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*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)

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Rubi [A]  time = 0.257508, antiderivative size = 320, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, 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 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]

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 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 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 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 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 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]

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*}

Mathematica [C]  time = 0.125682, 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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Maple [C]  time = 0.118, size = 510, normalized size = 1.6 \begin{align*} \text{result too large to display} \end{align*}

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*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))/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))/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)*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)+si
n(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)*E
llipticPi(((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)

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

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

[Out]

Timed out

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

[Out]

Timed out

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

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)