Sqrt[a + a*Sin[e + f*x]])
- (2*a*B*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n)*Sqrt[a + a*Sin[e + f*x]])/(d*f*(5 + 2*n))

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Rubi [A]  time = 0.493589, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2976, 2981, 2776, 67, 65} \[ -\frac{2 a^2 \left (A \left (8 n^2+30 n+25\right )+2 B \left (4 n^2+13 n+9\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};1-\sin (e+f x)\right )}{f (2 n+3) (2 n+5) \sqrt{a \sin (e+f x)+a}}-\frac{2 a^2 (A (2 n+5)+2 B (n+3)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) (2 n+5) \sqrt{a \sin (e+f x)+a}}-\frac{2 a B \cos (e+f x) \sqrt{a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(2*B*(9 + 13*n + 4*n^2) + A*(25 + 30*n + 8*n^2))*Cos[e + f*x]*Hypergeometric2F1[1/2, -n, 3/2, 1 - Sin[
e + f*x]]*(d*Sin[e + f*x])^n)/(f*(3 + 2*n)*(5 + 2*n)*Sin[e + f*x]^n*Sqrt[a + a*Sin[e + f*x]]) - (2*a^2*(2*B*(3
+ n) + A*(5 + 2*n))*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(d*f*(3 + 2*n)*(5 + 2*n)*Sqrt[a + a*Sin[e + f*x]])
- (2*a*B*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n)*Sqrt[a + a*Sin[e + f*x]])/(d*f*(5 + 2*n))

Rule 2976

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 2981

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

Rule 2776

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

Rule 67

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

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx &=-\frac{2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt{a+a \sin (e+f x)}}{d f (5+2 n)}+\frac{2 \int (d \sin (e+f x))^n \sqrt{a+a \sin (e+f x)} \left (\frac{1}{2} a d \left (2 B (1+n)+2 A \left (\frac{5}{2}+n\right )\right )+\frac{1}{2} a d (2 B (3+n)+A (5+2 n)) \sin (e+f x)\right ) \, dx}{d (5+2 n)}\\ &=-\frac{2 a^2 (2 B (3+n)+A (5+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) \sqrt{a+a \sin (e+f x)}}-\frac{2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt{a+a \sin (e+f x)}}{d f (5+2 n)}+\frac{\left (a \left (2 B \left (9+13 n+4 n^2\right )+A \left (25+30 n+8 n^2\right )\right )\right ) \int (d \sin (e+f x))^n \sqrt{a+a \sin (e+f x)} \, dx}{(3+2 n) (5+2 n)}\\ &=-\frac{2 a^2 (2 B (3+n)+A (5+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) \sqrt{a+a \sin (e+f x)}}-\frac{2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt{a+a \sin (e+f x)}}{d f (5+2 n)}+\frac{\left (a^3 \left (2 B \left (9+13 n+4 n^2\right )+A \left (25+30 n+8 n^2\right )\right ) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(d x)^n}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{f (3+2 n) (5+2 n) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 a^2 (2 B (3+n)+A (5+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) \sqrt{a+a \sin (e+f x)}}-\frac{2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt{a+a \sin (e+f x)}}{d f (5+2 n)}+\frac{\left (a^3 \left (2 B \left (9+13 n+4 n^2\right )+A \left (25+30 n+8 n^2\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n\right ) \operatorname{Subst}\left (\int \frac{x^n}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{f (3+2 n) (5+2 n) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 a^2 \left (2 B \left (9+13 n+4 n^2\right )+A \left (25+30 n+8 n^2\right )\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};1-\sin (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{f (3+2 n) (5+2 n) \sqrt{a+a \sin (e+f x)}}-\frac{2 a^2 (2 B (3+n)+A (5+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) \sqrt{a+a \sin (e+f x)}}-\frac{2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt{a+a \sin (e+f x)}}{d f (5+2 n)}\\ \end{align*}

Mathematica [B]  time = 15.4048, size = 478, normalized size = 2.09 \[ \frac{2^{n+1} \tan \left (\frac{1}{2} (e+f x)\right ) \sec \left (\frac{1}{2} (e+f x)\right ) (a (\sin (e+f x)+1))^{3/2} \sin ^{-n}(e+f x) \left (\frac{\tan \left (\frac{1}{2} (e+f x)\right )}{\tan ^2\left (\frac{1}{2} (e+f x)\right )+1}\right )^n \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )^n (d \sin (e+f x))^n \left (\tan \left (\frac{1}{2} (e+f x)\right ) \left (\frac{(3 A+2 B) \, _2F_1\left (\frac{n+2}{2},n+\frac{7}{2};\frac{n+4}{2};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{n+2}+\tan \left (\frac{1}{2} (e+f x)\right ) \left (\frac{2 (2 A+3 B) \, _2F_1\left (\frac{n+3}{2},n+\frac{7}{2};\frac{n+5}{2};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{n+3}+\tan \left (\frac{1}{2} (e+f x)\right ) \left (\frac{2 (2 A+3 B) \, _2F_1\left (n+\frac{7}{2},\frac{n+4}{2};\frac{n+6}{2};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{n+4}+\tan \left (\frac{1}{2} (e+f x)\right ) \left (\frac{(3 A+2 B) \, _2F_1\left (n+\frac{7}{2},\frac{n+5}{2};\frac{n+7}{2};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{n+5}+\frac{A \tan \left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (n+\frac{7}{2},\frac{n+6}{2};\frac{n+8}{2};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{n+6}\right )\right )\right )\right )+\frac{A \, _2F_1\left (\frac{n+1}{2},n+\frac{7}{2};\frac{n+3}{2};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{n+1}\right )}{f \sqrt{\sec ^2\left (\frac{1}{2} (e+f x)\right )} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(1 + n)*Sec[(e + f*x)/2]*(d*Sin[e + f*x])^n*(a*(1 + Sin[e + f*x]))^(3/2)*Tan[(e + f*x)/2]*(Tan[(e + f*x)/2]
/(1 + Tan[(e + f*x)/2]^2))^n*(1 + Tan[(e + f*x)/2]^2)^n*((A*Hypergeometric2F1[(1 + n)/2, 7/2 + n, (3 + n)/2, -
Tan[(e + f*x)/2]^2])/(1 + n) + Tan[(e + f*x)/2]*(((3*A + 2*B)*Hypergeometric2F1[(2 + n)/2, 7/2 + n, (4 + n)/2,
-Tan[(e + f*x)/2]^2])/(2 + n) + Tan[(e + f*x)/2]*((2*(2*A + 3*B)*Hypergeometric2F1[(3 + n)/2, 7/2 + n, (5 + n
)/2, -Tan[(e + f*x)/2]^2])/(3 + n) + Tan[(e + f*x)/2]*((2*(2*A + 3*B)*Hypergeometric2F1[7/2 + n, (4 + n)/2, (6
+ n)/2, -Tan[(e + f*x)/2]^2])/(4 + n) + Tan[(e + f*x)/2]*(((3*A + 2*B)*Hypergeometric2F1[7/2 + n, (5 + n)/2,
(7 + n)/2, -Tan[(e + f*x)/2]^2])/(5 + n) + (A*Hypergeometric2F1[7/2 + n, (6 + n)/2, (8 + n)/2, -Tan[(e + f*x)/
2]^2]*Tan[(e + f*x)/2])/(6 + n)))))))/(f*Sqrt[Sec[(e + f*x)/2]^2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sin[
e + f*x]^n)

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Maple [F]  time = 0.417, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sin \left ( fx+e \right ) \right ) ^{n} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( A+B\sin \left ( fx+e \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}} \left (d \sin \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (B a \cos \left (f x + e\right )^{2} -{\left (A + B\right )} a \sin \left (f x + e\right ) -{\left (A + B\right )} a\right )} \sqrt{a \sin \left (f x + e\right ) + a} \left (d \sin \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out