3.1.8 \(\int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx\) [8]

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

[Out]

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

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Rubi [A]
time = 0.34, antiderivative size = 229, normalized size of antiderivative = 1.00, 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 = {3055, 3060, 2855, 69, 67} \begin {gather*} -\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)} \end {gather*}

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 67

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

Rule 69

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 2855

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 3055

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)))
*Sin[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 3060

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)*Sqrt
[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]

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 ) \text {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 ) \text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(478\) vs. \(2(229)=458\).
time = 26.74, size = 478, normalized size = 2.09 \begin {gather*} \frac {2^{1+n} \sec \left (\frac {1}{2} (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n (a (1+\sin (e+f x)))^{3/2} \tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}\right )^n \left (1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^n \left (\frac {A \, _2F_1\left (\frac {1+n}{2},\frac {7}{2}+n;\frac {3+n}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{1+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(3 A+2 B) \, _2F_1\left (\frac {2+n}{2},\frac {7}{2}+n;\frac {4+n}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{2+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {2 (2 A+3 B) \, _2F_1\left (\frac {3+n}{2},\frac {7}{2}+n;\frac {5+n}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{3+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {2 (2 A+3 B) \, _2F_1\left (\frac {7}{2}+n,\frac {4+n}{2};\frac {6+n}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{4+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(3 A+2 B) \, _2F_1\left (\frac {7}{2}+n,\frac {5+n}{2};\frac {7+n}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{5+n}+\frac {A \, _2F_1\left (\frac {7}{2}+n,\frac {6+n}{2};\frac {8+n}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{6+n}\right )\right )\right )\right )\right )}{f \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \end {gather*}

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 = 0.21, size = 0, normalized size = 0.00 \[\int \left (d \sin \left (f x +e \right )\right )^{n} \left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (A +B \sin \left (f x +e \right )\right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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