\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx\) [324]

Optimal result

Integrand size = 25, antiderivative size = 126 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {(3 A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {(3 A+5 B) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}} \]

[Out]

-1/4*(A-B)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(5/2)-1/16*(3*A+5*B)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)-1/32*(3*A+
5*B)*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(5/2)/f*2^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2829, 2729, 2728, 212} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {(3 A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(3 A+5 B) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}}-\frac {(A-B) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}} \]

[In]

Int[(A + B*Sin[e + f*x])/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

-1/16*((3*A + 5*B)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[2]*a^(5/2)*f) - (
(A - B)*Cos[e + f*x])/(4*f*(a + a*Sin[e + f*x])^(5/2)) - ((3*A + 5*B)*Cos[e + f*x])/(16*a*f*(a + a*Sin[e + f*x
])^(3/2))

Rule 212

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

Rule 2728

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 2729

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

Rule 2829

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/(a*f*(2*m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {(3 A+5 B) \int \frac {1}{(a+a \sin (e+f x))^{3/2}} \, dx}{8 a} \\ & = -\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {(3 A+5 B) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {(3 A+5 B) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{32 a^2} \\ & = -\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {(3 A+5 B) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(3 A+5 B) \text {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 )}{16 a^2 f} \\ & = -\frac {(3 A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {(3 A+5 B) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.44 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.80 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (8 (A-B) \sin \left (\frac {1}{2} (e+f x)\right )+4 (-A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (3 A+5 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-(3 A+5 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+(1+i) (-1)^{3/4} (3 A+5 B) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{16 f (a (1+\sin (e+f x)))^{5/2}} \]

[In]

Integrate[(A + B*Sin[e + f*x])/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(A - B)*Sin[(e + f*x)/2] + 4*(-A + B)*(Cos[(e + f*x)/2] + Sin[(e + f
*x)/2]) + 2*(3*A + 5*B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (3*A + 5*B)*(Cos[(e + f*x)/
2] + Sin[(e + f*x)/2])^3 + (1 + I)*(-1)^(3/4)*(3*A + 5*B)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4
])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4))/(16*f*(a*(1 + Sin[e + f*x]))^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(278\) vs. \(2(107)=214\).

Time = 1.91 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.21

method	result	size
default	\(-\frac {\left (-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \left (3 A +5 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sin \left (f x +e \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \left (3 A +5 B \right )+20 A \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {5}{2}}-6 A \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a^{\frac {3}{2}}+12 B \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {5}{2}}-10 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a^{\frac {3}{2}}+6 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}+10 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{32 a^{\frac {11}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(279\)
parts	\(-\frac {A \left (-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \left (\cos ^{2}\left (f x +e \right )\right )+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (f x +e \right )+6 \sqrt {a -a \sin \left (f x +e \right )}\, \sin \left (f x +e \right ) a^{\frac {3}{2}}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+14 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}-\frac {B \left (-5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \left (\cos ^{2}\left (f x +e \right )\right )+10 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (f x +e \right ) a^{3}+12 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {5}{2}}-10 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a^{\frac {3}{2}}+10 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{32 a^{\frac {11}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(394\)

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/32*(-2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*(3*A+5*B)*cos(f*x+e)^2+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^3*(3*A+5*B)+20*A*(a-a*sin(f*x+e))^(1/2)*a^(5/2)-6*
A*(a-a*sin(f*x+e))^(3/2)*a^(3/2)+12*B*(a-a*sin(f*x+e))^(1/2)*a^(5/2)-10*B*(a-a*sin(f*x+e))^(3/2)*a^(3/2)+6*A*2
^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3+10*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)
*2^(1/2)/a^(1/2))*a^3)*(-a*(sin(f*x+e)-1))^(1/2)/a^(11/2)/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 392 vs. \(2 (107) = 214\).

Time = 0.27 (sec) , antiderivative size = 392, normalized size of antiderivative = 3.11 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (3 \, A + 5 \, B\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (3 \, A + 5 \, B\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (3 \, A + 5 \, B\right )} \cos \left (f x + e\right ) + {\left ({\left (3 \, A + 5 \, B\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (3 \, A + 5 \, B\right )} \cos \left (f x + e\right ) - 12 \, A - 20 \, B\right )} \sin \left (f x + e\right ) - 12 \, A - 20 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left ({\left (3 \, A + 5 \, B\right )} \cos \left (f x + e\right )^{2} + {\left (7 \, A + B\right )} \cos \left (f x + e\right ) + {\left ({\left (3 \, A + 5 \, B\right )} \cos \left (f x + e\right ) - 4 \, A + 4 \, B\right )} \sin \left (f x + e\right ) + 4 \, A - 4 \, B\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{64 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/64*(sqrt(2)*((3*A + 5*B)*cos(f*x + e)^3 + 3*(3*A + 5*B)*cos(f*x + e)^2 - 2*(3*A + 5*B)*cos(f*x + e) + ((3*A
+ 5*B)*cos(f*x + e)^2 - 2*(3*A + 5*B)*cos(f*x + e) - 12*A - 20*B)*sin(f*x + e) - 12*A - 20*B)*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*cos(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*((3*A + 5*B)*cos(f*x + e)^2 + (7*A + B)*cos(f*x + e) + ((3*A + 5*B)*cos(f*x + e) - 4*A + 4*B)*sin
(f*x + e) + 4*A - 4*B)*sqrt(a*sin(f*x + e) + a))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(
f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))

Sympy [F]

\[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {A + B \sin {\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**(5/2),x)

[Out]

Integral((A + B*sin(e + f*x))/(a*(sin(e + f*x) + 1))**(5/2), x)

Maxima [F]

\[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int { \frac {B \sin \left (f x + e\right ) + A}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)/(a*sin(f*x + e) + a)^(5/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (107) = 214\).

Time = 0.32 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.87 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\frac {\sqrt {2} {\left (3 \, A \sqrt {a} + 5 \, B \sqrt {a}\right )} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (3 \, A \sqrt {a} + 5 \, B \sqrt {a}\right )} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {2 \, {\left (3 \, \sqrt {2} A \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 5 \, \sqrt {2} B \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, \sqrt {2} A \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, \sqrt {2} B \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{64 \, f} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

1/64*(sqrt(2)*(3*A*sqrt(a) + 5*B*sqrt(a))*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*f
*x + 1/2*e))) - sqrt(2)*(3*A*sqrt(a) + 5*B*sqrt(a))*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^3*sgn(cos(-1/4
*pi + 1/2*f*x + 1/2*e))) - 2*(3*sqrt(2)*A*sqrt(a)*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 5*sqrt(2)*B*sqrt(a)*sin(-
1/4*pi + 1/2*f*x + 1/2*e)^3 - 5*sqrt(2)*A*sqrt(a)*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 3*sqrt(2)*B*sqrt(a)*sin(-1/
4*pi + 1/2*f*x + 1/2*e))/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[In]

int((A + B*sin(e + f*x))/(a + a*sin(e + f*x))^(5/2),x)

[Out]

int((A + B*sin(e + f*x))/(a + a*sin(e + f*x))^(5/2), x)