Optimal. Leaf size=211 \[ -\frac {i \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 b \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (5 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]
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Rubi [A]
time = 0.47, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3649, 3730,
3697, 3696, 95, 209, 212} \begin {gather*} -\frac {2 b \left (5 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}-\frac {2 b \sqrt {\tan (c+d x)}}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {i \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 95
Rule 209
Rule 212
Rule 3649
Rule 3696
Rule 3697
Rule 3730
Rubi steps
\begin {align*} \int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^{5/2}} \, dx &=-\frac {2 b \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 \int \frac {-\frac {b}{2}-\frac {3}{2} a \tan (c+d x)+b \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 \left (a^2+b^2\right )}\\ &=-\frac {2 b \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (5 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {4 \int \frac {-\frac {3 a^2 b}{2}-\frac {3}{4} a \left (a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a \left (a^2+b^2\right )^2}\\ &=-\frac {2 b \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (5 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {i \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}+\frac {i \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}\\ &=-\frac {2 b \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (5 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {i \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d}+\frac {i \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}\\ &=-\frac {2 b \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (5 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {i \text {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a-i b)^2 d}+\frac {i \text {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^2 d}\\ &=-\frac {i \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 b \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (5 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 3.87, size = 194, normalized size = 0.92 \begin {gather*} \frac {-\frac {3 \sqrt [4]{-1} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{5/2}}+\frac {3 \sqrt [4]{-1} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{5/2}}+\frac {2 b \sqrt {\tan (c+d x)} \left (-6 a^3+\left (-5 a^2 b+b^3\right ) \tan (c+d x)\right )}{a \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^{3/2}}}{3 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 1.30, size = 1489210, normalized size = 7057.87 \[\text {output too large to display}\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\tan {\left (c + d x \right )}}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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