3.57 $$\int \frac{1}{(a+b \sin (c+d x))^{7/2}} \, dx$$

Optimal. Leaf size=292 $\frac{2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 d \left (a^2-b^2\right )^3 \sqrt{a+b \sin (c+d x)}}+\frac{16 a b \cos (c+d x)}{15 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}+\frac{2 b \cos (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}-\frac{16 a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (23 a^2+9 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}$

[Out]

(2*b*Cos[c + d*x])/(5*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^(5/2)) + (16*a*b*Cos[c + d*x])/(15*(a^2 - b^2)^2*d*(a
+ b*Sin[c + d*x])^(3/2)) + (2*b*(23*a^2 + 9*b^2)*Cos[c + d*x])/(15*(a^2 - b^2)^3*d*Sqrt[a + b*Sin[c + d*x]])
+ (2*(23*a^2 + 9*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(15*(a^2 - b^2)^3
*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (16*a*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c
+ d*x])/(a + b)])/(15*(a^2 - b^2)^2*d*Sqrt[a + b*Sin[c + d*x]])

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Rubi [A]  time = 0.349682, antiderivative size = 292, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {2664, 2754, 2752, 2663, 2661, 2655, 2653} $\frac{2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 d \left (a^2-b^2\right )^3 \sqrt{a+b \sin (c+d x)}}+\frac{16 a b \cos (c+d x)}{15 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}+\frac{2 b \cos (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}-\frac{16 a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (23 a^2+9 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*Sin[c + d*x])^(-7/2),x]

[Out]

(2*b*Cos[c + d*x])/(5*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^(5/2)) + (16*a*b*Cos[c + d*x])/(15*(a^2 - b^2)^2*d*(a
+ b*Sin[c + d*x])^(3/2)) + (2*b*(23*a^2 + 9*b^2)*Cos[c + d*x])/(15*(a^2 - b^2)^3*d*Sqrt[a + b*Sin[c + d*x]])
+ (2*(23*a^2 + 9*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(15*(a^2 - b^2)^3
*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (16*a*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c
+ d*x])/(a + b)])/(15*(a^2 - b^2)^2*d*Sqrt[a + b*Sin[c + d*x]])

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 2754

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

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

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

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+b \sin (c+d x))^{7/2}} \, dx &=\frac{2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}-\frac{2 \int \frac{-\frac{5 a}{2}+\frac{3}{2} b \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx}{5 \left (a^2-b^2\right )}\\ &=\frac{2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac{16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac{4 \int \frac{\frac{3}{4} \left (5 a^2+3 b^2\right )-2 a b \sin (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx}{15 \left (a^2-b^2\right )^2}\\ &=\frac{2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac{16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac{2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt{a+b \sin (c+d x)}}-\frac{8 \int \frac{-\frac{1}{8} a \left (15 a^2+17 b^2\right )-\frac{1}{8} b \left (23 a^2+9 b^2\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=\frac{2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac{16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac{2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt{a+b \sin (c+d x)}}-\frac{(8 a) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^2}+\frac{\left (23 a^2+9 b^2\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=\frac{2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac{16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac{2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt{a+b \sin (c+d x)}}+\frac{\left (\left (23 a^2+9 b^2\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{15 \left (a^2-b^2\right )^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\left (8 a \sqrt{\frac{a+b \sin (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{15 \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}\\ &=\frac{2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac{16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac{2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (23 a^2+9 b^2\right ) E\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (c+d x)}}{15 \left (a^2-b^2\right )^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{16 a F\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}{15 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.29142, size = 198, normalized size = 0.68 $\frac{2 \left (\frac{b \cos (c+d x) \left (b^2 \left (23 a^2+9 b^2\right ) \sin ^2(c+d x)+2 a b \left (27 a^2+5 b^2\right ) \sin (c+d x)-5 a^2 b^2+34 a^4+3 b^4\right )}{\left (a^2-b^2\right )^3}-\frac{\left (\frac{a+b \sin (c+d x)}{a+b}\right )^{5/2} \left (\left (23 a^2+9 b^2\right ) E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+8 a (b-a) F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )\right )}{(a-b)^3}\right )}{15 d (a+b \sin (c+d x))^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*Sin[c + d*x])^(-7/2),x]

[Out]

(2*(-((((23*a^2 + 9*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] + 8*a*(-a + b)*EllipticF[(-2*c + Pi -
2*d*x)/4, (2*b)/(a + b)])*((a + b*Sin[c + d*x])/(a + b))^(5/2))/(a - b)^3) + (b*Cos[c + d*x]*(34*a^4 - 5*a^2*
b^2 + 3*b^4 + 2*a*b*(27*a^2 + 5*b^2)*Sin[c + d*x] + b^2*(23*a^2 + 9*b^2)*Sin[c + d*x]^2))/(a^2 - b^2)^3))/(15*
d*(a + b*Sin[c + d*x])^(5/2))

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Maple [A]  time = 0.412, size = 584, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(d*x+c))^(7/2),x)

[Out]

(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*(2/5/(a^2-b^2)/b^2*(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)/(sin(d*x+c)
+a/b)^3+16/15*a/(a^2-b^2)^2/b*(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)/(sin(d*x+c)+a/b)^2+2/15*b*cos(d*x+c)^2/(
a^2-b^2)^3*(23*a^2+9*b^2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)+2*(15*a^3+17*a*b^2)/(15*a^6-45*a^4*b^2+45*a^
2*b^4-15*b^6)*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-1-sin(d*x+c))*b/(a-b))^
(1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+2/
15*b*(23*a^2+9*b^2)/(a^2-b^2)^3*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-1-sin
(d*x+c))*b/(a-b))^(1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*((-a/b-1)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(
1/2),((a-b)/(a+b))^(1/2))+EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))))/cos(d*x+c)/(a+b*sin(
d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^(-7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{b^{4} \cos \left (d x + c\right )^{4} + a^{4} + 6 \, a^{2} b^{2} + b^{4} - 2 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left (a b^{3} \cos \left (d x + c\right )^{2} - a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sin(d*x + c) + a)/(b^4*cos(d*x + c)^4 + a^4 + 6*a^2*b^2 + b^4 - 2*(3*a^2*b^2 + b^4)*cos(d*x +
c)^2 - 4*(a*b^3*cos(d*x + c)^2 - a^3*b - a*b^3)*sin(d*x + c)), 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(1/(a+b*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^(-7/2), x)