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

Optimal. Leaf size=138 \[ -\frac{(c-d) (c+7 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}+\frac{d (c-5 d) \cos (e+f x)}{2 a f \sqrt{a \sin (e+f x)+a}} \]

[Out]

-((c - d)*(c + 7*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(2*Sqrt[2]*a^(3/2)*f)
+ ((c - 5*d)*d*Cos[e + f*x])/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x]))/(2
*f*(a + a*Sin[e + f*x])^(3/2))

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Rubi [A]  time = 0.215946, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2760, 2751, 2649, 206} \[ -\frac{(c-d) (c+7 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}+\frac{d (c-5 d) \cos (e+f x)}{2 a f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c - d)*(c + 7*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(2*Sqrt[2]*a^(3/2)*f)
+ ((c - 5*d)*d*Cos[e + f*x])/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x]))/(2
*f*(a + a*Sin[e + f*x])^(3/2))

Rule 2760

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

Rule 2751

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

Rule 2649

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.332217, size = 239, normalized size = 1.73 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left ((1+i) (-1)^{3/4} \left (c^2+6 c d-7 d^2\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )+2 (c-d)^2 \sin \left (\frac{1}{2} (e+f x)\right )-(c-d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-4 d^2 \cos \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+4 d^2 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2\right )}{2 f (a (\sin (e+f x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(c - d)^2*Sin[(e + f*x)/2] - (c - d)^2*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2]) + (1 + I)*(-1)^(3/4)*(c^2 + 6*c*d - 7*d^2)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(C
os[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 4*d^2*Cos[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 4*d^2
*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2))/(2*f*(a*(1 + Sin[e + f*x]))^(3/2))

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Maple [B]  time = 0.78, size = 316, normalized size = 2.3 \begin{align*} -{\frac{1}{4\,f\cos \left ( fx+e \right ) } \left ( \sin \left ( fx+e \right ) \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ) a{c}^{2}+6\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) acd-7\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) a{d}^{2}+8\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a}{d}^{2} \right ) +\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ) a{c}^{2}+6\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) acd-7\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) a{d}^{2}+2\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a}{c}^{2}-4\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a}cd+10\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a}{d}^{2} \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }{a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a^(5/2)*(sin(f*x+e)*(2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*c^2+6*2^(1/2)*arctanh(
1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*c*d-7*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2)
)*a*d^2+8*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*d^2)+2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*c^
2+6*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*c*d-7*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(
1/2)*2^(1/2)/a^(1/2))*a*d^2+2*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*c^2-4*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*c*d+10*(a-a*
sin(f*x+e))^(1/2)*a^(1/2)*d^2)*(-a*(-1+sin(f*x+e)))^(1/2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.69076, size = 953, normalized size = 6.91 \begin{align*} -\frac{\sqrt{2}{\left ({\left (c^{2} + 6 \, c d - 7 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c^{2} - 12 \, c d + 14 \, d^{2} -{\left (c^{2} + 6 \, c d - 7 \, d^{2}\right )} \cos \left (f x + e\right ) -{\left (2 \, c^{2} + 12 \, c d - 14 \, d^{2} +{\left (c^{2} + 6 \, c d - 7 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (f x + e\right )^{2} + 2 \, \sqrt{2} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \,{\left (4 \, d^{2} \cos \left (f x + e\right )^{2} + c^{2} - 2 \, c d + d^{2} +{\left (c^{2} - 2 \, c d + 5 \, d^{2}\right )} \cos \left (f x + e\right ) +{\left (4 \, d^{2} \cos \left (f x + e\right ) - c^{2} + 2 \, c d - d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{8 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(sqrt(2)*((c^2 + 6*c*d - 7*d^2)*cos(f*x + e)^2 - 2*c^2 - 12*c*d + 14*d^2 - (c^2 + 6*c*d - 7*d^2)*cos(f*x
+ e) - (2*c^2 + 12*c*d - 14*d^2 + (c^2 + 6*c*d - 7*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x +
e)^2 + 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*co
s(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) -
4*(4*d^2*cos(f*x + e)^2 + c^2 - 2*c*d + d^2 + (c^2 - 2*c*d + 5*d^2)*cos(f*x + e) + (4*d^2*cos(f*x + e) - c^2
+ 2*c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f -
(a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))

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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((c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 3.34741, size = 911, normalized size = 6.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*(d^2*tan(1/2*f*x + 1/2*e)/(a*sgn(tan(1/2*f*x + 1/2*e) + 1)) - d^2/(a*sgn(tan(1/2*f*x + 1/2*e) + 1)))/sq
rt(a*tan(1/2*f*x + 1/2*e)^2 + a) + sqrt(2)*(c^2 + 6*c*d - 7*d^2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*f*x + 1/
2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a) + sqrt(a))/sqrt(-a))/(sqrt(-a)*a*sgn(tan(1/2*f*x + 1/2*e) + 1)) + 2*
(3*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^3*c^2 - 6*(sqrt(a)*tan(1/2*f*x + 1/2*e)
- sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^3*c*d + 3*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2
+ a))^3*d^2 + (sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*sqrt(a)*c^2 - 2*(sqrt(a)*
tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*sqrt(a)*c*d + (sqrt(a)*tan(1/2*f*x + 1/2*e) - sqr
t(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*sqrt(a)*d^2 - (sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2
+ a))*a*c^2 + 2*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*a*c*d - (sqrt(a)*tan(1/2*
f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*a*d^2 + a^(3/2)*c^2 - 2*a^(3/2)*c*d + a^(3/2)*d^2)/(((sqrt(
a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2 + 2*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan
(1/2*f*x + 1/2*e)^2 + a))*sqrt(a) - a)^2*a*sgn(tan(1/2*f*x + 1/2*e) + 1)))/f