3.498 \(\int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=155 \[ \frac{a b \sqrt{a+b \sin (c+d x)}}{2 d}-\frac{(a-b)^{3/2} (2 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{4 d}+\frac{(2 a-3 b) (a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{4 d}+\frac{\sec ^2(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{2 d} \]

[Out]

-((a - b)^(3/2)*(2*a + 3*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(4*d) + ((2*a - 3*b)*(a + b)^(3/2)*
ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(4*d) + (a*b*Sqrt[a + b*Sin[c + d*x]])/(2*d) + (Sec[c + d*x]^2*
(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(3/2))/(2*d)

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Rubi [A]  time = 0.269716, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2668, 739, 825, 827, 1166, 206} \[ \frac{a b \sqrt{a+b \sin (c+d x)}}{2 d}-\frac{(a-b)^{3/2} (2 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{4 d}+\frac{(2 a-3 b) (a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{4 d}+\frac{\sec ^2(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

-((a - b)^(3/2)*(2*a + 3*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(4*d) + ((2*a - 3*b)*(a + b)^(3/2)*
ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(4*d) + (a*b*Sqrt[a + b*Sin[c + d*x]])/(2*d) + (Sec[c + d*x]^2*
(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(3/2))/(2*d)

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 825

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g*(d + e*x)^m)/
(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d + c*e*f)*x, x])/(a + c*x^2), x], x] /
; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && FractionQ[m] && GtQ[m, 0]

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

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

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

Mathematica [A]  time = 0.824758, size = 147, normalized size = 0.95 \[ \frac{-\sqrt{a-b} \left (2 a^2+a b-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )+\sqrt{a+b} \left (2 a^2-a b-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )+2 \sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (\left (a^2+b^2\right ) \sin (c+d x)+2 a b\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-(Sqrt[a - b]*(2*a^2 + a*b - 3*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]]) + Sqrt[a + b]*(2*a^2 - a*b
- 3*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + 2*Sec[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]]*(2*a*b + (
a^2 + b^2)*Sin[c + d*x]))/(4*d)

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Maple [B]  time = 0.446, size = 356, normalized size = 2.3 \begin{align*}{\frac{1}{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}d} \left ( 2\,\sin \left ( dx+c \right ) \sqrt{a+b\sin \left ( dx+c \right ) }\sqrt{-a+b}\sqrt{a+b} \left ({a}^{2}+{b}^{2} \right ) - \left ( -2\,{\it Artanh} \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{a+b}}} \right ){a}^{3}\sqrt{-a+b}-b{\it Artanh} \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a+b}}}} \right ){a}^{2}\sqrt{-a+b}+4\,{b}^{2}{\it Artanh} \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{a+b}}} \right ) a\sqrt{-a+b}+3\,{b}^{3}{\it Artanh} \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{a+b}}} \right ) \sqrt{-a+b}-2\,\arctan \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{-a+b}}} \right ){a}^{3}\sqrt{a+b}+b\arctan \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-a+b}}}} \right ){a}^{2}\sqrt{a+b}+4\,{b}^{2}\arctan \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{-a+b}}} \right ) a\sqrt{a+b}-3\,{b}^{3}\arctan \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{-a+b}}} \right ) \sqrt{a+b} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4\,\sqrt{a+b\sin \left ( dx+c \right ) }ba\sqrt{-a+b}\sqrt{a+b} \right ){\frac{1}{\sqrt{-a+b}}}{\frac{1}{\sqrt{a+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*sin(d*x+c))^(5/2),x)

[Out]

1/4*(2*sin(d*x+c)*(a+b*sin(d*x+c))^(1/2)*(-a+b)^(1/2)*(a+b)^(1/2)*(a^2+b^2)-(-2*arctanh((a+b*sin(d*x+c))^(1/2)
/(a+b)^(1/2))*a^3*(-a+b)^(1/2)-b*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a^2*(-a+b)^(1/2)+4*b^2*arctanh((a
+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a*(-a+b)^(1/2)+3*b^3*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*(-a+b)^(1/2
)-2*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a^3*(a+b)^(1/2)+b*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*
a^2*(a+b)^(1/2)+4*b^2*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a*(a+b)^(1/2)-3*b^3*arctan((a+b*sin(d*x+c))^
(1/2)/(-a+b)^(1/2))*(a+b)^(1/2))*cos(d*x+c)^2+4*(a+b*sin(d*x+c))^(1/2)*b*a*(-a+b)^(1/2)*(a+b)^(1/2))/(-a+b)^(1
/2)/(a+b)^(1/2)/cos(d*x+c)^2/d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.17225, size = 4980, normalized size = 32.13 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/32*((2*a^2 - a*b - 3*b^2)*sqrt(a + b)*cos(d*x + c)^2*log((b^4*cos(d*x + c)^4 + 128*a^4 + 256*a^3*b + 320*a
^2*b^2 + 256*a*b^3 + 72*b^4 - 8*(20*a^2*b^2 + 28*a*b^3 + 9*b^4)*cos(d*x + c)^2 - 8*(16*a^3 + 24*a^2*b + 20*a*b
^2 + 8*b^3 - (10*a*b^2 + 7*b^3)*cos(d*x + c)^2 - (b^3*cos(d*x + c)^2 - 24*a^2*b - 28*a*b^2 - 8*b^3)*sin(d*x +
c))*sqrt(b*sin(d*x + c) + a)*sqrt(a + b) + 4*(64*a^3*b + 112*a^2*b^2 + 64*a*b^3 + 14*b^4 - (8*a*b^3 + 7*b^4)*c
os(d*x + c)^2)*sin(d*x + c))/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) +
(2*a^2 + a*b - 3*b^2)*sqrt(a - b)*cos(d*x + c)^2*log((b^4*cos(d*x + c)^4 + 128*a^4 - 256*a^3*b + 320*a^2*b^2 -
256*a*b^3 + 72*b^4 - 8*(20*a^2*b^2 - 28*a*b^3 + 9*b^4)*cos(d*x + c)^2 + 8*(16*a^3 - 24*a^2*b + 20*a*b^2 - 8*b
^3 - (10*a*b^2 - 7*b^3)*cos(d*x + c)^2 - (b^3*cos(d*x + c)^2 - 24*a^2*b + 28*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt
(b*sin(d*x + c) + a)*sqrt(a - b) + 4*(64*a^3*b - 112*a^2*b^2 + 64*a*b^3 - 14*b^4 - (8*a*b^3 - 7*b^4)*cos(d*x +
c)^2)*sin(d*x + c))/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 16*(2*a*
b + (a^2 + b^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/(d*cos(d*x + c)^2), -1/32*(2*(2*a^2 - a*b - 3*b^2)*sqr
t(-a - b)*arctan(-1/4*(b^2*cos(d*x + c)^2 - 8*a^2 - 8*a*b - 2*b^2 - 2*(4*a*b + 3*b^2)*sin(d*x + c))*sqrt(b*sin
(d*x + c) + a)*sqrt(-a - b)/(2*a^3 + 3*a^2*b + 2*a*b^2 + b^3 - (a*b^2 + b^3)*cos(d*x + c)^2 + (3*a^2*b + 4*a*b
^2 + b^3)*sin(d*x + c)))*cos(d*x + c)^2 + (2*a^2 + a*b - 3*b^2)*sqrt(a - b)*cos(d*x + c)^2*log((b^4*cos(d*x +
c)^4 + 128*a^4 - 256*a^3*b + 320*a^2*b^2 - 256*a*b^3 + 72*b^4 - 8*(20*a^2*b^2 - 28*a*b^3 + 9*b^4)*cos(d*x + c)
^2 + 8*(16*a^3 - 24*a^2*b + 20*a*b^2 - 8*b^3 - (10*a*b^2 - 7*b^3)*cos(d*x + c)^2 - (b^3*cos(d*x + c)^2 - 24*a^
2*b + 28*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(a - b) + 4*(64*a^3*b - 112*a^2*b^2 + 64*a*
b^3 - 14*b^4 - (8*a*b^3 - 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x
+ c)^2 - 2)*sin(d*x + c) + 8)) - 16*(2*a*b + (a^2 + b^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/(d*cos(d*x +
c)^2), -1/32*(2*(2*a^2 + a*b - 3*b^2)*sqrt(-a + b)*arctan(1/4*(b^2*cos(d*x + c)^2 - 8*a^2 + 8*a*b - 2*b^2 - 2
*(4*a*b - 3*b^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(-a + b)/(2*a^3 - 3*a^2*b + 2*a*b^2 - b^3 - (a*b^2
- b^3)*cos(d*x + c)^2 + (3*a^2*b - 4*a*b^2 + b^3)*sin(d*x + c)))*cos(d*x + c)^2 + (2*a^2 - a*b - 3*b^2)*sqrt(
a + b)*cos(d*x + c)^2*log((b^4*cos(d*x + c)^4 + 128*a^4 + 256*a^3*b + 320*a^2*b^2 + 256*a*b^3 + 72*b^4 - 8*(20
*a^2*b^2 + 28*a*b^3 + 9*b^4)*cos(d*x + c)^2 - 8*(16*a^3 + 24*a^2*b + 20*a*b^2 + 8*b^3 - (10*a*b^2 + 7*b^3)*cos
(d*x + c)^2 - (b^3*cos(d*x + c)^2 - 24*a^2*b - 28*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(a
+ b) + 4*(64*a^3*b + 112*a^2*b^2 + 64*a*b^3 + 14*b^4 - (8*a*b^3 + 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(cos(d
*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 16*(2*a*b + (a^2 + b^2)*sin(d*x + c
))*sqrt(b*sin(d*x + c) + a))/(d*cos(d*x + c)^2), -1/16*((2*a^2 + a*b - 3*b^2)*sqrt(-a + b)*arctan(1/4*(b^2*cos
(d*x + c)^2 - 8*a^2 + 8*a*b - 2*b^2 - 2*(4*a*b - 3*b^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(-a + b)/(2
*a^3 - 3*a^2*b + 2*a*b^2 - b^3 - (a*b^2 - b^3)*cos(d*x + c)^2 + (3*a^2*b - 4*a*b^2 + b^3)*sin(d*x + c)))*cos(d
*x + c)^2 + (2*a^2 - a*b - 3*b^2)*sqrt(-a - b)*arctan(-1/4*(b^2*cos(d*x + c)^2 - 8*a^2 - 8*a*b - 2*b^2 - 2*(4*
a*b + 3*b^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(-a - b)/(2*a^3 + 3*a^2*b + 2*a*b^2 + b^3 - (a*b^2 + b
^3)*cos(d*x + c)^2 + (3*a^2*b + 4*a*b^2 + b^3)*sin(d*x + c)))*cos(d*x + c)^2 - 8*(2*a*b + (a^2 + b^2)*sin(d*x
+ c))*sqrt(b*sin(d*x + c) + a))/(d*cos(d*x + c)^2)]

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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(sec(d*x+c)**3*(a+b*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [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(sec(d*x+c)^3*(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out