3.1277 \(\int \frac{(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{5/2} \sqrt{d \sin (e+f x)}} \, dx\)

Optimal. Leaf size=159 \[ \frac{\left (2 a^2-b^2\right ) \sqrt{\sin (2 e+2 f x)} F\left (\left .\frac{1}{4} (4 e-\pi )+f x\right |2\right )}{3 f g^2 \sqrt{d \sin (e+f x)} \sqrt{g \cos (e+f x)}}+\frac{2 \left (a^2+b^2\right ) \sqrt{d \sin (e+f x)}}{3 d f g (g \cos (e+f x))^{3/2}}+\frac{4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}} \]

[Out]

(2*(a^2 + b^2)*Sqrt[d*Sin[e + f*x]])/(3*d*f*g*(g*Cos[e + f*x])^(3/2)) + (4*a*b*(d*Sin[e + f*x])^(3/2))/(3*d^2*
f*g*(g*Cos[e + f*x])^(3/2)) + ((2*a^2 - b^2)*EllipticF[(4*e - Pi)/4 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]])/(3*f*g^2
*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]])

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Rubi [A]  time = 0.427843, antiderivative size = 161, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.243, Rules used = {2911, 2563, 3202, 457, 329, 237, 335, 275, 232} \[ -\frac{2 \left (2 a^2-b^2\right ) \left (1-\csc ^2(e+f x)\right )^{3/4} (d \sin (e+f x))^{3/2} F\left (\left .\frac{1}{2} \csc ^{-1}(\sin (e+f x))\right |2\right )}{3 d^2 f g (g \cos (e+f x))^{3/2}}+\frac{2 \left (a^2+b^2\right ) \sqrt{d \sin (e+f x)}}{3 d f g (g \cos (e+f x))^{3/2}}+\frac{4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^2/((g*Cos[e + f*x])^(5/2)*Sqrt[d*Sin[e + f*x]]),x]

[Out]

(2*(a^2 + b^2)*Sqrt[d*Sin[e + f*x]])/(3*d*f*g*(g*Cos[e + f*x])^(3/2)) + (4*a*b*(d*Sin[e + f*x])^(3/2))/(3*d^2*
f*g*(g*Cos[e + f*x])^(3/2)) - (2*(2*a^2 - b^2)*(1 - Csc[e + f*x]^2)^(3/4)*EllipticF[ArcCsc[Sin[e + f*x]]/2, 2]
*(d*Sin[e + f*x])^(3/2))/(3*d^2*f*g*(g*Cos[e + f*x])^(3/2))

Rule 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

Rule 2563

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[((a*Sin[e +
f*x])^(m + 1)*(b*Cos[e + f*x])^(n + 1))/(a*b*f*(m + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2,
0] && NeQ[m, -1]

Rule 3202

Int[(cos[(e_.) + (f_.)*(x_)]*(c_.))^(m_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[(ff*c^(2*IntPart[(m - 1)/2] + 1)*
(c*Cos[e + f*x])^(2*FracPart[(m - 1)/2]))/(f*(Cos[e + f*x]^2)^FracPart[(m - 1)/2]), Subst[Int[(d*ff*x)^n*(1 -
ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x
] &&  !IntegerQ[m]

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 237

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[(x^3*(1 + a/(b*x^4))^(3/4))/(a + b*x^4)^(3/4), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rubi steps

\begin{align*} \int \frac{(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{5/2} \sqrt{d \sin (e+f x)}} \, dx &=\frac{(2 a b) \int \frac{\sqrt{d \sin (e+f x)}}{(g \cos (e+f x))^{5/2}} \, dx}{d}+\int \frac{a^2+b^2 \sin ^2(e+f x)}{(g \cos (e+f x))^{5/2} \sqrt{d \sin (e+f x)}} \, dx\\ &=\frac{4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}}+\frac{\cos ^2(e+f x)^{3/4} \operatorname{Subst}\left (\int \frac{a^2+b^2 x^2}{\sqrt{d x} \left (1-x^2\right )^{7/4}} \, dx,x,\sin (e+f x)\right )}{f g (g \cos (e+f x))^{3/2}}\\ &=\frac{2 \left (a^2+b^2\right ) \sqrt{d \sin (e+f x)}}{3 d f g (g \cos (e+f x))^{3/2}}+\frac{4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}}-\frac{\left (\left (-2 a^2+b^2\right ) \cos ^2(e+f x)^{3/4}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d x} \left (1-x^2\right )^{3/4}} \, dx,x,\sin (e+f x)\right )}{3 f g (g \cos (e+f x))^{3/2}}\\ &=\frac{2 \left (a^2+b^2\right ) \sqrt{d \sin (e+f x)}}{3 d f g (g \cos (e+f x))^{3/2}}+\frac{4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}}-\frac{\left (2 \left (-2 a^2+b^2\right ) \cos ^2(e+f x)^{3/4}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{x^4}{d^2}\right )^{3/4}} \, dx,x,\sqrt{d \sin (e+f x)}\right )}{3 d f g (g \cos (e+f x))^{3/2}}\\ &=\frac{2 \left (a^2+b^2\right ) \sqrt{d \sin (e+f x)}}{3 d f g (g \cos (e+f x))^{3/2}}+\frac{4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}}-\frac{\left (2 \left (-2 a^2+b^2\right ) \left (1-\csc ^2(e+f x)\right )^{3/4} (d \sin (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{d^2}{x^4}\right )^{3/4} x^3} \, dx,x,\sqrt{d \sin (e+f x)}\right )}{3 d f g (g \cos (e+f x))^{3/2}}\\ &=\frac{2 \left (a^2+b^2\right ) \sqrt{d \sin (e+f x)}}{3 d f g (g \cos (e+f x))^{3/2}}+\frac{4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}}+\frac{\left (2 \left (-2 a^2+b^2\right ) \left (1-\csc ^2(e+f x)\right )^{3/4} (d \sin (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1-d^2 x^4\right )^{3/4}} \, dx,x,\frac{1}{\sqrt{d \sin (e+f x)}}\right )}{3 d f g (g \cos (e+f x))^{3/2}}\\ &=\frac{2 \left (a^2+b^2\right ) \sqrt{d \sin (e+f x)}}{3 d f g (g \cos (e+f x))^{3/2}}+\frac{4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}}+\frac{\left (\left (-2 a^2+b^2\right ) \left (1-\csc ^2(e+f x)\right )^{3/4} (d \sin (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-d^2 x^2\right )^{3/4}} \, dx,x,\frac{\csc (e+f x)}{d}\right )}{3 d f g (g \cos (e+f x))^{3/2}}\\ &=\frac{2 \left (a^2+b^2\right ) \sqrt{d \sin (e+f x)}}{3 d f g (g \cos (e+f x))^{3/2}}+\frac{4 a b (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}}-\frac{2 \left (2 a^2-b^2\right ) \left (1-\csc ^2(e+f x)\right )^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}(\csc (e+f x))\right |2\right ) (d \sin (e+f x))^{3/2}}{3 d^2 f g (g \cos (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.441046, size = 127, normalized size = 0.8 \[ \frac{2 \tan (e+f x) \left (15 a^2 \cos ^2(e+f x)^{3/4} \, _2F_1\left (\frac{1}{4},\frac{7}{4};\frac{5}{4};\sin ^2(e+f x)\right )+b \sin (e+f x) \left (10 a+3 b \sin (e+f x) \cos ^2(e+f x)^{3/4} \, _2F_1\left (\frac{5}{4},\frac{7}{4};\frac{9}{4};\sin ^2(e+f x)\right )\right )\right )}{15 f g^2 \sqrt{d \sin (e+f x)} \sqrt{g \cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])^2/((g*Cos[e + f*x])^(5/2)*Sqrt[d*Sin[e + f*x]]),x]

[Out]

(2*(15*a^2*(Cos[e + f*x]^2)^(3/4)*Hypergeometric2F1[1/4, 7/4, 5/4, Sin[e + f*x]^2] + b*Sin[e + f*x]*(10*a + 3*
b*(Cos[e + f*x]^2)^(3/4)*Hypergeometric2F1[5/4, 7/4, 9/4, Sin[e + f*x]^2]*Sin[e + f*x]))*Tan[e + f*x])/(15*f*g
^2*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]])

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Maple [B]  time = 0.375, size = 371, normalized size = 2.3 \begin{align*}{\frac{\sqrt{2}\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) }{3\,f \left ( -1+\cos \left ( fx+e \right ) \right ) } \left ( -2\,\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) \sqrt{-{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\it EllipticF} \left ( \sqrt{-{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},1/2\,\sqrt{2} \right ){a}^{2}+\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) \sqrt{-{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\it EllipticF} \left ( \sqrt{-{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{\sqrt{2}}{2}} \right ){b}^{2}+2\,\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) \sqrt{2}ab+\cos \left ( fx+e \right ) \sqrt{2}{a}^{2}+\cos \left ( fx+e \right ) \sqrt{2}{b}^{2}-2\,\sin \left ( fx+e \right ) \sqrt{2}ab-\sqrt{2}{a}^{2}-\sqrt{2}{b}^{2} \right ) \left ( g\cos \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{d\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x)

[Out]

1/3/f*2^(1/2)*(-2*cos(f*x+e)*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x
+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(
1/2),1/2*2^(1/2))*a^2+cos(f*x+e)*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin
(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e
))^(1/2),1/2*2^(1/2))*b^2+2*cos(f*x+e)*sin(f*x+e)*2^(1/2)*a*b+cos(f*x+e)*2^(1/2)*a^2+cos(f*x+e)*2^(1/2)*b^2-2*
sin(f*x+e)*2^(1/2)*a*b-2^(1/2)*a^2-2^(1/2)*b^2)*cos(f*x+e)*sin(f*x+e)/(-1+cos(f*x+e))/(g*cos(f*x+e))^(5/2)/(d*
sin(f*x+e))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}{\left (g \cos \left (f x + e\right )\right )^{\frac{5}{2}} \sqrt{d \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e) + a)^2/((g*cos(f*x + e))^(5/2)*sqrt(d*sin(f*x + e))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)*sqrt(g*cos(f*x + e))*sqrt(d*sin(f*x + e))/(d*g
^3*cos(f*x + e)^3*sin(f*x + e)), 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((a+b*sin(f*x+e))**2/(g*cos(f*x+e))**(5/2)/(d*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}{\left (g \cos \left (f x + e\right )\right )^{\frac{5}{2}} \sqrt{d \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e) + a)^2/((g*cos(f*x + e))^(5/2)*sqrt(d*sin(f*x + e))), x)