3.6.33 \(\int \frac {(a+a \sin (e+f x))^{3/2}}{c+d \sin (e+f x)} \, dx\) [533]

Optimal. Leaf size=98 \[ \frac {2 a^{3/2} (c-d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{d^{3/2} \sqrt {c+d} f}-\frac {2 a^2 \cos (e+f x)}{d f \sqrt {a+a \sin (e+f x)}} \]

[Out]

2*a^(3/2)*(c-d)*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/d^(3/2)/f/(c+d)^(1/2)-2
*a^2*cos(f*x+e)/d/f/(a+a*sin(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 = {2842, 21, 2852, 214} \begin {gather*} \frac {2 a^{3/2} (c-d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{3/2} f \sqrt {c+d}}-\frac {2 a^2 \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 214

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

Rule 2842

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Dist[1/(
d*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d
*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], 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] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m,
2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2852

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(233\) vs. \(2(98)=196\).
time = 1.64, size = 233, normalized size = 2.38 \begin {gather*} \frac {\left (-2 \sqrt {d} \sqrt {c+d} \cos \left (\frac {1}{2} (e+f x)\right )+(c-d) \left (\log \left (-\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (c+d+\sqrt {d} \sqrt {c+d} \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {d} \sqrt {c+d} \sin \left (\frac {1}{2} (e+f x)\right )\right )\right )-\log \left ((c+d) \sec ^2\left (\frac {1}{4} (e+f x)\right )+\sqrt {d} \sqrt {c+d} \left (-1+2 \tan \left (\frac {1}{4} (e+f x)\right )+\tan ^2\left (\frac {1}{4} (e+f x)\right )\right )\right )\right )+2 \sqrt {d} \sqrt {c+d} \sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{3/2}}{d^{3/2} \sqrt {c+d} f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[d]*Sqrt[c + d]*Cos[(e + f*x)/2] + (c - d)*(Log[-(Sec[(e + f*x)/4]^2*(c + d + Sqrt[d]*Sqrt[c + d]*Cos
[(e + f*x)/2] - Sqrt[d]*Sqrt[c + d]*Sin[(e + f*x)/2]))] - Log[(c + d)*Sec[(e + f*x)/4]^2 + Sqrt[d]*Sqrt[c + d]
*(-1 + 2*Tan[(e + f*x)/4] + Tan[(e + f*x)/4]^2)]) + 2*Sqrt[d]*Sqrt[c + d]*Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*
x]))^(3/2))/(d^(3/2)*Sqrt[c + d]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

________________________________________________________________________________________

Maple [A]
time = 4.38, size = 137, normalized size = 1.40

method	result	size
default	\(-\frac {2 a \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (-\arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a c +a \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) d +\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\right )}{d \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(137\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(-arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a*c+a*a
rctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*d+(-a*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2))/d/(a*(c+d
)*d)^(1/2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (86) = 172\).
time = 0.44, size = 685, normalized size = 6.99 \begin {gather*} \left [-\frac {{\left (a c - a d + {\left (a c - a d\right )} \cos \left (f x + e\right ) + {\left (a c - a d\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {a}{c d + d^{2}}} \log \left (\frac {a d^{2} \cos \left (f x + e\right )^{3} - a c^{2} - 2 \, a c d - a d^{2} - {\left (6 \, a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left (c^{2} d + 4 \, c d^{2} + 3 \, d^{3} - {\left (c d^{2} + d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (c^{2} d + 3 \, c d^{2} + 2 \, d^{3}\right )} \cos \left (f x + e\right ) - {\left (c^{2} d + 4 \, c d^{2} + 3 \, d^{3} + {\left (c d^{2} + d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {\frac {a}{c d + d^{2}}} - {\left (a c^{2} + 8 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right ) + {\left (a d^{2} \cos \left (f x + e\right )^{2} - a c^{2} - 2 \, a c d - a d^{2} + 2 \, {\left (3 \, a c d + 4 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{d^{2} \cos \left (f x + e\right )^{3} + {\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )^{2} - c^{2} - 2 \, c d - d^{2} - {\left (c^{2} + d^{2}\right )} \cos \left (f x + e\right ) + {\left (d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \cos \left (f x + e\right ) - c^{2} - 2 \, c d - d^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (a \cos \left (f x + e\right ) - a \sin \left (f x + e\right ) + a\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{2 \, {\left (d f \cos \left (f x + e\right ) + d f \sin \left (f x + e\right ) + d f\right )}}, \frac {{\left (a c - a d + {\left (a c - a d\right )} \cos \left (f x + e\right ) + {\left (a c - a d\right )} \sin \left (f x + e\right )\right )} \sqrt {-\frac {a}{c d + d^{2}}} \arctan \left (\frac {\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) - c - 2 \, d\right )} \sqrt {-\frac {a}{c d + d^{2}}}}{2 \, a \cos \left (f x + e\right )}\right ) - 2 \, {\left (a \cos \left (f x + e\right ) - a \sin \left (f x + e\right ) + a\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{d f \cos \left (f x + e\right ) + d f \sin \left (f x + e\right ) + d f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((a*c - a*d + (a*c - a*d)*cos(f*x + e) + (a*c - a*d)*sin(f*x + e))*sqrt(a/(c*d + d^2))*log((a*d^2*cos(f*
x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2
+ d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d + 4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(
f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2)) - (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e
) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*cos(f*x + e))*sin(f*x + e))/(d^2*c
os(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e
)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(a*cos(f*x + e) - a*sin(f*x + e) + a)*sqrt(a*
sin(f*x + e) + a))/(d*f*cos(f*x + e) + d*f*sin(f*x + e) + d*f), ((a*c - a*d + (a*c - a*d)*cos(f*x + e) + (a*c
- a*d)*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(
-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*(a*cos(f*x + e) - a*sin(f*x + e) + a)*sqrt(a*sin(f*x + e) + a))/(d*f*cos
(f*x + e) + d*f*sin(f*x + e) + d*f)]

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 0.50, size = 130, normalized size = 1.33 \begin {gather*} \frac {\sqrt {2} {\left (\frac {2 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{d} + \frac {\sqrt {2} {\left (a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{\sqrt {-c d - d^{2}} d}\right )} \sqrt {a}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*(2*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)/d + sqrt(2)*(a*c*sgn(cos(-1/4*
pi + 1/2*f*x + 1/2*e)) - a*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2
*e)/sqrt(-c*d - d^2))/(sqrt(-c*d - d^2)*d))*sqrt(a)/f

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{c+d\,\sin \left (e+f\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________