3.282 \(\int \frac {(c \sin (a+b x))^{5/2}}{\sqrt {d \cos (a+b x)}} \, dx\)

Optimal. Leaf size=320 \[ -\frac {3 c^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{4 \sqrt {2} b \sqrt {d}}+\frac {3 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}+1\right )}{4 \sqrt {2} b \sqrt {d}}+\frac {3 c^{5/2} \log \left (-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+\sqrt {c} \tan (a+b x)+\sqrt {c}\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {3 c^{5/2} \log \left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+\sqrt {c} \tan (a+b x)+\sqrt {c}\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {c (c \sin (a+b x))^{3/2} \sqrt {d \cos (a+b x)}}{2 b d} \]

[Out]

-3/8*c^(5/2)*arctan(1-2^(1/2)*d^(1/2)*(c*sin(b*x+a))^(1/2)/c^(1/2)/(d*cos(b*x+a))^(1/2))/b*2^(1/2)/d^(1/2)+3/8
*c^(5/2)*arctan(1+2^(1/2)*d^(1/2)*(c*sin(b*x+a))^(1/2)/c^(1/2)/(d*cos(b*x+a))^(1/2))/b*2^(1/2)/d^(1/2)+3/16*c^
(5/2)*ln(c^(1/2)-2^(1/2)*d^(1/2)*(c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(1/2)+c^(1/2)*tan(b*x+a))/b*2^(1/2)/d^(1/
2)-3/16*c^(5/2)*ln(c^(1/2)+2^(1/2)*d^(1/2)*(c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(1/2)+c^(1/2)*tan(b*x+a))/b*2^(
1/2)/d^(1/2)-1/2*c*(c*sin(b*x+a))^(3/2)*(d*cos(b*x+a))^(1/2)/b/d

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Rubi [A]  time = 0.26, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2568, 2574, 297, 1162, 617, 204, 1165, 628} \[ -\frac {3 c^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{4 \sqrt {2} b \sqrt {d}}+\frac {3 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}+1\right )}{4 \sqrt {2} b \sqrt {d}}+\frac {3 c^{5/2} \log \left (-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+\sqrt {c} \tan (a+b x)+\sqrt {c}\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {3 c^{5/2} \log \left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+\sqrt {c} \tan (a+b x)+\sqrt {c}\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {c (c \sin (a+b x))^{3/2} \sqrt {d \cos (a+b x)}}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*c^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d]*Sqrt[c*Sin[a + b*x]])/(Sqrt[c]*Sqrt[d*Cos[a + b*x]])])/(4*Sqrt[2]*b*Sq
rt[d]) + (3*c^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d]*Sqrt[c*Sin[a + b*x]])/(Sqrt[c]*Sqrt[d*Cos[a + b*x]])])/(4*Sqrt
[2]*b*Sqrt[d]) + (3*c^(5/2)*Log[Sqrt[c] - (Sqrt[2]*Sqrt[d]*Sqrt[c*Sin[a + b*x]])/Sqrt[d*Cos[a + b*x]] + Sqrt[c
]*Tan[a + b*x]])/(8*Sqrt[2]*b*Sqrt[d]) - (3*c^(5/2)*Log[Sqrt[c] + (Sqrt[2]*Sqrt[d]*Sqrt[c*Sin[a + b*x]])/Sqrt[
d*Cos[a + b*x]] + Sqrt[c]*Tan[a + b*x]])/(8*Sqrt[2]*b*Sqrt[d]) - (c*Sqrt[d*Cos[a + b*x]]*(c*Sin[a + b*x])^(3/2
))/(2*b*d)

Rule 204

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

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2568

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

Rule 2574

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{k = Denomina
tor[m]}, Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Sin[e + f*x])^(1/k)/(b*Cos
[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rubi steps

\begin {align*} \int \frac {(c \sin (a+b x))^{5/2}}{\sqrt {d \cos (a+b x)}} \, dx &=-\frac {c \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2}}{2 b d}+\frac {1}{4} \left (3 c^2\right ) \int \frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}} \, dx\\ &=-\frac {c \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2}}{2 b d}+\frac {\left (3 c^3 d\right ) \operatorname {Subst}\left (\int \frac {x^2}{c^2+d^2 x^4} \, dx,x,\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}\right )}{2 b}\\ &=-\frac {c \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2}}{2 b d}-\frac {\left (3 c^3\right ) \operatorname {Subst}\left (\int \frac {c-d x^2}{c^2+d^2 x^4} \, dx,x,\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}\right )}{4 b}+\frac {\left (3 c^3\right ) \operatorname {Subst}\left (\int \frac {c+d x^2}{c^2+d^2 x^4} \, dx,x,\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}\right )}{4 b}\\ &=-\frac {c \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2}}{2 b d}+\frac {\left (3 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {c}{d}-\frac {\sqrt {2} \sqrt {c} x}{\sqrt {d}}+x^2} \, dx,x,\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}\right )}{8 b d}+\frac {\left (3 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {c}{d}+\frac {\sqrt {2} \sqrt {c} x}{\sqrt {d}}+x^2} \, dx,x,\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}\right )}{8 b d}+\frac {\left (3 c^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {c}}{\sqrt {d}}+2 x}{-\frac {c}{d}-\frac {\sqrt {2} \sqrt {c} x}{\sqrt {d}}-x^2} \, dx,x,\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}\right )}{8 \sqrt {2} b \sqrt {d}}+\frac {\left (3 c^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {c}}{\sqrt {d}}-2 x}{-\frac {c}{d}+\frac {\sqrt {2} \sqrt {c} x}{\sqrt {d}}-x^2} \, dx,x,\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}\right )}{8 \sqrt {2} b \sqrt {d}}\\ &=\frac {3 c^{5/2} \log \left (\sqrt {c}-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+\sqrt {c} \tan (a+b x)\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {3 c^{5/2} \log \left (\sqrt {c}+\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+\sqrt {c} \tan (a+b x)\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {c \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2}}{2 b d}+\frac {\left (3 c^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{4 \sqrt {2} b \sqrt {d}}-\frac {\left (3 c^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{4 \sqrt {2} b \sqrt {d}}\\ &=-\frac {3 c^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{4 \sqrt {2} b \sqrt {d}}+\frac {3 c^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{4 \sqrt {2} b \sqrt {d}}+\frac {3 c^{5/2} \log \left (\sqrt {c}-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+\sqrt {c} \tan (a+b x)\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {3 c^{5/2} \log \left (\sqrt {c}+\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+\sqrt {c} \tan (a+b x)\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {c \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2}}{2 b d}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 67, normalized size = 0.21 \[ \frac {2 \cos ^2(a+b x)^{3/4} \tan (a+b x) (c \sin (a+b x))^{5/2} \, _2F_1\left (\frac {3}{4},\frac {7}{4};\frac {11}{4};\sin ^2(a+b x)\right )}{7 b \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[3/4, 7/4, 11/4, Sin[a + b*x]^2]*(c*Sin[a + b*x])^(5/2)*Tan[a + b*x
])/(7*b*Sqrt[d*Cos[a + b*x]])

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fricas [B]  time = 55.06, size = 2074, normalized size = 6.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(32*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*c^2*sin(b*x + a) - 12*sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*d*a
rctan(((sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^13*cos(b*x + a) + sqrt(2)*(c^10/(b^4*d^2))^(3/4)*b^3*c^8*d*sin(b*x
+ a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) + sqrt(4*sqrt(c^10/(b^4*d^2))*b^2*c^11*d*cos(b*x + a)*sin(b*x
+ a) + c^16 - 2*(sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^13*cos(b*x + a) + sqrt(2)*(c^10/(b^4*d^2))^(3/4)*b^3*c^8*d
*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(2*c^8*cos(b*x + a)*sin(b*x + a) + sqrt(c^10/(b^4*d^
2))*b^2*c^3*d + (sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^5*sin(b*x + a) + sqrt(2)*(c^10/(b^4*d^2))^(3/4)*b^3*d*cos(
b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))))/(2*c^16*cos(b*x + a)^2 - c^16)) - 12*sqrt(2)*(c^10/(b^4*
d^2))^(1/4)*b*d*arctan(((sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^13*cos(b*x + a) + sqrt(2)*(c^10/(b^4*d^2))^(3/4)*b
^3*c^8*d*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) - sqrt(4*sqrt(c^10/(b^4*d^2))*b^2*c^11*d*cos(
b*x + a)*sin(b*x + a) + c^16 + 2*(sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^13*cos(b*x + a) + sqrt(2)*(c^10/(b^4*d^2)
)^(3/4)*b^3*c^8*d*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(2*c^8*cos(b*x + a)*sin(b*x + a) +
sqrt(c^10/(b^4*d^2))*b^2*c^3*d - (sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^5*sin(b*x + a) + sqrt(2)*(c^10/(b^4*d^2))
^(3/4)*b^3*d*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))))/(2*c^16*cos(b*x + a)^2 - c^16)) - 12*sq
rt(2)*(c^10/(b^4*d^2))^(1/4)*b*d*arctan(-1/2*(2*c^16*cos(b*x + a)*sin(b*x + a) - sqrt(4*sqrt(c^10/(b^4*d^2))*b
^2*c^11*d*cos(b*x + a)*sin(b*x + a) + c^16 + 2*(sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^13*cos(b*x + a) + sqrt(2)*(
c^10/(b^4*d^2))^(3/4)*b^3*c^8*d*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(sqrt(2)*(c^10/(b^4*d
^2))^(1/4)*b*c^5*sin(b*x + a) + sqrt(2)*(c^10/(b^4*d^2))^(3/4)*b^3*d*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c
*sin(b*x + a)) + (sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^13*sin(b*x + a) + sqrt(2)*(c^10/(b^4*d^2))^(3/4)*b^3*c^8*
d*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) - 4*(b^2*c^11*d*cos(b*x + a)^4 - b^2*c^11*d*cos(b*x
+ a)^2)*sqrt(c^10/(b^4*d^2)))/((2*c^16*cos(b*x + a)^3 - c^16*cos(b*x + a))*sin(b*x + a))) - 12*sqrt(2)*(c^10/(
b^4*d^2))^(1/4)*b*d*arctan(1/2*(2*c^16*cos(b*x + a)*sin(b*x + a) + sqrt(4*sqrt(c^10/(b^4*d^2))*b^2*c^11*d*cos(
b*x + a)*sin(b*x + a) + c^16 - 2*(sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^13*cos(b*x + a) + sqrt(2)*(c^10/(b^4*d^2)
)^(3/4)*b^3*c^8*d*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))*(sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c
^5*sin(b*x + a) + sqrt(2)*(c^10/(b^4*d^2))^(3/4)*b^3*d*cos(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))
- (sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^13*sin(b*x + a) + sqrt(2)*(c^10/(b^4*d^2))^(3/4)*b^3*c^8*d*cos(b*x + a)
)*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) - 4*(b^2*c^11*d*cos(b*x + a)^4 - b^2*c^11*d*cos(b*x + a)^2)*sqrt(c
^10/(b^4*d^2)))/((2*c^16*cos(b*x + a)^3 - c^16*cos(b*x + a))*sin(b*x + a))) + 3*sqrt(2)*(c^10/(b^4*d^2))^(1/4)
*b*d*log(2916*sqrt(c^10/(b^4*d^2))*b^2*c^11*d*cos(b*x + a)*sin(b*x + a) + 729*c^16 + 1458*(sqrt(2)*(c^10/(b^4*
d^2))^(1/4)*b*c^13*cos(b*x + a) + sqrt(2)*(c^10/(b^4*d^2))^(3/4)*b^3*c^8*d*sin(b*x + a))*sqrt(d*cos(b*x + a))*
sqrt(c*sin(b*x + a))) - 3*sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*d*log(2916*sqrt(c^10/(b^4*d^2))*b^2*c^11*d*cos(b*x
+ a)*sin(b*x + a) + 729*c^16 - 1458*(sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^13*cos(b*x + a) + sqrt(2)*(c^10/(b^4*d
^2))^(3/4)*b^3*c^8*d*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))) + 3*sqrt(2)*(c^10/(b^4*d^2))^(1/
4)*b*d*log(729/4*sqrt(c^10/(b^4*d^2))*b^2*c^11*d*cos(b*x + a)*sin(b*x + a) + 729/16*c^16 + 729/8*(sqrt(2)*(c^1
0/(b^4*d^2))^(1/4)*b*c^13*cos(b*x + a) + sqrt(2)*(c^10/(b^4*d^2))^(3/4)*b^3*c^8*d*sin(b*x + a))*sqrt(d*cos(b*x
+ a))*sqrt(c*sin(b*x + a))) - 3*sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*d*log(729/4*sqrt(c^10/(b^4*d^2))*b^2*c^11*d*
cos(b*x + a)*sin(b*x + a) + 729/16*c^16 - 729/8*(sqrt(2)*(c^10/(b^4*d^2))^(1/4)*b*c^13*cos(b*x + a) + sqrt(2)*
(c^10/(b^4*d^2))^(3/4)*b^3*c^8*d*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))))/(b*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\sqrt {d \cos \left (b x + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.15, size = 510, normalized size = 1.59 \[ -\frac {\left (3 i \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 i \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+2 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-2 \cos \left (b x +a \right ) \sqrt {2}\right ) \left (c \sin \left (b x +a \right )\right )^{\frac {5}{2}} \sqrt {2}}{8 b \left (-1+\cos \left (b x +a \right )\right ) \sqrt {d \cos \left (b x +a \right )}\, \sin \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/b*(3*I*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+co
s(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))-3*I
*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/s
in(b*x+a))^(1/2)*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-3*((1-cos(b*x+
a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1
/2)*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))-3*((1-cos(b*x+a)+sin(b*x+a)
)/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticP
i(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))+2*cos(b*x+a)^2*2^(1/2)-2*cos(b*x+a)*2^(1
/2))*(c*sin(b*x+a))^(5/2)/(-1+cos(b*x+a))/(d*cos(b*x+a))^(1/2)/sin(b*x+a)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\sqrt {d \cos \left (b x + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2}}{\sqrt {d\,\cos \left (a+b\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))**(5/2)/(d*cos(b*x+a))**(1/2),x)

[Out]

Timed out

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