3.146 \(\int \sec (c+d x) (a+a \sin (c+d x))^{7/2} \, dx\)
Optimal. Leaf size=110 \[ -\frac{8 a^3 \sqrt{a \sin (c+d x)+a}}{d}-\frac{4 a^2 (a \sin (c+d x)+a)^{3/2}}{3 d}+\frac{8 \sqrt{2} a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a (a \sin (c+d x)+a)^{5/2}}{5 d} \]
[Out]
(8*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*a^3*Sqrt[a + a*Sin[c + d*x]])/d
- (4*a^2*(a + a*Sin[c + d*x])^(3/2))/(3*d) - (2*a*(a + a*Sin[c + d*x])^(5/2))/(5*d)
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Rubi [A] time = 0.0852477, antiderivative size = 110, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used =
{2667, 50, 63, 206} \[ -\frac{8 a^3 \sqrt{a \sin (c+d x)+a}}{d}-\frac{4 a^2 (a \sin (c+d x)+a)^{3/2}}{3 d}+\frac{8 \sqrt{2} a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a (a \sin (c+d x)+a)^{5/2}}{5 d} \]
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]*(a + a*Sin[c + d*x])^(7/2),x]
[Out]
(8*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*a^3*Sqrt[a + a*Sin[c + d*x]])/d
- (4*a^2*(a + a*Sin[c + d*x])^(3/2))/(3*d) - (2*a*(a + a*Sin[c + d*x])^(5/2))/(5*d)
Rule 2667
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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/2])
Rule 50
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
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 (c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=\frac{a \operatorname{Subst}\left (\int \frac{(a+x)^{5/2}}{a-x} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{2 a (a+a \sin (c+d x))^{5/2}}{5 d}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{3/2}}{a-x} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{4 a^2 (a+a \sin (c+d x))^{3/2}}{3 d}-\frac{2 a (a+a \sin (c+d x))^{5/2}}{5 d}+\frac{\left (4 a^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+x}}{a-x} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{8 a^3 \sqrt{a+a \sin (c+d x)}}{d}-\frac{4 a^2 (a+a \sin (c+d x))^{3/2}}{3 d}-\frac{2 a (a+a \sin (c+d x))^{5/2}}{5 d}+\frac{\left (8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{8 a^3 \sqrt{a+a \sin (c+d x)}}{d}-\frac{4 a^2 (a+a \sin (c+d x))^{3/2}}{3 d}-\frac{2 a (a+a \sin (c+d x))^{5/2}}{5 d}+\frac{\left (16 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+a \sin (c+d x)}\right )}{d}\\ &=\frac{8 \sqrt{2} a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{8 a^3 \sqrt{a+a \sin (c+d x)}}{d}-\frac{4 a^2 (a+a \sin (c+d x))^{3/2}}{3 d}-\frac{2 a (a+a \sin (c+d x))^{5/2}}{5 d}\\ \end{align*}
Mathematica [A] time = 0.31554, size = 85, normalized size = 0.77 \[ \frac{120 \sqrt{2} a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a (\sin (c+d x)+1)}}{\sqrt{2} \sqrt{a}}\right )-2 a^3 \left (3 \sin ^2(c+d x)+16 \sin (c+d x)+73\right ) \sqrt{a (\sin (c+d x)+1)}}{15 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])^(7/2),x]
[Out]
(120*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqrt[a])] - 2*a^3*Sqrt[a*(1 + Sin[c + d*x])]*
(73 + 16*Sin[c + d*x] + 3*Sin[c + d*x]^2))/(15*d)
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Maple [A] time = 0.096, size = 83, normalized size = 0.8 \begin{align*} -2\,{\frac{a}{d} \left ( 1/5\, \left ( a+a\sin \left ( dx+c \right ) \right ) ^{5/2}+2/3\, \left ( a+a\sin \left ( dx+c \right ) \right ) ^{3/2}a+4\,{a}^{2}\sqrt{a+a\sin \left ( dx+c \right ) }-4\,{a}^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)*(a+a*sin(d*x+c))^(7/2),x)
[Out]
-2*a*(1/5*(a+a*sin(d*x+c))^(5/2)+2/3*(a+a*sin(d*x+c))^(3/2)*a+4*a^2*(a+a*sin(d*x+c))^(1/2)-4*a^(5/2)*2^(1/2)*a
rctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))/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)*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.66972, size = 274, normalized size = 2.49 \begin{align*} \frac{2 \,{\left (30 \, \sqrt{2} a^{\frac{7}{2}} \log \left (-\frac{a \sin \left (d x + c\right ) + 2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) +{\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 16 \, a^{3} \sin \left (d x + c\right ) - 76 \, a^{3}\right )} \sqrt{a \sin \left (d x + c\right ) + a}\right )}}{15 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")
[Out]
2/15*(30*sqrt(2)*a^(7/2)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c
) - 1)) + (3*a^3*cos(d*x + c)^2 - 16*a^3*sin(d*x + c) - 76*a^3)*sqrt(a*sin(d*x + c) + a))/d
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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)*(a+a*sin(d*x+c))**(7/2),x)
[Out]
Timed out
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Giac [B] time = 42.4063, size = 3368, normalized size = 30.62 \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)*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")
[Out]
1/60*(960*sqrt(2)*(a^4*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^2 + a^4*sgn(tan
(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1))*arctan(1/2*sqrt(2)*(sqrt((a*tan(1/2*c)^2 + a)*tan(1/2*d
*x)^2 + a*tan(1/2*c)^2 + a)*a*tan(1/2*c)^3 + sqrt((a*tan(1/2*c)^2 + a)*tan(1/2*d*x)^2 + a*tan(1/2*c)^2 + a)*a*
tan(1/2*c)^2 + 2*sqrt((a*tan(1/2*c)^2 + a)*tan(1/2*d*x)^2 + a*tan(1/2*c)^2 + a)*a*tan(1/2*c) + 2*sqrt((a*tan(1
/2*c)^2 + a)*tan(1/2*d*x)^2 + a*tan(1/2*c)^2 + a)*a + sqrt(a^3*tan(1/2*c)^8 + 2*a^3*tan(1/2*c)^7 + 6*a^3*tan(1
/2*c)^6 + 10*a^3*tan(1/2*c)^5 + 13*a^3*tan(1/2*c)^4 + 16*a^3*tan(1/2*c)^3 + 12*a^3*tan(1/2*c)^2 + 8*a^3*tan(1/
2*c) + 4*a^3)*tan(1/2*d*x) + sqrt(a^3*tan(1/2*c)^8 - 2*a^3*tan(1/2*c)^7 + 6*a^3*tan(1/2*c)^6 - 10*a^3*tan(1/2*
c)^5 + 13*a^3*tan(1/2*c)^4 - 16*a^3*tan(1/2*c)^3 + 12*a^3*tan(1/2*c)^2 - 8*a^3*tan(1/2*c) + 4*a^3))/sqrt(-a^3*
tan(1/2*c)^8 - 6*a^3*tan(1/2*c)^6 - 13*a^3*tan(1/2*c)^4 - 12*a^3*tan(1/2*c)^2 - 4*a^3))/sqrt(-a*tan(1/2*c)^4 -
2*a*tan(1/2*c)^2 - a) - ((((((73*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^
5 - 105*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^4 + 190*a^6*sgn(tan(1/2*d*
x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^3 - 190*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*
x) - tan(1/2*c) - 1)*tan(1/2*c)^2 + 105*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1
/2*c) - 73*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1))*tan(1/2*d*x)/(a^9*tan(1/2*c)^18 +
9*a^9*tan(1/2*c)^16 + 36*a^9*tan(1/2*c)^14 + 84*a^9*tan(1/2*c)^12 + 126*a^9*tan(1/2*c)^10 + 126*a^9*tan(1/2*c
)^8 + 84*a^9*tan(1/2*c)^6 + 36*a^9*tan(1/2*c)^4 + 9*a^9*tan(1/2*c)^2 + a^9) - 15*(7*a^6*sgn(tan(1/2*d*x)*tan(1
/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^5 - a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2
*c) - 1)*tan(1/2*c)^4 + 10*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^3 + 10*
a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^2 - a^6*sgn(tan(1/2*d*x)*tan(1/2*c
) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c) + 7*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c)
- 1))/(a^9*tan(1/2*c)^18 + 9*a^9*tan(1/2*c)^16 + 36*a^9*tan(1/2*c)^14 + 84*a^9*tan(1/2*c)^12 + 126*a^9*tan(1/2
*c)^10 + 126*a^9*tan(1/2*c)^8 + 84*a^9*tan(1/2*c)^6 + 36*a^9*tan(1/2*c)^4 + 9*a^9*tan(1/2*c)^2 + a^9))*tan(1/2
*d*x) + 10*(19*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^5 - 15*a^6*sgn(tan(
1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^4 + 22*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1
/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^3 - 22*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*t
an(1/2*c)^2 + 15*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c) - 19*a^6*sgn(tan(
1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1))/(a^9*tan(1/2*c)^18 + 9*a^9*tan(1/2*c)^16 + 36*a^9*tan(1/
2*c)^14 + 84*a^9*tan(1/2*c)^12 + 126*a^9*tan(1/2*c)^10 + 126*a^9*tan(1/2*c)^8 + 84*a^9*tan(1/2*c)^6 + 36*a^9*t
an(1/2*c)^4 + 9*a^9*tan(1/2*c)^2 + a^9))*tan(1/2*d*x) - 10*(19*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x)
- tan(1/2*c) - 1)*tan(1/2*c)^5 + 15*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c
)^4 + 22*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^3 + 22*a^6*sgn(tan(1/2*d*
x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^2 + 15*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x
) - tan(1/2*c) - 1)*tan(1/2*c) + 19*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1))/(a^9*tan
(1/2*c)^18 + 9*a^9*tan(1/2*c)^16 + 36*a^9*tan(1/2*c)^14 + 84*a^9*tan(1/2*c)^12 + 126*a^9*tan(1/2*c)^10 + 126*a
^9*tan(1/2*c)^8 + 84*a^9*tan(1/2*c)^6 + 36*a^9*tan(1/2*c)^4 + 9*a^9*tan(1/2*c)^2 + a^9))*tan(1/2*d*x) + 15*(7*
a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^5 + a^6*sgn(tan(1/2*d*x)*tan(1/2*c
) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^4 + 10*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*
c) - 1)*tan(1/2*c)^3 - 10*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^2 - a^6*
sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c) - 7*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) -
tan(1/2*d*x) - tan(1/2*c) - 1))/(a^9*tan(1/2*c)^18 + 9*a^9*tan(1/2*c)^16 + 36*a^9*tan(1/2*c)^14 + 84*a^9*tan(1
/2*c)^12 + 126*a^9*tan(1/2*c)^10 + 126*a^9*tan(1/2*c)^8 + 84*a^9*tan(1/2*c)^6 + 36*a^9*tan(1/2*c)^4 + 9*a^9*ta
n(1/2*c)^2 + a^9))*tan(1/2*d*x) - (73*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2
*c)^5 + 105*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^4 + 190*a^6*sgn(tan(1/
2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^3 + 190*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/
2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^2 + 105*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*t
an(1/2*c) + 73*a^6*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1))/(a^9*tan(1/2*c)^18 + 9*a^9*ta
n(1/2*c)^16 + 36*a^9*tan(1/2*c)^14 + 84*a^9*tan(1/2*c)^12 + 126*a^9*tan(1/2*c)^10 + 126*a^9*tan(1/2*c)^8 + 84*
a^9*tan(1/2*c)^6 + 36*a^9*tan(1/2*c)^4 + 9*a^9*tan(1/2*c)^2 + a^9))/(a*tan(1/2*d*x)^2*tan(1/2*c)^2 + a*tan(1/2
*d*x)^2 + a*tan(1/2*c)^2 + a)^(5/2))/d